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The math of saving the Enola Gay
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- Опубліковано 6 сер 2024
- How did the Enola Gay, the airplane that dropped the atomic bomb over Hiroshima, survive the strong shockwave from the nuclear explosion? The answer is a clever maneuver that can be derived using high-school physics and geometry. #some3
Erratum: @21:20 "1450 km" should be "1450 m"
Playlist Physics of Nuclear Weapons: • Physics of Nuclear Wea...
Code available on GitHub github.com/jsdiazpo/Saving-th...
Affiliate links (may earn a commission):
- How to Fall Slower Than Gravity amzn.to/3SQA7Dt
- The Making of the Atomic Bomb amzn.to/49ti6kX
Credits:
- all footage is of public domain
- problem adapted and presentation extended from "How to Fall Slower Than Gravity" by Paul Nahin
00:00 Introduction
01:10 dangerous mission
02:58 nuclear shockwave
05:43 the naive path
09:04 saving the Enola Gay
21:35 final solution
24:15 Hiroshima
I had to plot the course of the Enola Gay for my final in Fortran in 1968. My instructor was one of the team numbers on the Manhattan project.
wow, I think you own all the nerds in this channel a more complete version of your story, would you mind sharing more details? I am curious to know who you were working with and in which context you had to plot the plot of the Enola Gay.
@@jkzero Well the man was Colonel Robert Wright (bird Colonel) a retired army officer who was our instructor in computer science at Valley Forge. He did not give us a lot of specifics about what he had been doing on the Manhattan Project, and there were probably over 3000 people who were involved... but most of the cadets did recognize the name of that effort. Even as a retired officer he was still holding a top secret clearance....so I'm not sure what he did when he was not processing our 80 column computer punch cards over at Villanova University.
Although a second grade teacher had identified me as being in the realm of genius territory, which made me a target for well meaning teachers and a source of great exasperation for my parents, I did struggle with some of the concepts in mathematics.... mostly due to the personalities of the people who were presenting it, but Colonel Wright was able to change all that when I arrived as a summer camper in 1966 to take remedial algebra. For the first time in my life, I was being taught by someone that I could relate to..... which was a breath of fresh air for a young kid on the spectrum. For our final exam in Fortran IV, we used a calculus subroutine to choose the flight path that the Enola Gay should take once the bomb was released, to place as much distance as possible between the plane and the explosion. You'll have to forgive my lack of specifics...... because that was in 1968. HTH.
@@arthouston7361love it, fascinating story! Thanks for sharing. Sometimes I get hooked at the web of the Atomic Heritage Foundation where they have dozens of interviews to people from all the hierarchical level that participated in the Manhattan Project. Great to learn that you got the chance of god teaching, that can really change someone's life. Again, thanks for sharing.
My father was a flight instructor during WWII and had to learn (and teach) this math. Great job.
Wah. I learned Fortran in college in 1980 or 81. I really liked it... but never got to use it again.😮
If I had been given problems like this to solve in school, I would have loved my math classes. Well done!
You might like physics, then. My mechanics professor loved to put these types of problems on his exams.
I agree, though, that math is 100x better when it's applied to some problem in reality.
we really don't care this video is not about you
@@wolfumz Physics has always fascinated me, especially at the smallest scales where everything just gets weird.
That was the biggest problem I had with learning math in college. It was taught by the math department. They taught math for math's sake. If they had related it to engineering problems, I would have understood it much better.
I had the same problem in college, calculus, probability theory, differential equations, all taught by the Math Department so lots of theorems and proofs that were boring and of little use for non-mathematicians @@johns7734
The Enola Gay was one of a group of very special B-29s. They were called "Silverplate" B-29s. They had fuel injection engines only used on the Silverplate planes as well as aluminum reverseable propeller blades. As you know, all but one gun station was removed and very quick opening and closing bomb bay doors.
Didnt know about the fuel injection. The wing spar design was apparently very different also to permit a single long bomb bay. The silverplate b29s were significantly different from the regular b29, the amount of difference surprised me. I used to just think they just removed some turrets to free up weight but there's a lot more to it than that.
I always thought when I saw the footage of LB being dropped from the B29 that the bomb bay doors seemed very quick to open/close but wasn't sure if it was sped up or anything but it seems that was not the case.
Was the reason for this to do with the fast high G manouvere that needed to be carried out straight after the bomb release? I mean the may have been ripped from the plane or damaged if they weren't quick to close.
It's worth mentioning that aircrew practiced for the first atomic missions by dropping dummy bombs -- in military aviation we simply call them "shapes" -- that were built with the approximate size and aerodynamics of the real thing; no explosive, of course. And nobody knew what the real thing would look like until it showed up. These shapes were generally spherical and painted orange, so loaders and aircrew quickly began calling them "Pumpkins," while practice flights were called "Pumpkin flights." Pumpkin flights occurred over US practice ranges, plus in the South Pacific. And several over Japan, pre August 6. This must have lulled Japanese defenders into a false sense of security because when the three-ship formation finally came with the real bomb, it encountered no air defense fire or fighter planes.
"To cross the Sea without being caught by the Sky", how ironic
Well, by that point in the war, Japan had very few planes and trained pilots left and couldn’t afford to use AA fire against what they thought was a single conventional bomber.
This kinda sounds fake.
@@georgeofhamilton What do you mean?
@@RussellTeapot The original commenter might’ve made that information up.
I am French and I really appreciate your diction which allows me to understand almost every word, even if my level in math is not sufficient to understand everything.
Your videos are clear and captivating. Thank you very much
You're very welcome and I appreciate the comment, I am glad that my accent is not ruining everyone's experience, some have already complained :) Welcome to the channel and I hope you also check the other videos posted recently.
I may have missed it in the video, but the decision to turn right was because the rotation of the propellers slightly improved the turning performance.
You didn't miss it, the decision to go right instead of left was planned but apparently arbitrary. I am not aware of the propellers playing any role in this decision. The bombing mission included two other B29s for observations and measurements. Next to the Enola Gay was a B29 named "The Great Artiste" that dropped instrument packages suspended by parachutes for measurements of pressure, radioactivity, and other relevant quantities. A third B29 named "Necessary Evil" followed them with cameras. They planned that the Enola Gay would turn right, whereas The Great Artiste would perform the same maneuver but turning left.
Just to add here, actually a left turn will have a slightly greater turning performance due to the counter reaction of the propellers spinning clockwise.
*There should have been no preference if two engines were CW and two engines were CCW.*
The bomb blast shown at the opening is Nagasaki. There are no films of the detonation of Little Boy over Hiroshima. The plane assigned to film the attack did not arrive in time.
this is correct, this is why I explicitly included a caption indicating that the footage is from the bombing of Nagasaki. Unfortunately, most media outlets and even experienced people show the footage as if it were from the Hiroshima bombing.
As referenced by Jorge, this was motivated by Paul Nahin's writing. During Covid I purchased all of Paul's books. I have 20 of them and they are all simply outstanding. His topics include 3 books on stats, 2 books on Euler's equation and its usage in number theory and practical problems, 2 books on pure math (interesting integrals and the zeta function), 2 on time travel, 1 partial differential equation book on the heat equation, 6 physics puzzle books, a biography on Oliver Heaviside, 2 books on electrical engineering and a book on boolean algebra/information theory. He combines story telling and math/physics in a very entertaining fashion. If anyone is bored or does not get why we study math/physics I recommend getting any of his books and you will have an epiphany.
He is a Wonderful writer!!
I'm passing through from the SoME3 project-I really enjoyed this video, and this problem! Makes me wish I were still teaching, and more specifically teaching later high school or first year undergraduate physics. This is a _really_ nice problem that is accessible without advanced maths, physics or engineering knowledge, while demonstrating a) the practical application of theory and mathematical models of physics to the real world; b) the need of outside information (in this case aircraft performance and parameters of the mission, both as-designed and as-executed based on accounts and reports); c) the need to abstract the real-world problem to a modellable one, and the need to make simplifying assumptions to make it solvable; and d) the need to use tools (e.g. numerical computation) to make a problem tractable, and not get caught up in pride in analytical or "the book" results or the school standard of what is "cheating". (I was very guilty of the pride in obtaining analytical, "nice" expressions throughout my university career.)
If I were teaching this as a worked example, I might like to discuss how poor simplifying assumptions, like neglecting air resistance, can introduce error that may or may not be negligible. Ballistics topics in general might be very good for that, demonstrating how air resistance can matter to the success or failure of your objective.
As an aviation enthusiast, I also can't help but think how much more complicated this can be to analyse and execute from instrument data, given the effect of wind and IAS vs. TAS, if you don't have instrumentation to give you TAS or groundspeed. I was going to say this isn't something I'd teach, but... honestly, just the complication of a constant stiff wind could be worth introducing, it wouldn't be that complex and is a good demonstration of potentially bad assumptions. If they approached with a tailwind, they might maintain their airspeed after the manoeuvre (max airspeed for the aircraft) but have reduced their groundspeed by the headwind (tangential) component of the wind vector, because the plane's frame of reference is the air mass around it, not the ground.
I am glad you found the content of interest, make sure to check the several follow-up videos. Thanks for watching and welcome to the channel.
I think there was a fact that should have been considered. The bomb weighed almost 5 tons, so when the plane get released of such a weight, it will produce a direct vertical jump over the vertical to the ground. This abrupt altitude and speed change added to the beginning of the right turn maneuver, both results will produce a substantial change in the final trajectory. Very clear development, by the way. Kind regards.
The crew had to have trained meticulously for this maneuver, and should the pilot have wished to do so, he could have compensated for the excess lift. The release of the bomb made the plane about 10% lighter, so as per a rule of thumb, the plane then had about 10% more excess power to use. That amount would certainly have made a small difference for sure, but as they were going full pelt already, an extra 10% thrust will yield only a few percent of extra airspeed
Yes, that was one of several key factors that were ignored in this presentation, and perhaps the most important. “Little Boy” weighed 4400kg, and the B29 33800kg, so losing 13% of your weight will skew these numbers considerably.
Brilliant! As a physicist I hugely respect the way you presented and analyzed this problem, as well as how you did guide through the actual calculations. Another important thing that I noticed, was that you clearly identified all assumptions that you would be making. In my university years I had only a handful professors, who possessed these traits. Bravo.
thanks for watching and the positive feedback, I am glad you liked the content. Make sure to check the other videos and welcome to the channel.
@@jkzero Which I certainly will do, :) I actually started with the one about the history of the German atomic bomb development, and was totally flabbergasted having heard that Werner Heisenberg wanted to use random walk models to calculate neutron mean free path. What a gift to the World :) Cheers!
One slight flaw. It's a 3D world. There is a manoeuvre called a Chandelle. You pull into a climb turn, pull and dive out levelling the wings.
The reason why that might be better is that turning circle is a function of speed. Lower the speed, you turn tighter. Then on the way out, you give to speed up getting more distance.
you are right, a complete description of the problem requires a 3D characterization of the path; however, as indicated in the list of assumptions, I have kept the problem only in 2D for simplicity. Despite the assumptions, the final result agrees very well with the observations, which indicates that all the extra 3D details are quite negligible
@@jkzero I agree on the simplified solution. There’s a dog fighting trick called a yo-yo if you want more details.
On the calculation in practice you would chop it. Do every 20 degrees. The. Every 10, 5 and 1 degree near the peak
@@adenwellsmith6908 the numerical solution that I showed was only intended to reduce the mathematical complexity of the presentation that was already quite loaded for a general video. The fact is that the long and ugly equation that I programed can be solved analytically, it requires some rearrangements and it becomes a quadratic equation that we know how to solve with pen and paper. When I did this the video had almost 10 min extra just to simplify terms so I decided to cut this and use the easy route of just letting the computer do the number crunching.
You clearly gave a lot of effort into this to make it clear as possible. Really nice!
Thanks, I am glad you found it of interest. I did spend more time than I expected but learning to use Manim was the goal so I just picked a problem that I found interesting and challenged myself to animate it. More nuclear-related topics and physics in general coming soon, thanks for stopping by.
@@jkzero nice, you might even go viral because of the newly released movie Oppenheimer and its nuclear science related topics
@@pallaskedisiCokiyi yeah, on this topic you might want to check my latest video published last wee:, it is about Oppenheimer and the Trinity test ua-cam.com/video/GzfQY5FmURM/v-deo.html
The most amazing thing about physics in my opinion is that it gives us these tools to solve these questions that no one else has ever had to even think of before, and not just solve them "good enough" but often to the point where you can use the answer to solve question on stuff you know even less about as the answers can be accurate enough to be thought of almost on the level of truth. And any time the answers it gives us disagree with reality, it gives us the ability to quantify and identify the discrepancy and improve our understand, giving us more tools to solve even more complex questions.
I am glad you found the content of interest, make sure to check the several follow-up videos. Thanks for watching and welcome to the channel.
That was thoroughly enjoyable, thank you for spending the time putting this together. It's really inspiring.
I am glad you found the video of interest, you are welcome to check the other videos, share and subscribe, thanks
Nice presentation! I made the suggestion to Paul in a private email many years ago about this problem. I'd written up an analysis on my blog and recalled his book "Chases and escapes". I pinged him asking if anyone had done a detailed analysis in three dimensions (i.e., ideal path including acceleration from altitude change, etc.). He later added his analysis in the second edition of the book.
that's pretty cool, would you mind sharing the link to your blog post? I would be curious to check it out.
Thanks for mentioning that book - that looks super-interesting.
With this video I launched my UA-cam channel and won a Honorable Mention at the Summer of Math Exposition organized by @3b1b. You can support this newly created channel and growing community by liking, commenting, and subscribing. Thanks for watching and welcome to the channel.
You are welcome to watch the several follow-up videos of the series on the Physics of Nuclear Weapons ua-cam.com/play/PL_UV-wQj1lvUhNttvv4_KsYrQxHygj3Ey.html
Wow, excellent content here. Very careful and thorough. Great instruction on walking through the math and physics involved in this event. Many thanks.
Much appreciated! Thanks for watching and welcome to the channel
This is really great! The visuals and the explanations make it really easy to follow while still posing and answering a really interesting problem! Great job!
I am glad you found the content of interest, make sure to check the several follow-up videos. Thanks for watching and welcome to the channel.
"Does not focus on the morality" Lol, classic miltary engineering.
Very nice, clearly explained and reasoned. Well done!
Thanks, I am glad you found it of interest. More coming soon so you are welcome to stop by.
Great video! Optimization problems are always so interesting.
Thanks. I am glad you found it of interest. Optimization is what I do daily in my "day job" so it is nice to combine it with my personal interests.
@@jkzero Sounds like a fun job then. Would you be OK saying what is it about? I understand if you prefer to keep it private so no worries.
@@5eurosenelsuelo currently I work as a data scientist/statistical consultant in a chemical company, I analyze data from research labs, I design experiments, and then use optimization for finding the best compromise between conflicting objectives, such as production cost, sustainability goals, product performance, etc. There is a public video in which I describe what I do, in case you want to check it out: ua-cam.com/video/mfATFay9PhY/v-deo.html
Wow great video!!! i love kinematics and projectile motions, this one was great, you did an amazing job at explaining the physics and handling the equations, its insane how stuff that we take in high school can be used for something as complicated as dropping a bomb!!
thanks, glad you liked it. Make sure to check the other videos out and welcome to the channel.
Doctor, thank you for your videos! I have been obsessed with nuclear weapons since I was assigned to guard and protect these weapons in the Marine Corps, you awnser questions I've never had the courage to ask, I really like your videos, although they do make me wish I had paid more attention in math and physics class!
Quality informative content, expressed for the world's sake.
Thank you for your work and dedication.
I am glad you found the content of interest, make sure to check the several follow-up videos. Thanks for watching and welcome to the channel.
This video is a masterpiece! Congrats and thank you!
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Wow, excellent walk-through of the critical concepts to manage the airplane flight to account for shock wave impact, to safeguard the air flight crew in this critical event in history. Much respect, sir.
Glad it was helpful! Thanks for watching and welcome to the channel, there are already new videos and more coming soon. Subscription is highly appreciated ;)
It seems generally the strategy is to turn until the ground zero point is directly behind you and then keep going straight away from that point. If you keep turning then you're no longer going straight away, if you don't turn enough you're also not going straight away. To get far away from something, just keep it directly behind you and go as fast as you can. Don't need any math. It works regardless of what kind of airplane you have.
Yes, that is what I thought too after the circle with the tangent was drawn, but I would imagine it is easier to tell the pilot to turn whatever degrees instead of turn until the bomb is behind you.
No question it's easier to give a pilot instructions, but overlooking that basic principle while describing a process to find that ideal angle seems very strange to me.
I believe that you might be right and several others have pointed this out. I am planing a follow-up video to address this, so please reply to the question below. The whole calculation looks like an overkill when the final answer is simply "turn around until you face away from the target and then keep going straight." I will not attempt to defend my solution; however, I have the follow-up questions:
1. how do you prove that this is in fact the optimal path? I mean, imagine yourself in front of a military panel consulting your expertise: how would you justify this for real not just with "trust me, I don't need any math"
2. (related to 1.) the "turn around until you face away from the target" does not tell you anything about how far the airplane will be when the shockwave hits; therefore, how can you guarantee its survival?
1. A) Distance to the bomb is a monotonic function of horizontal distance, so ignore for now.
B) The second derivative of distance to the bomb over time is increasing as the angle goes from heading towards the bomb to going away from it. This is true for any location, speed and heading.
C) The second derivative of distance to the bomb over time is strictly decreasing along all straight paths not directly towards/away from the bomb.
D) Given B, C and the fact the distance is an integral of velocity away from the bomb it always increases the distance to the bomb to turn away from its point of explosion as fast as possible until you are facing away and the go straight.
2. Once you know you need to turn away from the bomb until you are facing away from it, you know TA and TB are tangent to the same circle and thus of equal length. With that established the angle is 180-2*arctan(R/TA) and the calculation of the distance can be done similarly to your video, but a bit quicker and with an analytical answer.
Thank you for your brilliant explanaition about solving this problem
thanks for the positive feedback, glad you found it of interest
Great exxplanation congratulations on!
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OMG, 58 years ago after 2 years of college physics, calculus, linear and differential equations, I could have solved this. Now I just marvel at the beauty of your analysis.
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Excellent analysis of the airplane's path. Thank you.
Glad you liked it! You are welcome to explore the new videos and welcome to the channel.
Fascinating! Thank you.
I am glad you found the video of interest, you are welcome to check the other videos, share and subscribe, thanks
Excellent! Subscribed.
thanks, glad that you found it of interest. More coming soon.
Interesting analysis - lots of steps. Fascinating! It seems obvious, though, that (1) you fly as fast as you can, and, (2) you stop turning when you are heading directly away from the target.
But could they calculate that in real time? Seems like you'd need eyes in the back of your head. :-)
you are right, these are obvious steps; however, the calculation of the optimal angle free the pilot from having to make a decision on the spot by continuously checking "are we aligned yet?" but instead made it easy: just turn 154° and "run" (flight?) like hell
there is no calculation to do in real time; once the maximum speed of the plane is known they just had have the desired speed and altitude when dropping the bomb. For this Tibbets carefully selected his crew and they practiced many times.
But you can calculate the angle necessary to go in the direction opposite impact very easily. Much easier and cleaner than the optimization. Draw a circle representing airplane turn. Draw line from impact that is tangent to circle. Angle between impact to drop point line and escape line is 2arctan R/d. I am curious how close this is to the optimal or if it is indeed optimal for some range of R and d. I wish you had discussed this intuitive heuristic but nevertheless great video!
you are right and several others have pointed this out. I am planing a follow-up video to address this, so please reply to the question below. The whole calculation looks like an overkill when the final answer is simply "turn around until you face away from the target and then keep going straight." I will not attempt to defend my solution; however, I have the follow-up questions:
1. how do you prove that this is in fact the optimal path? I mean, imagine yourself in front of a military panel consulting your expertise: how would you justify this for real not just with "trust me, just go tangent to the turning circle"
2. (related to 1.) the "turn around until you face away from the target" does not tell you anything about how far the airplane will be when the shockwave hits; therefore, how can you guarantee its survival? @@steveHoweisno1
Great video! Even better trig!! Thanks
I am glad you found the content of interest, make sure to check the several new videos. Thanks for watching and welcome to the channel.
Interesting problem! Well presented! Thanks! :)
I am glad you found the content of interest, make sure to check the several follow-up videos. Thanks for watching and welcome to the channel.
Thanks Jorge - great video!
It got me thinking and it seems like there is a simple geometric argument which gets basically the same answer:
theta = 2 * arctan(-x_T/R) = 155.6 deg
Assuming constant speed throughout, to maximise distance from the explosion we should end the turn manoeuvre when the plane is heading exactly radially away from the target (otherwise, we can always increase our radial speed component by turning a bit more).This completely fixes the geometry of the situation in terms of the target range when the bomb is released (-x_T) and the turn radius R.
Going through some more comments I now see that others have pointed this out as well!
yeah, several others have pointed this out: the whole calculation looks like an overkill when the final answer is simply "turn around until you face away from the target and then keep going straight." I will not attempt to defend my solution; however, I have two follow-up questions:
1. how do you prove that this is in fact the optimal path? I mean, imagine yourself in front of a military panel consulting your expertise: how would you justify this for real?
2. (related to 1.) the "turn around until you face away from the target" does not tell you anything about how far the airplane will be when the shockwave hits; therefore, how can you guarantee its survival?
Lets say r=-x_T is the target range at release, then the (horizontal) distance from target both at the start and the end of the turn manoeuvre is r. The length of the turning arc is theta * R. And between release and explosion the plane travels a distance r (same speed as the bomb). Since theta * R < r we know that the plane can complete the turn before the bomb explodes. This means after the turn the plane can travel another distance r - theta * R before the bomb explodes, giving it a total (horizontal) distance of 2 * r - theta * R from the bomb when it explodes.
To get the total distance when the shockwave hits we can then solve this equation for t (the time interval between explosion and shockwave hitting):
(v_s * t)^2 = (2 * r - theta * R + v * t)^2 + (Delta z)^2,
which finally gives the distance as v_s * t.
Doesn't fully answer 1) to be fair, but at least lets you ascertain that you can get to a save distance.
@@jkzero
I'd figure the radius of the turn isn't exactly even because in general the fastest way to turn an aircraft is to climb going into the turn and dive coming out. It evens out in terms of energy so the plane regains its initial indicated airspeed prior to the turn, but it does allow for a tighter turn. In the scheme of what there is to work with in this scenario, it's probably not amounting to that much in terms of quick calculations. The crew however would be taking any little extra they could get.
you are right, a complete description of the problem requires a 3D characterization of the path; however, as indicated in the list of assumptions, I have kept the problem only in 2D for simplicity. Despite the assumptions, the final result agrees very well with the observations, which indicates that all the extra details are quite negligible.
At 30000 feet altitude the indicated airspeed is going to be a lot slower than the actual speed of the airplane over the ground. Probably well over 400 mph.
Great video 😊
excellent video!
Thank you very much!
Thank you for your explanation. I am not a math buff but your video made the manoeuvre of the plane very logical.
thanks for the comment, I hope to not scare people away due to the math; unfortunately, it is impossible to explain details without using a few equations. But I plan to keep telling physics stories but always sprinkled with some math in a way that the equations can be ignored if the viewer doesn't feel comfortable with them.
The unspoken premis here was that the escape path should be on the line crossing the detonation point, which is probably true but not so obvious.
Why its not obvious?
Great explanation. I think it would add significantly if you included a brief discussion on estimating the likely magnitude and direction of sources of error when the simplified 2D model is compared to the real maneuver.
this is a great piece of advice, I will make sure in the future to point out the weak points of the assumptions. Thanks so much for the constructive feedback!
Ah, the good old "spherical cows in a vacuum" issue.
@@jkzero There would have been a small vertical distance component from the center of the bomb blast to the aircraft when the shock wave hit the aircraft, but since the point was to figure out the minimum maneuver required to keep from destroying the aircraft, that extra distance would be just so much gravy in the escape maneuver.
Once the principles are explained, it's obvious that the aircraft needed to turn back with a quickness from the point of bomb release. Kamikaze was not our game.
So… the optimal escape angle is that which is tangent to the turning circle… That way, when you complete the turning maneuver, you are headed directly away from the bomb.
you are right and several others have pointed this out. I am planing a follow-up video to address this, so please reply to the question below. The whole calculation looks like an overkill when the final answer is simply "turn around until you face away from the target and then keep going straight." I will not attempt to defend my solution; however, I have the follow-up questions:
1. how do you prove that this is in fact the optimal path? I mean, imagine yourself in front of a military panel consulting your expertise: how would you justify this for real not just with "trust me, just go tangent to the turning circle"
2. (related to 1.) the "turn around until you face away from the target" does not tell you anything about how far the airplane will be when the shockwave hits; therefore, how can you guarantee its survival?
@@jkzero You have to plan for bad weather and you have to act fast. You have to know what to do from the start. . All is well prepared when flying by instruments, they can be easily read from the pilot alone. Then, the explosion position is best visible only after the turn. Better to know how to fly based on instruments.I guess.
@@jkzero 1. I think it's simple to prove that flight directly away from the blast in the "run" phase of the escape maneuver is optimal. At any other angle, you can increase your ds/dt by turning towards the vector from the blast to the plane's current location. Since an aircraft can't make discontinuous changes in heading, the turn (or *a* turn) has to be tangent to the final straight "run" trajectory. I don't know how to prove the optimality of the path before the end of the turn.
2. It does - the circumference of [a segment of] a circle is easy, the distance the plane covers here is just R*theta (in radians). So we know at what time the plane ends the turn. At that points it's the same distance from the target as it was when it began its turn (two right triangles mirrored around each other's hypotenuses), and after that the plan distance just increases linearly. That's "just" a matter of solving (for t, then s) a sqrt(1+vt) = wt type of system, no?
@@jkzero If you are flying away from the explosion point but at an angle to it, turning farther away from the bomb is obviously beneficial (with the assumption of constant speed and altitude). Turning towards it is obviously bad. That naturally leads to the right answer.
"turn around until you face away from the target" makes the calculation of the ideal turning angle very simple, you can then calculate the distance after you found the angle and check if that distance is safe.
@@whocares2277 Additionally: Having the plane fly directly away from the blast point, presents the least amount of surface to the heat radiation.
Apparently, that was not their #1 concern. But if it were a real issue, they could have (theoretically) reduced the irradiated surface area to the very minimum by taking a slight climb angle just before the calculated detonation time, so that also the in the vertical orientation the plane is pointing directly away from the detonation point, that was 9 km below the plane's altitude.
phenomenal video one of the more interesting ive seen
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Very interesting analysis !
Glad you think so! You are welcome to explore the new videos and welcome to the channel.
I think you can solve this a lot easier than what’s in the video. The main characteristic of the path that maximizes your distance from the bomb detonation is “start turning when you drop the bomb, then turn until you are facing directly away from the bomb”. The only info you need to do that is the distance away from the target the bomb is dropped, and the turning radius of the plane. You can construct a right triangle from the drop position, target position, and center of the turn. If you reflect that over the hypotenuse, it will show the optimal escape path, and to get there you just ride the arc defined by your radius. So the angle to turn is just twice the angle made by the target/turn center/drop locations.
several others have pointed this out: the whole calculation looks like an overkill when the final answer is simply "turn around until you face away from the target and then keep going straight." I will not attempt to defend my solution; however, I have two follow-up questions:
1. how do you prove that this is in fact the optimal path? I mean, imagine yourself in front of a military panel consulting your expertise: how would you justify this for real?
2. (related to 1.) the "turn around until you face away from the target" does not tell you anything about how far the airplane will be when the shockwave hits; therefore, how can you guarantee its survival?
@jkzero well the first question is simple, just use goofy maths terms and have self confidence while speeching overcomplicated stuff to impress military guys
@@Matt_ctn this is just brilliant
@@jkzero 1) It seems obvious that as long as the plane is flying in direction where all the movement is away from the target the direction is not optimal and as long as turning toward that direction you are making better and better progress
2) this where the analysis of the turning radius and basic geometry (I think the use of the term physics is kind of wrong in this video as most is math and geometry) gets you the result
BTW I would have thought things like minimum turning radius for an airplane would be in the operating manuals...
@@Axel_Andersen I agree here with Axel. First get the optimal geometry (turn until facing exactly away from target), then solve for all relevant segments to find the exact distance.
There is a much simpler way to obtain the true value: for the maximum distance the plane needs to face away from the bomb after the maneuver. The turning circle is then tangent to AT and BT. This creates two identical right angle triangles where the two sides are R and AT and the angle is theta/2.
Substituting your values into theta = 2 arctan(AT/R) gives 154.46°
You are right; several others have pointed this out, I am planing a follow-up video to address this. The whole calculation looks like an overkill when the final answer is simply "turn around until you face away from the target and then keep going straight." I will not attempt to defend my solution; however, I have two follow-up questions:
1. how do you prove that this is in fact the optimal path? I mean, imagine yourself in front of a military panel consulting your expertise: how would you justify this for real?
2. (related to 1.) the "turn around until you face away from the target" does not tell you anything about how far the airplane will be when the shockwave hits; therefore, how can you guarantee its survival?
@@jkzero 1. True, your approach is more straight forward and justifiable to non-experts. I havent put too much thought into a proof, but I guess you could suppose an optimal angle larger or smaller than the true optimum and show that varying the angle towards heading straight away is beneficial.
2. Since the speed of the shockwave has not been used at all in my calculation, I guess youd have to repeat your calculations but maybe exploit some shortcuts only applicable for the optimal angle.
For the application of actually saving the aircraft, your approach is the most appropriate, but I guess playing with geometry and symmetry might give more insights into the math.
I love the language, math. I don't speak it well, but it certainly is beautiful. Excellent presentation.
thanks for the feedback, I am glad you liked the content. In case you want to learn the math check the link below, the lessons are fun and interactive; and the first 30 days is free. In case you want a full subscription the link gets you 20% off and also helps supporting the channel ;) brilliant.org/JKzero/
You must take into consideration that the speeds given for the airplane are airspeeds and must be corrected for pressure (altitude) to give a true air speed (tas) and for pressure and wind to get a ground speed 8:08
Very well done.
Thank you! Glad you found it of interest, make sure to check newly posted videos on the same topic: ua-cam.com/play/PL_UV-wQj1lvUhNttvv4_KsYrQxHygj3Ey.html
Jorge. Excelente y muy didáctico.
Gracias, Humberto, me alegro que te haya gustado. Dale un vistazo al resto de los videos y se agradece el like y subscribe ;)
All airplanes have a 'maneuvering speed' in he airplane manual. It varies with gross weight and other factors, but this is the maximum rate of turn ( G force) the airplane is capable of without stalling. Structurally, it is the air speed for a particular gross weight where the stall speed matches the maximum g force the plane is designed for. Pilots use this in heavy weather and turbulence. You don't want to be flying fast enough that you fold the wings by entering an updraft. If the airplane stalls when entering an updraft, the lift (G force) is reduced aerodynamically keeping the wings intact. You want to be fast at the point of bomb release to throw the bomb the maximum distance to the target, then you would enter a climbing wing over maneuver to get to maneuvering speed for the remainder of the turn, then a shallow dive to increase speed to VNE (velocity never exceed) Full emergency power until just before the shock wave is calculated to catch up and then slow to maneuvering speed for the shock wave impact. Airplanes aren't designed around shock waves coming from the rear, but you cannot outrun these things, so being too fast when they catch up is not a good situation.
You made a number of errors -
1) bombs do not follow a parabolic course (because air drag is a thing and the nuclear weapons were specifically designed to have extremely high drag to maximise the escape time - later bombs used parachutes to further slow the fall).
2) the optimum course does place the bomb directly behind the aircraft - because the aircraft doesn’t have a zero turn radius. Basically you need to find the tangent of the curve to identify the optimum roll out point. You also need to be high enough at bomb release to complete the turn before detonation or else you are toast.
Knowing the distance and radius of the B-29’s turn you can solve the problem with a compass and ruler. 2.5G is a 67degree bank angle (maximum bank angle is limited by the stalling speed of the aircraft at the release altitude and G load (called an accelerated stall) - and stalling in this operation would be BAD). My guess is Tibbets had probably practiced this manoeuvre repeatedly to nail down the absolute minimum turning radius possible without stalling and spinning the B-29 (probably significantly tighter than the book figures for the aircraft).
I invite you to check the 05:24 mark in the video, where the assumptions are presented and explicitly stated; number one is no air resistance.
To ensure hitting the target (albeit at 1500 ft above) a very heavy bomb with an aerodynamic shape will almost free-fall. The actual fall time is so close to the theoretical that it not only fell at almost the zero-air resistance acceleration, i.e. no terminal velocity, but also broke the sound barrier in its fall. Larger bombs tested in later years needed parachutes as you say, but aiming them had also improved considerably.
@@karhukivi The first generation were specifically designed to be subsonic (precisely to enable escape).
The British “Grand Slam” and “Tallboy” were supersonic by design but only shared the drop mechanisms and weight.
The published release height (31,000 ft), detonation height (2000) and the fall time (43 sec)for the Hiroshima bomb fit very closely to the theoretical time for a zero air resistance fall under normal gravitational acceleration = 0.5gt^2 = 16x43^2 = 29,600 ft. I read elsewhere that the detonation height was 1500, in which case it is a near-perfect fit. So little or no evidence of air resistance. @@allangibson8494
Dr Diaz gets the right answer (of course) but he does it the hard way. The easy way is to observe that the optimal angle of turn is that which ends when the aircraft's direction points 180 degrees away from the point of detonation. At that point, the component of the aircraft's speed away from the blast is maximised (at 100%) and any further turning will reduce it. The angle can be calculated by elementary trigonometry as twice arctan(D/R), where D is the distance from the point of bomb release to the target, and R is the radius of turn. For the numbers given in the video, namely D = 6268m, R = 1450m, this gives a turn angle of slightly under 154 degrees. This is a much simpler approach and in addition, gives some insight into the geometry of the problem.
You are totally right, your solution works and you find the correct angle. Other viewers have pointed this out: the whole calculation looks like an overkill when the final answer is simply "turn around until you face away from the target and then keep going straight." I included the whole long calculation instead for two reasons: firstly, the "turn around until you face away from the target and then keep going straight" solution does not guarantee that this is in fact the optimal path. I mean, imagine yourself in front of a military panel consulting your expertise: how would you justify this for real? The other reason is that the "turn around until you face away from the target" does not tell you anything about how far the airplane will be when the shock wave hits; therefore, its survival how cannot be guaranteed.
Amazing video.
Glad you liked it, more coming soon
One of the best I have ever seen! Thank you
I am glad you found the content of interest, make sure to check the several new videos. Thanks for watching and welcome to the channel.
Prof Diaz, thank you for sharing this mathematical excursion. I had always wondered how the escape angle was calculated.
thanks, I am glad that you found the video of interest. I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?
@@jkzero I have a rusty Master's Degree in Math with a minor in physics that I received in 1974. I like presentations like yours, where I can review my math and physics. The turn-away angle was a question that I've been curious about for years, ever since I found out about it as a teenager. At one time, I was working on a M.S. in Operations Research, where I like to find optimum solutions to problems. Certainly, the turn-away angle is an optimum problem. This was many years after my first master's degree. I got distracted and never completed that degree as I moved away from that college to be with the woman who eventually became my wife.
By the way, it was the UA-cam 'mighty algorithm' that showed your presentation as one of many options to look at while perusing YT.
@@rmandra thanks fro sharing your story and I am glad that the algorithm found you. More videos coming soon!
Thanks for sharing and I am glad the algorithm is working and brought you here. These days you can take free online courses; the MIT OCW is a fantastic source for all levels of math and physics, you should check it out. I hope you find the other videos of interest too and welcome to the channel.
The crew that dropped the Tsar Bomba was only given a 50% chance of survival.
almost a kamikaze mission
This was only a theoretical figure, and subject to many biases and interpretations.
Just because you have read somewhere that there was a '50% change of survival', doesn't mean it should be taken litterally.
@@sailorman8668
"Both aircraft were painted with special reflective paint to minimize heat damage. Despite this, Durnovtsev and his crew were given only a 50% chance of surviving the test.[42][43]"
From Wikipedia. For what it's worth, reference 42 appears to be an outdated link, and reference 43 contains this paragraph:
"In order to give the two planes a chance to survive - and this was calculated as no more than a 50% chance - Tsar Bomba was deployed by a giant parachute weighing nearly a tonne. The bomb would slowly drift down to a predetermined height - 13,000ft (3,940m) - and then detonate. By then, the two bombers would be nearly 50km (30 miles) away. It _should_ be far enough away for them to survive."
Awesome video. I always wonder about the return path after drop. Didn't know if it they could pretty much any path, or they were not sure, and optimized it.
Thanks for the feedback, glad you liked it. I honestly never wondered too much about this, until some years ago when I read a paper about this. But, yeah, I shouldn't be surprised that they calculated the optimal path in advance.
Now I’m going to need you to solve for String Theory, remember to show your work & also provide proof. Thx 🙏 😊
no pressure :) I did have the experience to work in string-theory topics during my undergrad thesis. Beautiful mathematics but found it boring in the end, it felt like pure mathematics with little connection to the real world so for my M.Sc. and Ph.D. I moved on to more close-to-earth topics, you know, physics.
The graphics remind me very much of the excellent series "The Mechanical Universe" from Cal Tech.
for the nice graphics I thank @3b1b that made his mathematical animation library open, learning to use it requires some dedication but I think it is worth it
I wonder if the B-29 would've tolerated a toss-bombing manouever - dive to Vmax, pull sharply into a 45° climb (hopefully, keeping your wings), release bomb, turn rapidly ~160°
Very good video.
one detail, on time 21:20, the radius of the plane is 1450m and not 1450Km (it is write kilometers)
good catch, and thanks for letting me know, other users have also noticed. I did my best to catch these minor mistakes but still some went through. I hope this didn't ruin the general flow and the final result
@@jkzero You could add errata to the video description-that is editable, right? Some people also correct errors in Klingon subtitles. This one is pretty obvious, especially with the miles also being there, but it's always nice when one can double-check. Great video👌
@@OnDragi thanks for the suggestion, I should have done this earlier. I added a correction comment in the description.
One of the funniest things in being a physicist is doing stuff like that :D You can do some hard science daily but gathering data from everywhere and crushing numbers is so relaxing. I also teach kids in the secondary and we estimated the velocity of falcon 9 at Mico taking masses from wiki and reading time from telemetry from yt :D We got 1400 m/s. They loved it!
This is a fantastic applied problem, and great that get your students involved. Congrats. I really love those simple projects that get the students to try their math and physics intuition. A friend of mine once did an online session with students in different geographic locations, a few thousand kilometers apart, and use the shadow of a stick to calculate Earth's radius. Kids loved it.
The escape angle also seems to minimise the plane cross section exposed to the incoming blast waves.
this could be; however, this is probably just a nice-to-have result. If closer to the explosion center, no matter the cross section the aircraft would have likely been crushed by the strong pressure wave.
Slant range could have been further increased by pitching down to accelerate - running away from the bomb faster. Exactly when to pitch down and by how much would be a trivial calculus problem for any one of the folks working on the bomb. Why they didn't bother is a curious question. Did they assume 20km was good enough, and just turned their attention back to working on the next bomb (Fat Man)? Concerned the added complexity of a three dimensional (rather than two dimensional) maneuver might mentally overload the pilots? Left the maneuver calculation to someone less skilled at math? Wanted to keep the altitude high so the pilots could recover from a temporary loss of control? Something else? We'll probably never know.
Anyway, thanks for a nice video!
I am glad you found the content of interest. To be honest, I do not know about the details of the actual calculation, I have searched for the original for years with no success. I can only guess that a 3D treatment adds little value to the simplicity of just "turn around this angle and run like hell." Another guess regarding your valid point of gaining speed by pitching down, maybe they did want to risk the aircraft at lower altitudes, but again, I have not been able to find the details. Make sure to check the several follow-up videos. Thanks for watching and welcome to the channel.
Great video. I am still surprised the aircraft survived because the shock wave arrived from the rear and the flight controls were not designed for this load case. I would have expected rudder, elevators and ailerons to be very vulnerable and likely to be seriously damaged unless special precautions were taken. E.g were the controls strengthened or momentarily locked in some way to survive these "reversed" loads
An interesting applied physics problem, with a fascinating historical context, and that can be solved just using school math. Very well done! 😀👍
You got all the intended characteristics in the video. I plan to continue at different levels of technicality in both the physics and mathematics, I hope to get feedback in case people want more advanced details and topics.
Looking forward for more material like this, already subscribed 🙂 I'm a physicist too, I believe the explanation was very pedagogical. Don't be afraid to tackle more advanced physics/math in the channel, many on YT are doing it with popular success. I guess the key is that every one gets something from a video, even f they don't understand everything. @@jkzero
@@luisdaumas I appreciate the encouragement. A new video is already underway and I found in the middle of its production that a complete explanation requires solving the diffusion equation; since I do not want to loose the audience I might create a separate video just for the solution for those interested. Will see.
@@jkzero That's what some science, and especially medical youtubers, do when they tackle some topics. They will make a video like you did here but then make a separate video that goes into more technical detail for the advanced members of their audience who want to get into the finer technical or mathematical points.
Was the bombing run plotted so that after the turn the plane would have a tailwind prior to the detonation? It would help them get a bit further away before the blast wave hit them.
The whole process, including the shock wave, is taking place inside the mass of air. If the air was moving, it would have made no difference to the survival problem. They might have had to take account of it to get the bomb to explode exactly over the target.
I commented on wind as well. I assume they used what they knew about local wind patterns in their withdrawal plan.
@@sylviaelse5086 Yes the whole process took place inside the same (moving) air mass. But if the wind is behind the airplane, thereby moving it at a faster ground speed, it would be fractionally farther down range from the explosion.
The shock wave is supersonic and omnidirectional, and very likely cancelled any effect of relative wind at the site of the explosion on the shock wave itself. I'd gladly take being farther away in the airplane as benefited by a tailwind, even if only a little.
@@2whl4re If the wind was blowing in the direction of retreat, that means the wind would have also slowed the bomb's forward velocity down.
the shock wave moves at the speed of sound, is not supersonic, the speed varies by the pressure and temperature of the air gas@@2whl4re
Is maintaining constant altitude the best strategy? Depending on what's limiting your speed you might be able to trade some altitude for horizontal speed, or now that the plane is lighter, climb while maintaining airspeed.
they did in fact dive down to gain speed, they lost about 500 meters during the maneuver, which put them closer to the explosion center. In the video I made the assumption of constant altitude to keep the problem in 2D. I am not aware of a 3D description of the problem, but I suspect that the speed gain would not be enough to put the Enola Gay at a safe distance. I also think that they did not want to risk the B29 at lower altitudes when turning a specific angle was enough to keep the crew and airplane safe.
@@jkzero They would not have been able to maintain constant altitude with a 60° bank angle at their high altitude. It would be interesting to see a 3D calculation of what the ideal bank angle would have been 😄
I was interested that you didn't say "assumed safe distance" since I didn't assume that the resistance of the B-29 to such a shockwave would have been well known. It's also intetesting to consider how well known was it that the plane would easily handle the forces of a full throttle 60 degree banked turn. An interesting follow on exercise would be calculating the G-force experienced by the plane and the crew during the turn. Very intriguing problem and solution. Secondary shockwave next?
Very Interesting! I'd say a bit too unnecessarily technical, except from deriving the radius,
as everything boils down to two times: the time to turn(make running more efficient) vs the time to run.
I suspect the resulting function is much simpler with the proper abstractions.
Thanks for watching and the comment. Several viewers have complained about the high degree of complexity but I think that this is consequence of showing every single step of the calculation. I have received suggestions on how to do it simpler but so far nobody has provided a concrete solution that is simpler and convey the same information. What I mean by this is the following: it is not just about finding the optimal angle but also determine how far from the explosion center the aircraft will be when the shockwave hits. In front of a military panel deciding the details of such a mission I am sure that high degree of technical details would be required.
The optimal angle with these assumptions can be found analytically to be 180-2*arctan(Radius/43s/bomb horizontal speed) as that is when the plane will be moving directly away from the explosion. Until that moment turning increases the rate of increase of the distance from the Enola Gay to the explosion.
You are completely right, your account provides a correct qualitative description of the optimal angle; however, in such a delicate mission the military would always request a quantitative description too. In fact, finding the angle is just half of the problem, you also need to prove that using that angle the aircraft would be beyond the safe distance. Moreover, the solution presented is not an unsolvable equation, in fact, it is a trivial quadratic equation leading to an analytic solution for the slant range. The use of a computer code was only to avoid more ugly equations on a already long video.
incredible
I am glad you found the content of interest, make sure to check the several follow-up videos. Thanks for watching and welcome to the channel.
with slightly different parameters/inputs, 159 may have been the true answer. there is, "what the plane theoretically can do", and "what a particular plane really CAN do". And the boiler plate B-29 were not standard aircraft.
Great and detailed explanation. I like this kind of practical problem than a persion buying 50 watermelons.
However the way you do the transition for the equations makes it difficult to follow.
For example, at 13:46, the sole xB equation morph into all xB, yB and zB, with the minus sign morph into zB, which makes no sense.
Or at 18:14, theta morph into ys, ys morph into = R, minus sign morph into parenthesis (why delete ys then create a new ys at the exact same place).
The transition looks cool and good to keep attention, but it backfires a lot when I (watcher) want to follow a variable and see where it ends up after the manipulation to be realize that it morph into a completely unrelated thing.
I think if the relative position in the equation of an element is not change, it should only move around, not morphing into other thing (yep, that's why I use the word 'morph' since the beginning).
thanks for the extensive comment and extra thanks fro the constructive feedback, this is the kind of details that really contribute to make it better next time. I see your points, to me the 'morphing' of equations looked OK as a why to 'clear' the screen of already used expressions and replace them with partial or final results but from your comment I get that different people might interpret them in completely unexpected ways. The transitions might be overdone and I take it as a learning point for future videos. This was my first video using Manim (the mathematical animation tool under the hood) and maybe I went overboard in some points. Once again, I really appreciate the feedback!
4:25 - “works with conventional weapons”
True-ish
Requires *some* altitude. Too low / too slow and you’ll get rocked by your own blast. Or - in the case of the dambuster bouncing bombs, there is dramatic war-era footage {seen in the documentary (NOVA maybe?) of the Canadian prof/student recreation} of a bouncing bomb test popping up and severing the tail of its own bomber (fatally I assume - they were at no more than 100 feet above the surface)
this is crazy man. The math is super involved, but I had to stop mid way caues I cannot focus on this problem given the history behind it
I fully understand; I am surprised my video was selected for a prize, I really doubted about making it in the first place but convinced myself that it was worth for the math lesson but you are right, dropping the bomb was a horrible act. Anyway, thanks for watching
Lovely presentation but you may need one more edit.
You mainly articulate that the Enola Gay maintained altitude after bomb-release. She did not.
Mentioned early in your presentation then ignored, was that the right turn was a DIVING right turn.
Col Tibbetts was trading altitude for speed and of course distance.
The Silverplate Squadron had been practicing that turn for nearly a year stateside.
As long as Enola Gay didn't exceed the speed of sound (Tibbetts DID know better) it was a good swap.
THAT math is every bit as fascinating as what you put into this excellent presentation.
It will be a lovely visual adding the 3rd dimension to your graphics, and so very worth it.
I get your point, it is true that the presentation involves several assumptions; however, the calculated distance to the target fits perfectly with the actual distance determined by the crew, which indicates that despite all the assumptions, the results holds well. This means that relaxing the assumptions introduce negligible corrections to the final result. I hope this is more satisfactory.
@@jkzero
Distance from release point to point of detonation is accurate.
It is what happens after release of the gadget that needs one more data point - or so I feel - the effects of the diving turn to the Right
.
Thank you by the way. I sort of understood what the diving 159-degree turn was all about - but not perfectly
Your explanation is super helpful.
Further Kudos for the film clips from 1945.
The B29 had a cruise speed of 220mph/350kpg.
Max speed is listed as 357mph/575kph (you quoting Steven Walker at 7:41 uses this number too)
However I don't think 357/575 is the max for the airframe. I think it is the max in level flight under its own power. Thems were big engines.
Descending, any aircraft will pick up speed, trading altitude for velocity.
This is vital stuff in aerial combat with fighters, but it impacts bombers too.
At 2:34 you again quote Steven Walker: "This has long been my understanding too.
Descending under full power will translate to more ground distance but less altitude.
I would certainly make that trade running from a nuke - I'm confident Tibbets and the Silverplate boffins (including VanKirk) modeled it many many times stateside.
It's just yet another triangle for you to diagram. If you don't want to do it 3-D a 2-D view from release altitude, graphing points B-D should do the job.
For extra credit you could make point A - to B a straight line. Easy to calculate time and distance. Imperfections would be background noise.
BTW: I posted this vid of Theodore Van Kirk the navigator you quote some years ago. Its from a talk I attended in July 2010: ua-cam.com/video/T9C_SOQLfow/v-deo.html
That's part 1 of 14. I still haven't learned to sew video segments together. REALLY liked a couple of punchlines in part 2: ua-cam.com/video/xWQFbRfPmWw/v-deo.html
I got a problem like that in my final exam at course Shock Waves in Real Media. Bonus question was to calculate the hole deep after explosion. My professor asked me many questions about fire ball radiation after I solved numerical problems. The class was a part of major in chemical physic’s including combustion and detonation.
I wish I had a course on Shock Waves in Real Media, I didn't have any formal course on blast physics, I took several courses on fluid dynamics, some lectures online, and then read many scientific publications and books to teach this topic myself. Great experience that ended up in a few scientific publications, check my videos on the radius and energy of a nuclear blast for details ua-cam.com/play/PL_UV-wQj1lvUhNttvv4_KsYrQxHygj3Ey.html
@@jkzeroDr Jorge, I enjoy your videos. I’m going to watch all of them! You do very good work!
I am thinking about this and Ive got an idea which i dont know if is correct. At 11:14 If I know turning radius of that plane, could I draw that as circle with appropiate size and just draw a line from target in a way it just touches the circle on a side and get fairly educated guess instead of actually calculating vectors? Or would it be wrong?
you are totally right, your solution would work and you would find the correct angle. Other viewers have pointed this out: the whole calculation looks like an overkill when the final answer is simply "turn around until you face away from the target and then keep going straight." I included the whole long calculation instead for two reasons: firstly, the "turn around until you face away from the target and then keep going straight" solution does not guarantee that this is in fact the optimal path. I mean, imagine yourself in front of a military panel consulting your expertise: how would you justify this for real? The other reason is that the "turn around until you face away from the target" does not tell you anything about how far the airplane will be when the shockwave hits; therefore, its survival how cannot be guaranteed.
Incredible animation !
How do you do it ?
thanks, I am glad you value the effort, I spent several hours on the animations. I used Manim, the Mathematical Animation software created by @3b1b. It is open source and the community is very welcoming, you check it out here docs.manim.community/en/stable/examples.html
Seems like the easier solution would be to put a parachute on the bomb, therefore giving the plane more time to escape. Which is exactly what they did as bombs got more powerful and no manoeuvres would never give the plane enough time to escape.
The bombs dropped over Hiroshima and Nagasaki had no parachutes because one of the objectives of these missions was precision bombing, even radar bombing was not allowed. A standard parachute would have slowed the bomb down enough for a safe escape without the need of the special maneuver; however, any potential wind could have made the bomb drift far from the carefully selected target. One sort of parachute was indeed used: if you look at pictures of Little Boy and Fat Man you will notice a boxy tail fin called "California parachute." Their main objective was to stabilize the drop by minimizing spinning of the bombs but they could also slow them a bit down. Any footage of nuclear bombs with parachute are likely thermonuclear (H-bombs), in which case the yield is so high that you must slow the drop down to escape; and the yield is so high that any drift from target is probably irrelevant, everything will be destroyed anyway.
@@jkzero Thanks for the response and the additional detail!
The computation is much simpler if you work it out in the target-B-C-D line. You just need the turning radius and airspeed to determine when the plane reaches B. The distance target-B is the same as target-A. From there the plane is just flying away from the target, so your computation from 6:40 applies.
You're assuming that T, B, C, and D are all collinear, which they are in most of the animations, but not in the general case where theta is variable
@@engineer_cat They are collinear for the optimal turning angle (with the assumptions made here, like constant speed). You stop turning when you are flying exactly away from the explosion point.
@@whocares2277 intuitively, that makes sense, but I'd like to see a mathematical proof that it always matches even if you vary some parameters (eg speed, turning rate, etc)
@engineer_cat just take a path from B to C which isn't colinear with TB (call the new final point E). |BC|=|BE|, |TE|
@@phire442 and here you're assuming B is a fixed point, which it isn't. Imagine a circle of the aircraft's turning radius tangent to the line TA at point A. Point B can be anywhere on that circle, depending on theta.
Thanks
Thank you so much for supporting the channel. Thanks again for using this material with your students and I look forward for them visiting here. Greeting to the other side of the globe.
this is why i think math is by far the most important tool humanity has ever created
All you really need to know to solve this is what is the turning radius of the plane. As soon as the detonation location is directly behind you, straighten the aircraft so you're heading away as quickly as possible. Turning more would decrease your escape velocity as would turning less.
several others have pointed this out: the whole calculation looks like an overkill when the final answer is simply "turn around until you face away from the target and then keep going straight." I will not attempt to defend my solution; however, I have two follow-up questions:
1. how do you prove that this is in fact the optimal path? I mean, imagine yourself in front of a military panel consulting your expertise: how would you justify this for real?
2. (related to 1.) the "turn around until you face away from the target" does not tell you anything about how far the airplane will be when the shockwave hits; therefore, how can you guarantee its survival?
@@jkzero You calculate the arclength of that angle, then do the calculations he showed for the straightline distance. That will give you the position in space and you can use pretty basic trig to determine the distance.
@@fireburner81 I agree with you, I think that the "you can use pretty basic trig to determine the distance" is exactly what I did but apparently showing all the details of the calculation made it look harder than it really is
Great explanation and visulalisation. I just have one doubt. Is the roll angle of the bomber during the maneuver related to the radius of the circle? If then how would they have determined which roll angle to go with?
Thanks, I am glad you found the content of interest. Answering your question: yes, the roll angle fully determines the radius of the circle, check the explanation at the 18:19 mark for the details of this derivation. I skipped the part of how was the 60° roll angle chosen: this angle gives something like 2g of net acceleration and you don't want this to be too high. Thanks again for watching, make sure to check the several follow-up videos, and welcome to the channel.
@@jkzeroOhhh ok, that makes sense. Thanks a lot for clearing my doubt
Since no one knew till the, " TRNITY TEST " what the yield of the bomb would produce, can we assume that the calculation for the escape path wasn't calculated till afterwards since a yield factor would be needed ? Or was it simply a matter of turn to the limits of the plane till a centerline thru the long axis of the plane would meet at the point of detonation ? Also was a last seconds climb considered since the plane made a diving turn,... in order to gain altitude into thinner air to have less shock vs. greater shock at lower altitude ? Thanks, great math work and video !
Thanks for the comment and the positive feedback! You are right, the actual yield was unknown until the Trinity test; however, I doubt that they waited until after Trinity to calculate the path. It probably makes more sense to calculate many paths and then pick the right one. Although the exact yield was unclear, the 20 kt estimate was well in many of the scientists' minds due to some easy back-of-the-envelope calculation that allows relating the critical mass, bomb efficiency, and yield. The rule of thumb was 1 kg of fissionable material ~ 20 kt.
The speed of the shock wave would have been the same at any yield.
However, her speed was higher, because she dived from 31000 feet, giving her spare distance too.
they did in fact dive down to gain speed, they lost about 500 meters during the maneuver. In the video I made the assumption of constant altitude to keep the problem in 2D. I am not aware of a 3D description of the problem, but I suspect that the speed gain would not be enough to put the Enola Gay at a safe distance. I also think that they did not want to risk the B29 at lower altitudes when turning a specific angle was enough to keep the crew and airplane safe.
The way you covered the history was really nicely done, and it's a cool idea for a video. However, this seems to be a _really_ overcomplicated approach. If your goal is to run away, then you should proceed directly away from the center of the blast. Then sketching a single triangle gives the answer as \theta = 2 asin(R/X) where X is the distance from A to T. Can you explain what's wrong with this simple approach?
thanks for posting a good question in such a candid way, several viewers have pointed out the same observation; however, most do it is such an aggressive and condescending way that provide little value.
Yes, you are right: the whole calculation looks like an overkill when the final answer is simply "turn around until you face away from the target and then keep going straight." I will not attempt to defend my solution; however, this way of apparently solving the problem leads to two follow-up questions:
1. how do you prove that this is in fact the optimal path? I mean, imagine yourself in front of a military panel consulting your expertise: how would you justify this for real?
2. (related to 1.) the "turn around until you face away from the target" does not tell you anything about how far the airplane will be when the shockwave hits; therefore, how can you guarantee its survival?
I wrote that the "turn around until you face away from the target and then keep going straight" method gives you the correct angle but it does not necessarily answer the original question: will the aircraft survive? I am planing a follow-up to address this point, but again, thanks so much for asking the question in such humble and respectful manner.
I would imagine the aircraft gained a lot of elevation and airspeed when it suddenly lost 10,000+ pounds of weight!
Forgive me if i'm naive but would it not be easier to just find the tangent from the blastpoint to the circle formed by the Enola Gay's maximum turning circle?
thanks for posting a good question in such a candid way, several viewers have pointed out the same observation; however, most do it is such an aggressive and condescending way that provide little value.
Yes, you are right: the whole calculation looks like an overkill when the final answer is simply "turn around until you face away from the target and then keep going straight." I will not attempt to defend my solution; however, this way of apparently solving the problem leads to two follow-up questions:
1. how do you prove that this is in fact the optimal path? I mean, imagine yourself in front of a military panel consulting your expertise: how would you justify this for real?
2. (related to 1.) the "turn around until you face away from the target" does not tell you anything about how far the airplane will be when the shockwave hits; therefore, how can you guarantee its survival?
I wrote that the "turn around until you face away from the target and then keep going straight" method gives you the correct angle but it does not necessarily answer the original question: will the aircraft survive? I am planing a follow-up to address this point, but again, thanks so much for asking the question in such humble and respectful manner.
If you know the speed, with which the airplane makes the smallest possible turn, you should be able to anticipate the distance between the plane and the explosion, when the axis of the plane in its direction of flight points to the explosion. From there on its just running strait away. Isn't all that matters the smallest and fastest turning radius of the plane?
Let me try to explane this idea to that imaginary military panel. The fastest escape route would be an instant turn around of 180° without changing the speed of the plane. Therefore, the nearer you get to this ideal turn around, the closer you get to that maximum distance from the explosion that you could achive with the ideal instant turn around. So, smallest radius possible with the highest speed possible. Both values result in a force of acceleration. Limiting factors for speed an turning radius is the capability of plane an crew to withstand this G-force. So, the turning angle is also a result of that capability of plane and crew. I wonder if this aproach to the problem would be proof enough. Maybe I would have to do the same calculations as you did in the first place.
It might be a good idea for a following video to describe the capabilities and the limits of the plane in handling g-forces in turns. I wonder if, back then, those numbers have been at the starting point of the whole calculation.
В Final Solution могли использовать производную от этой функции и определение вершин максимума и минимума чтобы узнать максимальное расстояние (км)? (on Russia language )
You are totally, right; the numerical solution that I found was only intended to reduce the mathematical complexity of the presentation that was already quite loaded for a general video. The fact is that the long and ugly equation that I programed can be solved analytically, it requires some rearrangements and it becomes a quadratic equation that we know how to solve with pen and paper. When I did this the video had almost 10 min extra just to simplify terms so I decided to cut this and use the easy route of just letting the computer do the number crunching.
I know it's off topic, but the noises and sensations described by the crew when the shockwave hit, sound violent enough to cause structural damage to the airframe. Does anyone know whether Enola Gay was tested for such damage after the flight?
I am not aware of any type of testing of the Enola Gay after the bombing of Hiroshima; however, it appears that the aircraft was in good condition as it served as a weather reconnaissance airplane for the second nuclear bombing over Kokura (which had to be aborted due to cloud coverage, Fat Man was dropped over Nagasaki instead). It was later selected to drop "Gilda," the bomb of the first nuclear test over the Pacific as part of Operation Crossroads. However, in the end another aircraft was used. After that, the Enola Gay remained in storage until put in exhibition at the Smithsonian.
25:32 "practical problem"
but seriously great video
yeah... I could have used better words; thanks for the comment and make sure to check the several follow-up videos. Thanks for watching and welcome to the channel.
Was any consideration given to slowing the bomb’s rate of descent either via parachute or a less than efficient aeronautical shape of the bomb?
The bombs dropped over Hiroshima and Nagasaki had no parachutes because one of the objectives of these missions was precision bombing, even radar bombing was not allowed. A standard parachute would have slowed the bomb down enough for a safe escape without the need of the special maneuver; however, any potential wind could have made the bomb drift far from the carefully selected target. One sort of parachute was indeed used: if you look at pictures of Little Boy and Fat Man you will notice a boxy tail fin called "California parachute." Their main objective was to stabilize the drop by minimizing spinning of the bombs but they could also slow them a bit down. Any footage of nuclear bombs with parachute are likely thermonuclear (H-bombs), in which case the yield is so high that you must slow the drop down to escape; and the yield is so high that any drift from target is probably irrelevant, everything will be destroyed anyway.