Wow adding drag really does make the whole situation much more complex. I’ve just finished second order differential equations in school and it’s really interesting to see them used inside a problem not just by themselves. High quality video as always!
Thanks. Since acceleration is the second derivative of position with respect to time, Newton's second law itself is really a second order ODE. So if you continue studying Physics you'll be seeing a lot of them in context!
For 0:20, I remember the drag force should be proportional to the the square of its velocity which gives us F = bv². As velocity, v constantly changes can't be included in the constant, b. May I know what I'm missing? Or could you how to solve the equation will be if we use F = bv²?
Indeed, quadratic drag would be a more realistic model for a typical projectile moving through air, unless perhaps the projectile is very small or launched with a low speed. However, the equivalent problem with quadratic drag is not solvable analytically, other than in 1 dimension (i.e. if the projectile is launched directly upwards and falls back down along the same line).
@@DrBenYelverton Amazing video, very informative and straight to the point. Yet I still wonder how you'd solve the quadratic drag problem in one dimension. What function would you choose so that the first derivative is linearly dependent with the square of the second derivative? I'm very curious and would appreciate if you could give a hint or solve this, please
@@katkollare809 Thanks! I don't remember the solution off the top of my head, but the approach is a little different to what I did in this video - instead of guessing a sensible solution for x and substituting it into the equation of motion, it's much easier to integrate twice, i.e. first solve for v and then get x from there. Since quite a few people have asked about this I will probably end up making a video showing the full solution soon.
this is also totally consistent with the previous video without the drag, i.e. if b is nearly zero... then the Taylor expansion of ln(1-ax) will eliminate the extra linear term and the second term becomes the second term in the original formula without the drag! Very cool to see this!
If the initial speed is u and the launch angle is θ, then uₓ = ucosθ and uᵧ = usinθ. So the equation does depend on θ implicitly, via the components uₓ and uᵧ.
Thank you very much, this tutorial was very helpful for my internal assessment. Any clues on how to calculate the constant "b" or average values for this constant? Thank you!
I'm glad it was helpful. Perhaps you could calculate a value for b using Stokes' law and the viscosity of air - this would work for a sphere moving at low velocity, which won't really be the case for a typical projectile but will at least give an appropriate order of magnitude estimate. In reality the drag force experienced by a projectile is more likely to be proportional to v², but that case becomes impossible to solve analytically.
This is pretty advanced stuff for your age! If the initial speed is u and the launch angle is θ, you can resolve the velocity into horizontal and vertical components to get uₓ = ucosθ and uᵧ = usinθ. You can then substitute these expressions into the trajectory equation to get the whole thing in terms of u and θ directly.
Thank you so much, i'm going to do some recearch about how acurate the equation is (and more importantly: how good i can replicate the specific circumstances 😅) for school.
Very nice. It's actually a more realistic model to have the drag force proportional to v², but it's impossible to solve the equations that you get for that case.
Great video! Do you think it would be possible to add the effect of wind using F=lw^2 (where w - constant wind velocity colinear with horizontal velocity, l - coefficient of the resistance of the medium with wind) into the equation?
Thanks! Yes, the equation of motion for x would have a right hand side of ±F/m instead of zero, where the sign depends on the direction of the wind. If F is constant then the solution for the x equation would mostly be the same, but with the addition of a particular integral term, which you can work out using the same way we handled the -g term in the y equation. If the wind is purely horizontal then there is no change to the y equation.
@@DrBenYelverton Just curious, how would you go about removing the t from the equation of motion for x to form the cartesian equation, given that it appears twice in both x and y equations of motion? Would you say the taylor series has any use in solving this (such as to expand the expression e−bm/t), or the lambert function?
@@soojandie The problem with Taylor expanding is that even if b is small, bt/m won't be small when t is large. I suppose you could do this to get an approximate trajectory if the total time of flight is known to be small, though (i.e. much less than m/b). The Lambert W function may be helpful but I'm not really familiar with it, probably worth researching if you're interested!
if I wanted to calculate the horizontal displacement of the projectile, how would I calculate the value of b? The resistance coefficient? Also do you think this model is sufficient for a small projectile like a nerf draft? Since you base the model on the fact that F drag is proportional to velocity so the equation is analytically solvable? And would there perhaps be any way to mathematically evaluate this model for a project?
You can quantify how appropriate or inappropriate the linear drag model is by calculating the Reynolds number; you'll need to look this up if you haven't come across it before. I suspect that quadratic drag is a better model for your case though. If you use the linear drag model, you could come up with an order of magnitude value for b using Stokes' law, technically this only applies for a sphere but it would be a starting point. You can do some research and try to find out if anyone has come up with an appropriate value of b for the shape of your projectile, either numerically or experimentally.
@@DrBenYelverton Alright, yes thank you so much, I was actually going to do three conditions, 1 without drag 2 linear drag 3 quadratic drag. But yes I will take your advice into consideration.
That's an interesting topic - I don't think I've ever actually worked through the calculation for a rotating projectile, but in principle it should be possible starting from an expression for the Magnus force. I will have a go when I get a chance and see if it can be solved analytically!
Isn't the drag force proportional to the square of the velocity instead of just the velocity? (taking into account the drag equation for fluids) (thanks for the video! You've made it really simple to understand)
Thanks for watching, glad it was helpful! Yes, quadratic drag would be a better model for a typical projectile, but the problem then becomes unsolvable analytically, other than the 1D case where the projectile moves straight up and back down along the same straight line.
If you square the components of the velocity vector, the system remains uncoupled and the two equations can still be solved in closed form. Or is that illegal?@@DrBenYelverton
The overall displacement would come from solving y = 0. Since x appears in both the linear and logarithmic terms, we can't solve for this exactly. You'd either have to do it numerically or use an approximation for the log term.
I was looking to make a ballistics program in a game called SimpleRockets2, but I am unable to find a simpler equation to calculate the trajectory (bomb release from a plane flying horizontally). I have the distance from the plane to the target on the y-axis, the straight distance from the plane to the target (not y-axis or x-axis, but in a straight line from the plane to the target), the coefficient of drag for the bomb, an equation to calculate the ballistics coefficient and lastly, a ballistics program in the form of: x= sqrt(2v²*y/g) how can I introduce the coefficient of drag (0.30) into my equation without having to work with unknown variables?
Proportional to own velocity would be great for low Reynolds numbers like in viscous fluids or small projectiles. But proportional to velocity squared is more realistic for golf balls and baseballs and bullets etc. so if we integrate each independent directional equation, we should have velocity components for both x and y for a given time. And if you integrate again you get distance By time. If you solve for time on the x equation and plug into time for the y equation, you get dy vs dx for the trajectory profile. If you solve for dx and then derive dx/dtheta you can find optimal angle for a given velocity. Yes?
In principle that would be the correct method, but in general it's not possible to solve analytically for x(t) and y(t) in the case of quadratic drag. The reason is that the square of the speed is v² = v_x² + v_y², so if the drag force is proportional to v² then the equations of motion in x and y are no longer independent.
@@DrBenYelverton how do you go about plotting the trajectory then? If you do it in excel with a 0.01 time increment, can’t you correctly plot the curve as everything adjusts?
Sure, it won't be exact but you can integrate the equations of motion numerically and plot the results. I'd probably use python/scipy for this but Excel should work too.
@@DrBenYelverton thanks for the replies. Could you do a video on solving a bar being lifted up with chains attached underneath? Like F -g(M+3x)=m d2x/dt2 + dm/dt dx/dt ?
Actually, one of the videos I've been meaning to make for a while is on a problem involving a ball thrown into the air with a chain attached, which sounds like a very similar system but without the extra F term in your equation. Out of interest, did you come up with the idea or did you come across it somewhere else?
Thanks! No, b is the drag force per unit velocity, while the drag coefficient is usually defined in a different way. Wikipedia has an article on the drag coefficient that you might want to refer to.
If you take the limit of the x(t) equation as t becomes infinite, you find that x never exceeds muₓ/b. Therefore the argument of the log, 1 - bx/muₓ, is always positive. So if you're working with some data which give a negative argument then it sounds like it's inconsistent with this equation.
They're the complementary function (CF) and particular integral (PI), respectively. In brief, the CF is the solution to the differential equation but with the right hand side set to zero, while the PI is a solution chosen to make the right hand side actually equal to -g. The full solution is a superposition of both of these.
it does because i’ve hitted targets at that distance. 1000 meters away and 900m/s initial speed. The bullet is shooted half a degree over the x axis and the target is at y=0. the bullet bc is 0,350 in g7, wich result as a 0,255 drag for what i’ve found on the internet. maybe that’s wrong and screw off everything?
@@michelezeffiro3086if you're plugging in observational data, then if you're getting a hit, you can forget about the y component and focus on your initial velocity on the x-axis. I haven't tried yet, but I think you can calculate the correct drag if you have that initial x velocity, the distance to Target, and the time to reach the target. Or did you already find the correct drag value?
You'd need to set y = 0 and solve the resulting equation for x. Since x appears both inside and outside the logarithm, it can't be done analytically and you'd need to use numerical methods or perhaps solve it approximately using a Taylor expansion of the log term.
is it possible to explicitly calculate the time in terms of y, I'm trying to calculate the angle required to hit a target at any x and y distance, i have figured out how to do it for a target with a y distance of 0, but i cant account for the change in y height, any suggestions?
Finding t(y) is not straightforward because y(t) contains both exponential and linear terms. I have a feeling you may be able to do it using the Lambert W function but that's not something I'm particularly familiar with.
@@DrBenYelverton yeah that’s fair enough I’ve tried messing around with lambert w a bit and it’s still very confusing, what’s really annoying is lua doesn’t have a built in math function for it so I have to approximate it with my own function (what I’m doing requires me to use lua).
@@djmicrowave6073how did your attempt work out? I come back to this problem periodically because I'd like to get a version I feel confident in before I attempt to solve for target lead
It's just the parameter in the trial solution exp(λt) - by substituting this into the differential equation, you find that λ(λ+b/m) = 0 and therefore that 0 and -b/m are both possible values of λ. This is how we get to the solution x = A exp(0t) + B exp(-bt/m) = A + B exp(-bt/m).
@@tsunami870 The first couple of sections at the following link cover methods for solving this type of equation (second order ODEs) quite thoroughly: tutorial.math.lamar.edu/classes/de/introsecondorder.aspx
Wow adding drag really does make the whole situation much more complex. I’ve just finished second order differential equations in school and it’s really interesting to see them used inside a problem not just by themselves. High quality video as always!
Thanks. Since acceleration is the second derivative of position with respect to time, Newton's second law itself is really a second order ODE. So if you continue studying Physics you'll be seeing a lot of them in context!
@@DrBenYelverton I'm going crazy
For 0:20, I remember the drag force should be proportional to the the square of its velocity which gives us F = bv². As velocity, v constantly changes can't be included in the constant, b. May I know what I'm missing? Or could you how to solve the equation will be if we use F = bv²?
Indeed, quadratic drag would be a more realistic model for a typical projectile moving through air, unless perhaps the projectile is very small or launched with a low speed. However, the equivalent problem with quadratic drag is not solvable analytically, other than in 1 dimension (i.e. if the projectile is launched directly upwards and falls back down along the same line).
@@DrBenYelverton Amazing video, very informative and straight to the point. Yet I still wonder how you'd solve the quadratic drag problem in one dimension. What function would you choose so that the first derivative is linearly dependent with the square of the second derivative? I'm very curious and would appreciate if you could give a hint or solve this, please
@@katkollare809 Thanks! I don't remember the solution off the top of my head, but the approach is a little different to what I did in this video - instead of guessing a sensible solution for x and substituting it into the equation of motion, it's much easier to integrate twice, i.e. first solve for v and then get x from there. Since quite a few people have asked about this I will probably end up making a video showing the full solution soon.
im tryna figure out how to predict bomb paths
this is also totally consistent with the previous video without the drag, i.e. if b is nearly zero... then the Taylor expansion of ln(1-ax) will eliminate the extra linear term and the second term becomes the second term in the original formula without the drag! Very cool to see this!
Indeed, satisfying to see and also a useful check on the result!
This is amazingly instructive. Thank you so much for sharing!
Mate, thanks so much, I needed to understand this derivation for my project and it was explained really clearly step by step!
Excellent, I'm happy to hear that it helped!
How is it that the equation doesn't include an angle? Since wouldn't the angle of the projectile effect it's trajectory?
If the initial speed is u and the launch angle is θ, then uₓ = ucosθ and uᵧ = usinθ. So the equation does depend on θ implicitly, via the components uₓ and uᵧ.
Thank you very much, this tutorial was very helpful for my internal assessment. Any clues on how to calculate the constant "b" or average values for this constant?
Thank you!
I'm glad it was helpful. Perhaps you could calculate a value for b using Stokes' law and the viscosity of air - this would work for a sphere moving at low velocity, which won't really be the case for a typical projectile but will at least give an appropriate order of magnitude estimate. In reality the drag force experienced by a projectile is more likely to be proportional to v², but that case becomes impossible to solve analytically.
How could you add the angle and velocity of the previous to this equation so you don't need u_x and u_y? P.S. I'm only 13, so take it easy on me. 😉
This is pretty advanced stuff for your age! If the initial speed is u and the launch angle is θ, you can resolve the velocity into horizontal and vertical components to get uₓ = ucosθ and uᵧ = usinθ. You can then substitute these expressions into the trajectory equation to get the whole thing in terms of u and θ directly.
Thank you so much, i'm going to do some recearch about how acurate the equation is (and more importantly: how good i can replicate the specific circumstances 😅) for school.
Very nice. It's actually a more realistic model to have the drag force proportional to v², but it's impossible to solve the equations that you get for that case.
Hello, for the value of x, why do we utilise both values of lambda? Sorry if its a bad question.
Great video! Do you think it would be possible to add the effect of wind using F=lw^2 (where w - constant wind velocity colinear with horizontal velocity, l - coefficient of the resistance of the medium with wind) into the equation?
Thanks! Yes, the equation of motion for x would have a right hand side of ±F/m instead of zero, where the sign depends on the direction of the wind. If F is constant then the solution for the x equation would mostly be the same, but with the addition of a particular integral term, which you can work out using the same way we handled the -g term in the y equation. If the wind is purely horizontal then there is no change to the y equation.
@@DrBenYelverton Just curious, how would you go about removing the t from the equation of motion for x to form the cartesian equation, given that it appears twice in both x and y equations of motion? Would you say the taylor series has any use in solving this (such as to expand the expression e−bm/t), or the lambert function?
@@soojandie The problem with Taylor expanding is that even if b is small, bt/m won't be small when t is large. I suppose you could do this to get an approximate trajectory if the total time of flight is known to be small, though (i.e. much less than m/b). The Lambert W function may be helpful but I'm not really familiar with it, probably worth researching if you're interested!
if I wanted to calculate the horizontal displacement of the projectile, how would I calculate the value of b? The resistance coefficient? Also do you think this model is sufficient for a small projectile like a nerf draft? Since you base the model on the fact that F drag is proportional to velocity so the equation is analytically solvable? And would there perhaps be any way to mathematically evaluate this model for a project?
You can quantify how appropriate or inappropriate the linear drag model is by calculating the Reynolds number; you'll need to look this up if you haven't come across it before. I suspect that quadratic drag is a better model for your case though. If you use the linear drag model, you could come up with an order of magnitude value for b using Stokes' law, technically this only applies for a sphere but it would be a starting point. You can do some research and try to find out if anyone has come up with an appropriate value of b for the shape of your projectile, either numerically or experimentally.
@@DrBenYelverton Alright, yes thank you so much, I was actually going to do three conditions, 1 without drag 2 linear drag 3 quadratic drag. But yes I will take your advice into consideration.
bro wtf im in 9th grade why tf my science teacher making me do this
This is brilliant! Can you make a video about how the spin of an object affect it trajectory of projectile?
That's an interesting topic - I don't think I've ever actually worked through the calculation for a rotating projectile, but in principle it should be possible starting from an expression for the Magnus force. I will have a go when I get a chance and see if it can be solved analytically!
Isn't the drag force proportional to the square of the velocity instead of just the velocity? (taking into account the drag equation for fluids) (thanks for the video! You've made it really simple to understand)
Thanks for watching, glad it was helpful! Yes, quadratic drag would be a better model for a typical projectile, but the problem then becomes unsolvable analytically, other than the 1D case where the projectile moves straight up and back down along the same straight line.
If you square the components of the velocity vector, the system remains uncoupled and the two equations can still be solved in closed form. Or is that illegal?@@DrBenYelverton
Can we solve for the displacement?
The overall displacement would come from solving y = 0. Since x appears in both the linear and logarithmic terms, we can't solve for this exactly. You'd either have to do it numerically or use an approximation for the log term.
Thank you !!! Just one question, what is Uy equal to? Is it Vo*sin(phi) with Vo = initial speed and phi the angle ?
Yes, exactly.
@Amir Jalilvand That's correct!
I was looking to make a ballistics program in a game called SimpleRockets2, but I am unable to find a simpler equation to calculate the trajectory (bomb release from a plane flying horizontally). I have the distance from the plane to the target on the y-axis, the straight distance from the plane to the target (not y-axis or x-axis, but in a straight line from the plane to the target), the coefficient of drag for the bomb, an equation to calculate the ballistics coefficient and lastly, a ballistics program in the form of: x= sqrt(2v²*y/g)
how can I introduce the coefficient of drag (0.30) into my equation without having to work with unknown variables?
Proportional to own velocity would be great for low Reynolds numbers like in viscous fluids or small projectiles. But proportional to velocity squared is more realistic for golf balls and baseballs and bullets etc. so if we integrate each independent directional equation, we should have velocity components for both x and y for a given time. And if you integrate again you get distance
By time. If you solve for time on the x equation and plug into time for the y equation, you get dy vs dx for the trajectory profile. If you solve for dx and then derive dx/dtheta you can find optimal angle for a given velocity. Yes?
In principle that would be the correct method, but in general it's not possible to solve analytically for x(t) and y(t) in the case of quadratic drag. The reason is that the square of the speed is v² = v_x² + v_y², so if the drag force is proportional to v² then the equations of motion in x and y are no longer independent.
@@DrBenYelverton how do you go about plotting the trajectory then? If you do it in excel with a 0.01 time increment, can’t you correctly plot the curve as everything adjusts?
Sure, it won't be exact but you can integrate the equations of motion numerically and plot the results. I'd probably use python/scipy for this but Excel should work too.
@@DrBenYelverton thanks for the replies. Could you do a video on solving
a bar being lifted up with chains attached underneath? Like F -g(M+3x)=m d2x/dt2 + dm/dt dx/dt ?
Actually, one of the videos I've been meaning to make for a while is on a problem involving a ball thrown into the air with a chain attached, which sounds like a very similar system but without the extra F term in your equation. Out of interest, did you come up with the idea or did you come across it somewhere else?
Wow, amazing video. Just a query, is b the drag coefficient?
Thanks! No, b is the drag force per unit velocity, while the drag coefficient is usually defined in a different way. Wikipedia has an article on the drag coefficient that you might want to refer to.
I tried to use this formula to fin but then the natural log becomes negative and doesnt work, is my data wrong?
If you take the limit of the x(t) equation as t becomes infinite, you find that x never exceeds muₓ/b. Therefore the argument of the log, 1 - bx/muₓ, is always positive. So if you're working with some data which give a negative argument then it sounds like it's inconsistent with this equation.
just help me out here: what are y(cf) and y (pi) in this calculation at 17:14?
They're the complementary function (CF) and particular integral (PI), respectively. In brief, the CF is the solution to the differential equation but with the right hand side set to zero, while the PI is a solution chosen to make the right hand side actually equal to -g. The full solution is a superposition of both of these.
what if i try to put numbers in it and ln has an argument
That means you're plugging in a value of x that the projectile never actually reaches!
it does because i’ve hitted targets at that distance. 1000 meters away and 900m/s initial speed. The bullet is shooted half a degree over the x axis and the target is at y=0. the bullet bc is 0,350 in g7, wich result as a 0,255 drag for what i’ve found on the internet. maybe that’s wrong and screw off everything?
@@michelezeffiro3086if you're plugging in observational data, then if you're getting a hit, you can forget about the y component and focus on your initial velocity on the x-axis. I haven't tried yet, but I think you can calculate the correct drag if you have that initial x velocity, the distance to Target, and the time to reach the target. Or did you already find the correct drag value?
Very nice.
Nice.
How exactly would you solve for the horizontal range of the projectile?
You'd need to set y = 0 and solve the resulting equation for x. Since x appears both inside and outside the logarithm, it can't be done analytically and you'd need to use numerical methods or perhaps solve it approximately using a Taylor expansion of the log term.
You can use Euler method, using time step, as small as possible, b is the resistance coefficient.
is it possible to explicitly calculate the time in terms of y, I'm trying to calculate the angle required to hit a target at any x and y distance, i have figured out how to do it for a target with a y distance of 0, but i cant account for the change in y height, any suggestions?
Finding t(y) is not straightforward because y(t) contains both exponential and linear terms. I have a feeling you may be able to do it using the Lambert W function but that's not something I'm particularly familiar with.
@@DrBenYelverton yeah that’s fair enough I’ve tried messing around with lambert w a bit and it’s still very confusing, what’s really annoying is lua doesn’t have a built in math function for it so I have to approximate it with my own function (what I’m doing requires me to use lua).
I see - a few people have been asking me about this, perhaps I'll have to do some research and make a video on it at some point!
@@djmicrowave6073how did your attempt work out? I come back to this problem periodically because I'd like to get a version I feel confident in before I attempt to solve for target lead
What is lambda? And what is the relationship of it to b/m?
It's just the parameter in the trial solution exp(λt) - by substituting this into the differential equation, you find that λ(λ+b/m) = 0 and therefore that 0 and -b/m are both possible values of λ. This is how we get to the solution x = A exp(0t) + B exp(-bt/m) = A + B exp(-bt/m).
@@DrBenYelverton Thanks! Can you recommend any good resources for further learning this topic?
@@tsunami870 The first couple of sections at the following link cover methods for solving this type of equation (second order ODEs) quite thoroughly: tutorial.math.lamar.edu/classes/de/introsecondorder.aspx
Thanks!
Thanks so much, I really appreciate that!