Matching the aircraft, engine and propeller
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- Опубліковано 12 вер 2024
- This video hopes to show the importance of matching the engine and propeller to the aircraft to achieve maximum efficiency.
This is a re-recording as the original video had an error in it that just needed to be corrected.
Excellent video. Helped me understand the important parameters and how they are related. Thank you for making this video 🙏.... Just made me happy to see the video and get the insight in to the fundamentals of how the propeller, engine and aircraft are sized..
Love that accent. The difficulty I am having here is trying to convince these other people that when you overload the airplane the lift the drag ratio insist that the drag is more. If the propeller was designed for a lower drag you can't speed the engine up enough to get the power you need without huge amounts of slipping. If you throttle back you don't make enough power to pull the airplane with the increased drag. I did use the term cavitation which is technically incorrect because you'd have to have an incompressible fluid to have cavitation with the propeller. I think the airplane propeller is just a spinning example of a wing or lifting shape. Is it related to a stall when your propeller is pitched too high and you spend so many RPMs you can't get any thrust? I'm trying to dumb this down because we have a lot of pilots they're going out here in a mathematically impossible airplane ensure that their skills can compensate.I keep arguing with these people before they hit the ground hard and leave the gene pool and it's not doing much good because every time they see the last pilot fly the mathematically impossible airplane they are sure they are a better pilot and they can fly it. Thank you for the videos I subscribed as soon as I saw the first three equations come on in the first 3 minutes.lol
this video was very helpful, thank you !
Interesting and useful videos. I hope they will help me to design my human powered aircraft. In this case the propeller.
Please help me to understand and link those calculations to the Advance Ratio. As the Advance Ratio is optimum when Beta is 45 degree, then when we should apply that condition (to set the geometric pitch (GP) to 45 degree)?
Advance ratio = V / nD. The Blade Angle or pitch = Angle of advance (phi) + Angle of Attack (alpha). Then Tan (phi) = V / wr which is V/ ( pi. n d), which is very similar to the Advance ratio. So, if you divide the advance ratio by pi, and get the ATAN of it, you’ll get the angle of advance, add 4 degrees (typically) and you’ ll get the blade angle (pitch) of the prop.
Sir, Could you please upload a video regarding Ship-Propeller-Engine matching, with an example.
Sorry, I know very little about ships.
Great video.
I have been doing the math and I have a question on the equation at 4:23.
When I use the values given in the example, I get 10669.089.
What units of power does this represent?
Does it break down to Kw?
Or did I just get the math wrong?
The units are Watts, so if you got 10669 then that is 10669W or 10.7kW.
@@RoddyMcNamee Thanks for the Quick reply!
Now I can get back to the math.
Great Video!!!
At 5:17 when I plotted the range of various air speeds into the power formula, I am surprised that the values for the power required were so low. I expected the high airspeed calculations to be near the top end of the engine power available, but they were not even half of the HP that the engine could produce.
is that normal?
I am trying to convert the results from Kw to hp. maybe I am missing something in the conversion from watts to hp.
@@jeffery7756 The power increases very quickly at the higher airspeeds, so for example at V=50m/s power could be 70%, but at V=55m/s could be at 90%. To convert kW to HP, multiply the kW value by 1.341 and you will have the value in HP.
@@RoddyMcNamee Thanks again for the quick reply. I found your video extremely helpful for determining the performance of my plane.
I also appreciate your continued help with the math.
I don't think it's gonna be an easy fix to just multiply the result (although I hope you are right) .
I am using your formulas to figure out my plane's performance, so I have substituted Meters/sec to feet/sec, Kg for lbs and standard pressure from 1.225 to .0765 for units of measure.
I was using the following input;
p=.0765 lbs/ft3
V=146.66 ft/sec
S=237.9 sq/ft
weight= 2005 lbs (converted to 8918 newtons)
Cdo=.05
K=1.16
This makes the formula a bit tricky to be confident that the output is correct.
The final numbers I am getting from the formula at 5:17 is 1,504,389.34 ... way too high for kw or hp.
I may have to convert everything on my plane to metric, do the formula and then convert it back to get the answer.
One of the things I picked up in your video was to convert weight to newtons, using pounds gave the wrong answer but newtons was on the money. Very helpful.
I tried to understand the Vmd and the Vmr (the velocity with minimum drag and the velocity with maximum range), but I failed. I understood about the math and the formula you used, need its derivative to get the Vmd and the Vmr. But I could not understand the physic explanation about why does that aircraft will have its minimum drag at 32m/s and its maximum range will be achieved at 42 m/s? Can you please explain it in a simple English?
Yes, it does seem contradictory but let me try to explain with the analogy of a car. If I was driving a car in the city, my velocity would be low and hence the drag of the car would be low, but the car will be burning let's say 8L/100km of fuel. So, at that rate, I would only travel 100km if I had 8L of fuel. Now, when I get out of the city and onto a motorway, the car velocity can increase and therefore so will the drag, but the fuel burn will reduce to let's say 7L/100km, so if I had 8L of fuel I would travel 114km. The range is bigger but the drag is bigger too. It's all to do with the power required per km.
@@RoddyMcNamee Well, fine with the maximum range. But why does the drag is minimum at 32m/s, and not at 24m/s or even less? As you just mentioned in your car analogy that the slower we drive the less the drag is. This is absolutely make sense. But hence, the minimum drag should be as low speed as possible, not at 32m/s, which that speed is above the velocity of minimum power (Vmp).
@@StrsAmbrg The drag for a car is all profile drag, where as for an aircraft, the drag (in its simplest form) is a combination of lift induced drag and profile drag. Profile drag increases with velocity, where as lift induced drag decreases with velocity. At low airspeeds, lift induced drag is the dominant form of drag. So for an aircraft as you transition from low airspeed to higher airspeed, your lift induced drag slowly reduces and your induced drag increases. At some point the drag will be a minimum. Hence drag at 32m/s could be less than the drag at 24m/s.
@@RoddyMcNamee Quote: "For an aircraft, the drag (in its simplest form) is a combination of lift induced drag and profile drag." I got it. Thank you for your explanation.