Putting Wings or Winglets on a Mavic?

[Gernby]

Please elaborate?

I just looked back through the thread, and noticed that I missed a bunch of posts.

I believe that if the MP was hovering in front of an "ideal fan" with a ~5 MPH non-turbulent breeze, then I think it would be pretty easy to observe the difference in average RPM and electrical load while changing the heading of the MP in relation to the fan. I suspect that average RPM and electrical load will be lowest when the fan is producing a head wind, higher with a tail wind, and highest with a cross wind.

 

[AlanTheBeast]

I finally had a minute to analyze your image, and am pretty surprised by the numbers I calculated from it. As I mentioned before, it's been a LONG time since I did any trigonometry, so please help me verify this...

I think that the magnitude of the horizontal vectors (Fh) for each prop would be related to the vertical vectors (Fv) like this:
Front: TAN(3.9) = Fh(front) / Fv(front)
Rear: TAN(6.3) = Fh(rear) / Fv(rear)

Since we know that Fh(front) must equal Fh(rear) while hovering (opposite direction, but equal magnitude), we can substitute:

Fv(rear) * TAN(6.3) = Fv(front) * TAN(3.9)

Therefore:

Fv(rear) = Fv(front) * TAN(3.9) / TAN(6.3)

Which is:

Fv(rear) = 0.62 * Fv(front)

If this is correct, then the vertical component of thrust from the rear props is just 62% of the magnitude of the vertical component of thrust from the front props. Since the angles are relatively small for front and rear, the magnitude of the total thrust from the rear is also about 62% of the magnitude of the total thrust from the front. If we assume that the props produce thrust with a linear relationship to RPM, then it seems that the front will be spinning about 62% faster than the rear. However, I doubt the props will have the same efficiency across such a wide RPM range, so it may be an even greater difference.

Let's start by assuming the props are all doing the same amount of force just to see where we're at.

Fvf = Ff * sin (front)
front = 90 - 3.9 = 86.1°
Ff = 1
Fvf = sin(86.1) = 0.998 ie: 99.8% of the upward thrust is vertical.

Fvr = Ff * sin(back)
Fvr = sin (90-6.3)= 0.994 ....99.4%

So they're very close in vertical lift. From there you can calculate the forward or rear thrust, and then sort out the differences that would seem to be made up from different power (RPM) front and back.

Not sure if that's what you wanted?? (I'm tired and cranky right now).

 

[Gernby]

I think you'll feel different about that after a good night sleep.

 

[Wombat55]

I finally had a minute to analyze your image, and am pretty surprised by the numbers I calculated from it. As I mentioned before, it's been a LONG time since I did any trigonometry, so please help me verify this...

I think that the magnitude of the horizontal vectors (Fh) for each prop would be related to the vertical vectors (Fv) like this:
Front: TAN(3.9) = Fh(front) / Fv(front)
Rear: TAN(6.3) = Fh(rear) / Fv(rear)

Since we know that Fh(front) must equal Fh(rear) while hovering (opposite direction, but equal magnitude), we can substitute:

Fv(rear) * TAN(6.3) = Fv(front) * TAN(3.9)

Therefore:

Fv(rear) = Fv(front) * TAN(3.9) / TAN(6.3)

Which is:

Fv(rear) = 0.62 * Fv(front)

If this is correct, then the vertical component of thrust from the rear props is just 62% of the magnitude of the vertical component of thrust from the front props. Since the angles are relatively small for front and rear, the magnitude of the total thrust from the rear is also about 62% of the magnitude of the total thrust from the front. If we assume that the props produce thrust with a linear relationship to RPM, then it seems that the front will be spinning about 62% faster than the rear. However, I doubt the props will have the same efficiency across such a wide RPM range, so it may be an even greater difference.

The tilt of the propellors from vertical is supposed to be 3.9˚ and 6.3˚ from vertical? The immediate problem is that both of these angles have the same sign (positive), which means that the propellors are tilted in the same direction which means that their horizontal thrusts are in the same direction so that there is no way for the horizontal forces to balance out.

If, on the other hand, one of these angles (3.9˚ or 6.3˚) is supposed to be actually negative so that the horizontal thrusts are opposed to each other, then the problem is that that means that the relative tilt between the front and rear propellors is 10.2˚ (=3.9˚+6.3˚) which is obviously too large of a value.

Actually, in looking further down at your subsequent calculations after your first two equations, it appears you actually wrote down equations which assume that the 3.9˚ and 6.3˚ tilts were in opposite directions with opposite signs. So your final answer is correct but not for your originally stated problem of two sets of propellers with a relative tilt of a small 2.4˚ between them, but for two sets of propellers with a large 10.2˚ tilt between them (=3.9˚+6.3˚).

 

[Gernby]

Did you see the note that said "opposite direction, but equal magnitude"? The horizontal vectors are not cumulative. They cancel each other, since the front and rear props pitch toward each other. As AlanTheBeast pointed out from my video, the difference in [absolute] pitch is 2.4 degrees.

 

[Wombat55]

Did you see the note that said "opposite direction, but equal magnitude"? The horizontal vectors are not cumulative. They cancel each other, since the front and rear props pitch toward each other. As AlanTheBeast pointed out from my video, the difference in [absolute] pitch is 2.4 degrees.

I see the note now. But the problem is that the way you wrote down the equations does not reflect that. The equations reflect a situation in which one set of propellers is tilted 3.9˚ in one direction, and the other set is tilted 6.3˚ in the opposite direction, for a total relative angle between them of 10.2˚, not 2.4˚.

 

[AlanTheBeast]

I see the note now. But the problem is that the way you wrote down the equations does not reflect that. The equations reflect a situation in which one set of propellers is tilted 3.9˚ in one direction, and the other set is tilted 6.3˚ in the opposite direction, for a total relative angle between them of 10.2˚, not 2.4˚.

This is not formalized math just a sketch to discovery. If you saw the photo with the angles clearly illustrated you would not be bringing this up.

And of course your objections are irrelevant since I was only looking at lift for comparative reasons.

If it makes you feel better, let's correct the front vector so that:

front = 90 - (-3.9) = 93.9
and sin(93.9) = 0.998 or 99.8% of the propeller force in the vertical. Oh my same thing (and no surprise).

 

[AlanTheBeast]

The tilt of the propellors from vertical is supposed to be 3.9˚ and 6.3˚ from vertical? The immediate problem is that both of these angles have the same sign (positive), which means that the propellors are tilted in the same direction which means that their horizontal thrusts are in the same direction so that there is no way for the horizontal forces to balance out.

If, on the other hand, one of these angles (3.9˚ or 6.3˚) is supposed to be actually negative so that the horizontal thrusts are opposed to each other, then the problem is that that means that the relative tilt between the front and rear propellors is 10.2˚ (=3.9˚+6.3˚) which is obviously too large of a value.

Actually, in looking further down at your subsequent calculations after your first two equations, it appears you actually wrote down equations which assume that the 3.9˚ and 6.3˚ tilts were in opposite directions with opposite signs. So your final answer is correct but not for your originally stated problem of two sets of propellers with a relative tilt of a small 2.4˚ between them, but for two sets of propellers with a large 10.2˚ tilt between them (=3.9˚+6.3˚).

If you had looked at the annotated image you would know that the angles are opposing each other and that would have saved you a lot of windiness above.

If it makes you feel better, let's correct the front vector so that:

front = 90 - (-3.9) = 93.9
and sin(93.9) = 0.998 or 99.8% of the propeller force in the vertical. Oh my same thing (and no surprise).

 

[AlanTheBeast]

I just looked back through the thread, and noticed that I missed a bunch of posts.

I believe that if the MP was hovering in front of an "ideal fan" with a ~5 MPH non-turbulent breeze, then I think it would be pretty easy to observe the difference in average RPM and electrical load while changing the heading of the MP in relation to the fan. I suspect that average RPM and electrical load will be lowest when the fan is producing a head wind, higher with a tail wind, and highest with a cross wind.

You could add a gridbox to remove the rotation. Vortexes form at the exit of each grid hole edge, however, and the output won't be as smooth as you'd desire. If someone vapes you can probably visualize that.

There is no doubt that hover needs more power than hover with a 5 mph headwind. Just moving the drone from hover to a few mph and the required RPM to maintain altitude drops noticeably.

 

[Gernby]

Let's start by assuming the props are all doing the same amount of force just to see where we're at.

Fvf = Ff * sin (front)\
front = 90 - 3.9 = 86.1°
Ff = 1
Fvf = sin(86.1) = 0.998 ie: 99.8% of the upward thrust is vertical.

Fvr = Ff * sin(back)
Fvr = sin (90-6.3)= 0.994 ....99.4%

So they're very close in vertical lift. From there you can calculate the forward or rear thrust, and then sort out the differences that would seem to be made up from different power (RPM) front and back.

Not sure if that's what you wanted?? (I'm tired and cranky right now).

Your strategy here seems off in a couple ways. First, it starts with an assumption that all props are putting out similar levels of thrust, which I believe is 100% incorrect. Second, it ignores the one thing we know must be true, which is that the horizontal components will be equal in magnitude and opposite in direction.

However, it does seem like a worthy effort to use your approach to compare horizontal components IF the props were spinning at the same RPM. I think that would go something like this:

Fhf = Ff * cos(86.1)
Ff = 1
Fhf = .068

Fvr = Ff * cos(83.7)
Ff = 1
Fhr = .110

If this is correct, then the rear props would be producing 62% more horizontal thrust than the front, which we know can't be true while hovering.

 

[Gernby]

I hope to make some time to go buy a cheap fishing rod and weights this weekend, and experiment with a tail for the MP. I believe that a small weight at the end of a tail (of the right length) would improve efficiency while hovering, by reducing or eliminating the overall forward pitch of the props.

 

[Wombat55]

This is not formalized math just a sketch to discovery. If you saw the photo with the angles clearly illustrated you would not be bringing this up.

Oh, stop trying to BS me. I simply pointed out that the equations and solution as written were wrong.

 

[Wombat55]

If you had looked at the annotated image you would know that the angles are opposing each other and that would have saved you a lot of windiness above.

If it makes you feel better, let's correct the front vector so that:

front = 90 - (-3.9) = 93.9
and sin(93.9) = 0.998 or 99.8% of the propeller force in the vertical. Oh my same thing (and no surprise).

Very good as a first step. Now tell me and Gernby what happens when you balance the horizontal forces to get the relative lift forces of the front and rear props in the manner that he calculated.

 

[Wombat55]

.

However, it does seem like a worthy effort to use your approach to compare horizontal components IF the props were spinning at the same RPM. I think that would go something like this:

Fhf = Ff * cos(86.1)
Ff = 1
Fhf = .068

Fvr = Ff * cos(83.7)
Ff = 1
Fhr = .110

If this is correct, then the rear props would be producing 62% more horizontal thrust than the front, which we know can't be true while hovering.

You didn't read my comments on your calculation, did you?

 

[AlanTheBeast]

Very good as a first step. Now tell me and Gernby what happens when you balance the horizontal forces to get the relative lift forces of the front and rear props in the manner that he calculated.

I didn't really look at his calculation as it, offhand, didn't make sense to me - I was tired at the time so I didn't delve into it.

As to my calc, it's rather simple to get the other components and figure out the different amount of force required - even you could solve it. The remaining component that is unknown is the pitch moment due to the slightly forward CofG. As I don't have the means to weigh the bird accurately that's not on. An accurate measurement of RPM's forward and aft could be a backdoor to that, perhaps - though the power/thrust/RPM relationship may not be linear enough so the result wouldn't be better than a WAG.

 

[AlanTheBeast]

Your strategy here seems off in a couple ways. First, it starts with an assumption that all props are putting out similar levels of thrust, which I believe is 100% incorrect. Second, it ignores the one thing we know must be true, which is that the horizontal components will be equal in magnitude and opposite in direction.

However, it does seem like a worthy effort to use your approach to compare horizontal components IF the props were spinning at the same RPM. I think that would go something like this:

Fhf = Ff * cos(86.1)
Ff = 1
Fhf = .068

Fvr = Ff * cos(83.7)
Ff = 1
Fhr = .110

If this is correct, then the rear props would be producing 62% more horizontal thrust than the front, which we know can't be true while hovering.

I never claimed the RPM's to be the same as it's clear they can't be. I just wanted to show you that the force vectors of the front and rear rotors have to be necessarily similar since the CofG is very close to ideal and the angle of the props are close.

(Edited) What I see is Fhf is -6.8% of total forward thrust and Fhr is 11.0% of total rear thrust. Given the difference in angles that doesn't seem so much (ie: don't compare them to each other. Compare them to the whole. These are the "smallest" vectors, and as such it's no surprise that you see a large difference between them (62%). But looking at them as part of the whole makes that pretty meaningless. (/Edited)

Oh, stop trying to BS me. I simply pointed out that the equations and solution as written were wrong.

You're the BS artist as you were in the range extension discussion. We're just hashing out observations here, not doing an AP physics white paper for proffie and even less so for your satisfaction.

 

[Wombat55]

You're the BS artist as you were in the range extension discussion. We're just hashing out observations here, not doing an AP physics white paper for proffie and even less so for your satisfaction.

You're being ridiculous. Gernby went off the tracks early with his calculations, and you didn't even notice.

 

[AlanTheBeast]

You're being ridiculous. Gernby went off the tracks early with his calculations, and you didn't even notice.

You're the one being ridiculous. You're fast to poke but void of any explanations of anything while playing Prof. Dimbulb-with-two-clues.

As I did explain, when I first saw his explanation I was too tired to go over it, but did offer my view of things. In the end (while I haven't re-done the math) I believe his math to be correct, just his interpretation of the results to not be so. No big deal at all - I certainly don't fault him in any way. In the end there are a few variables that are out of reach w/o the right measuring tools.

As always, he and I throw out observations and hypotheses for discussion and all you do is throw ineffectual rocks at them. Ridiculous and boring but perhaps your boat floats better for it.

 

[Wombat55]

OK, I apologize to Gernby. I saw the diagram of the Mavic with the angles of rotors marked and listed and they actually are tilted by a surprisingly large angle of 10.2˚ with respect to each other, not just 2.4˚. I didn't believe that it was that large before seeing the image. Gernby's calculation is correct. In order for the horizontal forces to balance, the rear propellers have to be putting out only about 62% as much vertical thrust as the forward propellers in level, stationary flight, as he said. The center-of-weight has to be a bit forward as a result.

 

[Gernby]

You didn't read my comments on your calculation, did you?

I did, but as you stated, it was a moot point.