What is the formula for this? PartII

[Gary Warner]

I’m’ intrigued about the fact that a plane in a constant angle/speed/height turn will have a different airspeed at the inside and outside wing tips. I thought I’d give the numbers a crunch to see just how much it was. Having NO experience in this kind of math, I welcome corrections to my approach.

To start off with, I’m keeping things simple. First I’ll model the effects at a very large scale, say a Cessna 150. I’m very interested to see the effects at this scale since I “care” if it effects my flying a real plane. Then I’ll plug in the numbers to a model sailplane on another post (assuming these calculations are correct – and I got to get back to work!!!).

For the Cessna 150, I’ll use a 45 deg. bank angle and 100 MPH. The plane has a (rough) 28 foot wingspan.

First the centerline circumference of the flight path:

(Thanks “Sail-N-Soar” )
Radius = .0668 * 1 (tan of 45 deg. bank angle) * speed squared

R=.0688 * 10000 = 668 feet

Centerline Circumference = 668 * 2 * PI = 4197 feet

At 100 MPH that’s 146.66666667 feet per second or 28.61590908 seconds for the 360 deg. turn (or in other words for the 4197 foot distance).

*Side note: When I fly a C150 at 45 deg. steep turns at 90 MPH, it seems like it's FAR less than 28 seconds for a 360 deg. turn. I might question the formula for the radius, but never the less I'll use what I've got for now.)

Now for the outer and inner wing tip circumferences:

(I know that the projected horizontal span is reduced by the angle of bank. If no bank, projection span is equal to actual span. At a 45 deg. bank, projected span is ½ of actual span. And at 90 deg. there is no projected span. Though I don’t know how to calculate the projected span for assorted angles, I do know the 45 deg. rule. That’s why I chose a 45 deg. bank angle.)

Tip Radius:
Outer = 668 + (span / 2) / 2 = 675 feet
Inner = 668 – (span / 2) / 2 = 661 feet

Since time remains constant for the whole wing making a 360 deg. turn, I’ll use the above mentioned time constant of 28.61590908 seconds.

4241 feet Outer wing tip circumference / time (28.61….) = 148.2042729 feet per second or 101.0483678 MPH.
4153 feet Inside wing tip circumference / time (28.61...) = .145.1290604 feet per second or 98.9516321 MPH

Is this right? Is there really a 2.0967357 MPH difference between the two wing tips?

That’s it for today.

Gary
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[Sail 'n Soar]

Gary, accept it, the formula is correct. One's perception of time while flying is off - seems shorter if you enjoy it and MUCH LONGER, when you don't. A minor point, you need to consider that your wings are at a 45 degree angle, so you need to multiply your span by the sine of the bank angle (.707 for 45 degrees. That will reduce your differential speed. Probably an easier way to get to your number is to just compare radii.

R, centerline = 668
R, outer tip = 668+28*sin(45)/2 = 677.9
R, inner tip = 668-28*sin(45)/2 = 658.1

V, outer = 677.9*90/668 = 91.33
V, inner = 658.1*90/668 = 89.67

delta = 2.67 mph

I see my number is a little larger than yours. Haven't gone through your calcs close enough to discover why the difference. In any case, I believe the speed. I fly gliders turning slower, tighter circles while thermaling. At times I have the ailerons deflected to roll out of the bank just to preserve the bank. Unless it is all in this pilot's imagination I believe this at least partially due to the slower speed and resulting lower lift of the inside wing over the relatively larger glider span. (lift ~ V^2)

[erik1938]

Maybe if you think of the difference in velocity relative to what position. There would be a difference relative to the center of the circle, but to an observer on the ground there would be no difference.

[Gary Warner]

(snip) A minor point, you need to consider that your wings are at a 45 degree angle, so you need to multiply your span by the sine of the bank angle (.707 for 45 degrees). That will reduce your differential speed.

I thought I did reduce the projected span when I used:

Outer = 668 + (span / 2) / 2 = 675 feet
Inner = 668 – (span / 2) / 2 = 661 feet

I reduced the total span by 1/2 for the 1/2 actual span (inside or outside wing) and 1/2 again for the projected span. Am I wrong on this? I thought that a wing at a 45 deg angle would produce a 1/2 size projected span. Nothing to do with flight per say, just normal projected geometry.

POST EDIT: Oops! I see what you are saying about the .707 - my bad. You are correct.

[Gary Warner]

V, outer = 677.9*90/668 = 91.33
V, inner = 658.1*90/668 = 89.67

delta = 2.67 mph

I see my number is a little larger than yours. Haven't gone through your calcs close enough to discover why the difference.

I see it - It looks like you used 90 MPH where I was using 100 MPH. And like you said, you used the .707 for the projected span.

[Gary Warner]

Maybe if you think of the difference in velocity relative to what position. There would be a difference relative to the center of the circle, but to an observer on the ground there would be no difference.

For this example, I'm interested in what the aircraft will "see" when it flies.

[Sail 'n Soar]

I see it - It looks like you used 90 MPH where I was using 100 MPH. And like you said, you used the .707 for the projected span.


The 668 ft is consistent with 100 mph. I goofed! :o

Numbers are:

V, outer tip = 677.9*100/668 = 101.5
V, inner tip = 658.1*100/668 = 98.5

Delta V = 3 mph.

[Gary Warner]

(I'll use Sail 'N Soar's abbreviated formula.)

Given: 20 MPH, 30 deg. bank, 10 foot wing span

Resolve Radius: R=(.0668*(tan(30)))*(20^2) = 15.4268

R, centerline 15.4268 feet
R, outer tip = 15.4268+(10*(sin(30)/2)) = 17.9268
R, inner tip = 15.4268-(10*(sin(30)/2)) = 12.9268

V, outer = (17.9268*20)/15.4268 = 23.24
V, inner = (12.9268*20)/15.4268 = 16.76

delta = 6.48 mph airspeed difference at the tips

WOW! That's a lot of speed difference for our sailplanes.

I'm assuming that the difference expressed in lift differential would need to take into account the average cord and lift distribution per side. This example would put the average airspeed over each wing panel at +/-1.62 MPH (tip speed + root speed / 2). That's an average differential speed of 3.24 MPH. Still, that's a significant number that we deal with.

I wonder if this effect is not taken into consideration on some of the model airplane computer flight sims, and if not, is this why the flight sims never quite seem to match real-world performance and handling?