- Jason McDowell
**Posts:**26**Joined:**

I understand that the overbanking tendency in a steep turn (~50 degree bank) is caused by the outside wing traveling more quickly than the inside wing, thus producing more lift than the slower inside wing.

My question is, if one were to install a pitot tube on each wingtip, how much faster (in mph or kts) is the outside wing really traveling?

-JM

My guess only........but I think the difference would be so insignificant that you wouldn't even see a difference on your airspeed indication.

Last year 1,944 New Yorkers saw something and said something.

If the turn is symetric, i mean well coordinated, there is no reason to have a wing faster than the other

Actually, there is. Think about an airplane doing a constant radius turn around a point (to the left, for example). The right tip of the wing must travel faster than the left tip of the wing (on a miniscule scale) to compensate for the fact that it must also travel farther in the same amount of time. The farther away something is from the center of a circle, the faster it must travel to complete the circle in the same amount of time. Think of a merry-go-round. The inner horses are traveling slower than the outer horses.

But like I said before, the difference is so small it is not significant.

Last year 1,944 New Yorkers saw something and said something.

Jhooper is absolutely correct.

23 victor, turn right heading 210, maintain 3000 till established, cleared ILS runwy 24.

The math guys could probably figure it out... you have all the tools.

Find the turn radius. Then figure the wingspan, and figure out where on that wingspan the pitot tube is located. Then figure the angular velocity of the location of the pitot tube and the would be location on the opposite wing.

The difference is probably not very much, but there is a difference!

Find the turn radius. Then figure the wingspan, and figure out where on that wingspan the pitot tube is located. Then figure the angular velocity of the location of the pitot tube and the would be location on the opposite wing.

The difference is probably not very much, but there is a difference!

.

The horizontal distance between the wingtips will also be reduced since the aircraft is banking.

What's a normal turning radius and speed when banking the aircraft 45-50 degrees??

Staffan

What's a normal turning radius and speed when banking the aircraft 45-50 degrees??

Staffan

Staffan,

say 100 knots. The turn rate and radii follows with a known bank angle.

I'll try to get back to the subject later if that does not provide the encouragement to figure the connection out - if nobody else has filled in. In a bit of a hurry right now.

Cheers,

Fred

say 100 knots. The turn rate and radii follows with a known bank angle.

I'll try to get back to the subject later if that does not provide the encouragement to figure the connection out - if nobody else has filled in. In a bit of a hurry right now.

Cheers,

Fred

I thought I was doing good trying to avoid those airport hotels... and look at me now.

Ok, here's a go at it..

If the center of a wing with a 10 meter span is travelling at 100 kts in a 45 degree bank (maintaining altitude), the inner tip will be travelling at 97,1922 kts and the outer wing at 101,08 kts.

If anyone is interested in the actual calculations I'll post them later..

Staffan

If the center of a wing with a 10 meter span is travelling at 100 kts in a 45 degree bank (maintaining altitude), the inner tip will be travelling at 97,1922 kts and the outer wing at 101,08 kts.

If anyone is interested in the actual calculations I'll post them later..

Staffan

- delta-flyer
**Posts:**2631**Joined:**

Steffan, I am curious as to why the incremental velocity between the center of the wing and its tips is not symmetrical about the center of the wing, which is traveling at 100 kts. Also, I get a difference of only 3 kts between the inner and outer wingtips.

Let's see your solution.

Pete

Let's see your solution.

Pete

"In God we trust, everyone else bring data"

Lateral acceleration - al

Bank angle - a = 50 degrees

Gravitational constant - g = 9,82 m/s^2

Velocity - v = 100 K = 51,4 m/s

Time for one full circle = T

ri/ry - Inner/outer radius

vi/vy - Inner/outer velocity

Wing span s = 10 m

al = g * tan a = 11,7 m/s^2

r = v^2/al = 226 m

T = 2*r*pi/v = 27,6 s

ri = r - s*cos a = 219,7 m

ry = r + s*cos a = 232,6 m

vi = 2*ri*pi/T = 50,0 m/s = 97,2 K

vy = 2*ry*pi/T = 52,9 m/s = 103 K

Cheers,

Fred

Bank angle - a = 50 degrees

Gravitational constant - g = 9,82 m/s^2

Velocity - v = 100 K = 51,4 m/s

Time for one full circle = T

ri/ry - Inner/outer radius

vi/vy - Inner/outer velocity

Wing span s = 10 m

al = g * tan a = 11,7 m/s^2

r = v^2/al = 226 m

T = 2*r*pi/v = 27,6 s

ri = r - s*cos a = 219,7 m

ry = r + s*cos a = 232,6 m

vi = 2*ri*pi/T = 50,0 m/s = 97,2 K

vy = 2*ry*pi/T = 52,9 m/s = 103 K

Cheers,

Fred

I thought I was doing good trying to avoid those airport hotels... and look at me now.

Bah, got the conversion to knots wrong at the end..

The results should be:

Center 100 kts

Inner 98,69 kts

Outer 101,31

With a 45 degree bank that is.

Anyway, my calculations were the same as Fred's, only that I was lazy using 45 degrees, so vertical and horizontal components were the same

Staffan

[Edited 2003-09-17 12:10:36]

The results should be:

Center 100 kts

Inner 98,69 kts

Outer 101,31

With a 45 degree bank that is.

Anyway, my calculations were the same as Fred's, only that I was lazy using 45 degrees, so vertical and horizontal components were the same

Staffan

[Edited 2003-09-17 12:10:36]

Staffan,

I have it all spreadsheeted so I plugged 45 degrees in as well and get 97,4/103K... mind verifying your calculations to confirm? Finding the source of the discrepancy would be nice.

Cheers,

Fred

I have it all spreadsheeted so I plugged 45 degrees in as well and get 97,4/103K... mind verifying your calculations to confirm? Finding the source of the discrepancy would be nice.

Cheers,

Fred

I thought I was doing good trying to avoid those airport hotels... and look at me now.

Fred,

I recalculated it again, same results, but I think i found where our calculations differ. Shouldn't your ri and ry be as follows:

ri = r - (s*cos a)/2

ry = r + (s*cos a)/2

since you only want to add and subtract half of the horizontal component of the wing?

Staffan

I recalculated it again, same results, but I think i found where our calculations differ. Shouldn't your ri and ry be as follows:

ri = r - (s*cos a)/2

ry = r + (s*cos a)/2

since you only want to add and subtract half of the horizontal component of the wing?

Staffan

Staffan,

yuppers. Thank you! Was in the paper version of the calcs but was lost in translation. Now our results match.

Cheers,

Fred

yuppers. Thank you! Was in the paper version of the calcs but was lost in translation. Now our results match.

Cheers,

Fred

I thought I was doing good trying to avoid those airport hotels... and look at me now.

- Jason McDowell
**Posts:**26**Joined:**

Impressive work. Thanks a lot, guys!

-JM

- ThirtyEcho
**Posts:**1409**Joined:**

I worked it out with *two* pencils.

We are only assuming that the overbanking tendency is the result of one wing travelling faster than the other; that is only partially the case.

This is easy to see if you use your old instructor's device, involving two pencils and a model airplane, with one of the pencils representing the lifting force perpendicular to the plane of the wings and the other perpendicular to the horizon. At over 45 degrees of bank, the pencil that represents lift perpendicular to the horizon is*under* the outside wing; this creates a lift component that actually serves to increase the banking force. If you have done steep turns very much, you'll notice a point slightly past 45 degrees where the airplane seems to "slip" toward the turn as the upward lift component moves under the upward wing.

We are only assuming that the overbanking tendency is the result of one wing travelling faster than the other; that is only partially the case.

This is easy to see if you use your old instructor's device, involving two pencils and a model airplane, with one of the pencils representing the lifting force perpendicular to the plane of the wings and the other perpendicular to the horizon. At over 45 degrees of bank, the pencil that represents lift perpendicular to the horizon is

- delta-flyer
**Posts:**2631**Joined:**

Steffan,

OK, that's better - now your answer matches mine. I also just used the fact that the horizontal acceleration force equals the weight.

Cheers,

Pete

OK, that's better - now your answer matches mine. I also just used the fact that the horizontal acceleration force equals the weight.

Cheers,

Pete

"In God we trust, everyone else bring data"

I'd like to refer you all to See How it Flies.

ThirtyEcho,

I'd like to have that one run by me again in more detail - I can't see it adding up, but I'm not sure about what you mean. In what way does the vertical component of lift go under the outside wing?

Cheers,

Fred

ThirtyEcho,

I'd like to have that one run by me again in more detail - I can't see it adding up, but I'm not sure about what you mean. In what way does the vertical component of lift go under the outside wing?

Cheers,

Fred

I thought I was doing good trying to avoid those airport hotels... and look at me now.

Lockheed Dragon Lady (U2) pilots will know all about this. At high altitude, with thin air, the "coffin corner" of the flight envelope is approached and turns have to be made very carefully. The wing on the outside can speed up enough for a danger of Mach buffet, while the wing on the inside can be near stall.

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