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Overbanking Tendency - C172  
User currently offlineJason McDowell From , joined Dec 1969, posts, RR:
Posted (11 years 3 months 2 weeks 2 days 15 hours ago) and read 4902 times:


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



19 replies: All unread, jump to last
 
User currently offlineJhooper From United States of America, joined Dec 2001, 6206 posts, RR: 12
Reply 1, posted (11 years 3 months 2 weeks 2 days 15 hours ago) and read 4898 times:

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.
User currently offlineSalim From Lebanon, joined Jun 2001, 303 posts, RR: 1
Reply 2, posted (11 years 3 months 2 weeks 2 days 14 hours ago) and read 4892 times:

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

User currently offlineJhooper From United States of America, joined Dec 2001, 6206 posts, RR: 12
Reply 3, posted (11 years 3 months 2 weeks 2 days 14 hours ago) and read 4886 times:

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.
User currently offlineNWA From United States of America, joined Jun 2001, 1200 posts, RR: 3
Reply 4, posted (11 years 3 months 2 weeks 2 days 14 hours ago) and read 4870 times:

Jhooper is absolutely correct.


23 victor, turn right heading 210, maintain 3000 till established, cleared ILS runwy 24.
User currently offlineSkyguy11 From , joined Dec 1969, posts, RR:
Reply 5, posted (11 years 3 months 2 weeks 2 days 1 hour ago) and read 4813 times:

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!


User currently offlineStaffan From , joined Dec 1969, posts, RR:
Reply 6, posted (11 years 3 months 2 weeks 1 day 19 hours ago) and read 4788 times:

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


User currently offlineFredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 7, posted (11 years 3 months 2 weeks 1 day 16 hours ago) and read 4764 times:

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



I thought I was doing good trying to avoid those airport hotels... and look at me now.
User currently offlineStaffan From , joined Dec 1969, posts, RR:
Reply 8, posted (11 years 3 months 2 weeks 1 day 12 hours ago) and read 4737 times:

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


User currently offlineDelta-flyer From United States of America, joined Jul 2001, 2676 posts, RR: 6
Reply 9, posted (11 years 3 months 2 weeks 1 day 6 hours ago) and read 4716 times:

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


User currently offlineFredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 10, posted (11 years 3 months 2 weeks 1 day 1 hour ago) and read 4699 times:

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



I thought I was doing good trying to avoid those airport hotels... and look at me now.
User currently offlineStaffan From , joined Dec 1969, posts, RR:
Reply 11, posted (11 years 3 months 2 weeks 23 hours ago) and read 4695 times:

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  Smile

Staffan

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

User currently offlineFredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 12, posted (11 years 3 months 2 weeks 19 hours ago) and read 4674 times:

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 thought I was doing good trying to avoid those airport hotels... and look at me now.
User currently offlineStaffan From , joined Dec 1969, posts, RR:
Reply 13, posted (11 years 3 months 2 weeks 18 hours ago) and read 4669 times:

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


User currently offlineFredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 14, posted (11 years 3 months 2 weeks 17 hours ago) and read 4665 times:

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

Cheers,
Fred



I thought I was doing good trying to avoid those airport hotels... and look at me now.
User currently offlineJason McDowell From , joined Dec 1969, posts, RR:
Reply 15, posted (11 years 3 months 2 weeks 17 hours ago) and read 4661 times:


Impressive work. Thanks a lot, guys!


-JM



User currently offlineThirtyEcho From United States of America, joined Dec 2001, 1662 posts, RR: 1
Reply 16, posted (11 years 3 months 2 weeks 13 hours ago) and read 4643 times:

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.


User currently offlineDelta-flyer From United States of America, joined Jul 2001, 2676 posts, RR: 6
Reply 17, posted (11 years 3 months 2 weeks 8 hours ago) and read 4630 times:

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


User currently offlineFredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 18, posted (11 years 3 months 2 weeks 1 hour ago) and read 4622 times:

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



I thought I was doing good trying to avoid those airport hotels... and look at me now.
User currently offlineNfield From United Kingdom, joined Dec 2002, 38 posts, RR: 0
Reply 19, posted (11 years 3 months 1 week 6 days 20 hours ago) and read 4623 times:

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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