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enter image description here
(YouTube) Wildcat undercarriage.

For planes with non-truck undercarriages, can they be rotated around the lateral axis of the main landing gear with the brakes applied without causing translation?

My answer is no, they must be locked and there must be translation (image below).

I had the following comment on a recent answer, and frankly I'm not quite sure, hence the question.

[They] are wheels. The whole structure is free to rotate around this 'locked' axle...

I don't agree, imagine chocks are placed forward of the main landing gear:

  • A) Frame is locked, blue man can't lift as he can't cause translation.
  • B) Frame is not locked, plane rotates around wheels (from the wheels' perspective, they're free moving).

enter image description here
(Original) Above shows what I mean by translation, the plane moves forward (blue distance) with the wheel—plane frame (red lines) being locked. (Brakes are applied.)

Note: This question has nothing to do with deceleration or a running engine.

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  • $\begingroup$ On those vintage aircraft the brakes were not very firm. They did not need to be better in order to make a headstand less easy, but in consequence it was comparatively easy to rotate a wheel where the brake was engaged. $\endgroup$ – Peter Kämpf Apr 7 '17 at 22:47
  • $\begingroup$ Is the blue man part of any sort of organization? $\endgroup$ – hobbs Apr 8 '17 at 1:17
  • $\begingroup$ @PeterKämpf that makes sense for taxiing / landing roll, very good info thanks. $\endgroup$ – ymb1 Apr 8 '17 at 23:26
  • $\begingroup$ Why wouldn't the blue man be able to cause translation? $\endgroup$ – Sean Mar 24 at 0:15
  • $\begingroup$ @Sean: before A and B there's the assumption of chocks being placed :-) $\endgroup$ – ymb1 Mar 24 at 3:05
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Yes, the parking brakes lock the wheel and the airframe together. If the brakes are not applied, the wheel can rotate round the axle.

If the wheel's position is fixed to the ground (e.g. by adding chocks, but the brakes are not applied, then the aircraft will rotate round the wheel axle when the tail is lifted by the blue man.

If the brakes are applied, but the wheel's position is not fixed to the ground, the aircraft will rotate round the wheel's point of contact with the ground. Since the contact point moves as the wheel rolls, this results in translation in addition to the rotation.

If both the brakes are applied and the wheels are on chocks, the wheels either have to roll over the chocks or something (the wheel, the chock or the brake) will have to slip.

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Let's look at the DHC-2 as an example. Most data is taken from the DHC-2 manual. Here is a basic diagram of the forces involved:

enter image description here

  • Fg is from the weight of the plane on the main gear
  • Ft is from the weight of the plane on the tail
  • FL is the force applied to lift the tail
  • Fx is the lateral force applied to the wheel from friction
  • FN is the normal force from the ground supporting the main gear

Gross weight is 5100 lb (2300 kg). The main gear is 25 in (0.64m) in front of the CG, the tail wheel is 244 in (6.2m) aft of the CG. This means that about 4630 lb (2100 kg) rests on the main gear, and 470 lb (215 kg) rests on the tail.

The friction force Fx can be up to Fgμs, where μs is the coefficient of static friction, which is about 0.9 (depending on conditions of course).

If the plane is prevented from translating forward, then the airplane will rotate around the wheel axle. The tires are 20.5 in (0.5m) diameter, so the moment on the wheel from friction can be up to 3540 lbft (4800 Nm).

The tail is about 270 in (6.9m) from the main gear. The force on the tail to overcome the moment on the main gear would be 160 lbf (700 N), which is the equivalent of lifting 160 lb (70 kg). This would be in addition to lifting the 470 lb (215kg) from the weight of the plane.

It's also worth noting that right before the tire overcomes static friction, you would need an equal and opposite force to oppose Fx, which would be 4160 lbf (18500 N). We would need the mythical frictionless massless (spherical, immovable) chocks to provide this force without adding rotational resistance to the tires.

The centerline of the engine is 53 in (1.3m) above the axle, so a thrust force of 830 lbf (3700 N) would be needed to overcome the wheel friction. The tail has 14500 Nm of moment to overcome, which means another 11200 N or another 2500 lb of thrust.

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If you're asking if an aircraft like a taildragger can be tipped forward rotating about the main landing gear axles, the answer is yes and you do have to be careful of this not to apply too much braking pressure during taxiing or during the landing roll.

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    $\begingroup$ @ymb1 I'm not sure what you're asking here. Of course there is nothing stopping the plane from rotating about the main gear. It doesn't matter whether it's inertia or a person pushing up on the tail that applies the moment. $\endgroup$ – fooot Apr 7 '17 at 14:46
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    $\begingroup$ @ymb1 I was referencing the blue person pushing up on the tail, if that's the image you mean. The only thing keeping the tail on the ground is the CG being aft of the main gear. That can be overcome by a force applied upward on the tail, or by forward inertia of the plane, or engine thrust that rotates the plane forward. All apply a moment that would counter the CG aft of the main gear. $\endgroup$ – fooot Apr 7 '17 at 14:52

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