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I'm working on a program that requires a taxiing passenger aircraft (of any model) to perform a semi-accurate turn given a steering angle of the front landing gear. For this I need to calculate the pivot/center point for the aircraft to rotate around. There seem to be a few factors that influence this (including some basic specifications of the aircraft) but I can't seem to find even a general equation that breaks it down.

I'm aware that there are also a number of more complex factors that would influence this, but for my purposes I'm assuming an ideal situation with minimal interference.

Thanks in advance!

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  • $\begingroup$ I think the easiest way to approximate this is to draw an imaginary line between the two rear gear. When the front wheel is turned 90 degrees left, the left tire is the pivot point. As the front tire straightens to 0 degrees, the pivot point linearly moves to the center of the line. Repeat as the front gear turns right. $\endgroup$
    – Ron Beyer
    Commented Jun 10, 2016 at 3:04

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The pivot point will be where an imaginary line through the rear wheels and an imaginary line perpendicular to the front wheel intersects.

pivot point illustration

So if θ is the steering angle and d is the distance between the front wheel and a point halfway between the rear wheels then

X = d • tan(θ)

will give you the distance from the point between the rear wheels to the pivot point.

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If you assume the main gear does not rotate (in the left-right sense), then the pivot point will always be somewhere on a line joining the two main gears.

You can then use trigonometry and the turn angle of the nose wheel to identify the centre of rotation. (Assume the nose tyre does not slip sideways.)

(By the way this is very similar to Ackermann steering used on cars - lots of standard mechanical texts talk about this.)

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  • $\begingroup$ Ah, seems simple enough. So, past my original question - if the main gear does rotate, how does this influence the center of rotation? What about wheel slip? $\endgroup$
    – ujm18
    Commented Jun 9, 2016 at 9:06
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    $\begingroup$ The main gear does not rotate significantly in the left/right direction. Like the rear wheels of a car, they just point forwards all the time and don't steer - so the centre of rotation must be on the same line. Also, if the tyres slipped sideways appreciably, the tyres would scuff at an appreciable rate. $\endgroup$
    – Andy
    Commented Jun 9, 2016 at 9:13
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    $\begingroup$ Also, consider that many larger passenger/freighter craft have more than one MLG on each side.(747, A380, etc). $\endgroup$
    – FreeMan
    Commented Jun 9, 2016 at 12:12
  • $\begingroup$ How does the distance change (adjacent leg d in TomMcW's diagram) with a more complex wheel arrangement? I'm guessing that for an aircraft with multiple sets of main landing gears, d extends to the half-way point between all main gears (and the pivot would be on a line that passes through that point). Or would the pivot be on a line that passes through a point between the main gears with the widest track? And when there are multiple bogey gears, I'm assuming from the middle of the gear? $\endgroup$
    – ujm18
    Commented Jun 11, 2016 at 7:13

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