Aeroplane geometry?

[Select Zone Five]

I wasn't too sure where to post this as it's comprehension, rather than technical, but here goes...

I'm working through some aviation related physics questions in preparation for the ATPL theory and have come across the following statement.
The fore and aft distance between the two main wheels and the single tail wheel of an aircraft is 8m. Maybe I'm missing something here but, what on earth does that mean? The question is about calculating forces on the wheels so I'm thinking the simplest configuration is most likely.

[Genghis the Engineer]

It means that when seen in side view, the aircraft is a taildragger with the tailwheel 8m behind the mains. This would be generally stated with the aircraft in the flying attitude.

G

[Lu Zuckerman]

To: Select Zone Five

The question is about calculating forces on the wheels so I'm thinking the simplest configuration is most likely.


If the question is about calculating the forces on the wheels then you will have to provide the entire question and all of the information provided in order to solve the problem.

1) In order to calculate the CG of an aircraft it is necessary to weigh it. In the case of the subject aircraft it would require the use of three scales. One under each of the main wheels and one under the tail wheel. This will show the weight distribution of the aircraft or the gravitational force on each wheel. Does the question provide the weight distribution on the three wheels?

2) Does the question entail a downward velocity of the aircraft during landing and does it reference a three-point landing? Any forces during landing are transitory and will vary from impact loads to roll out loads and then to the static loads indicated in 1) above.

3) Does it give the gross weight of the aircraft?

4) If the question implies the distance between the main wheels and the tail wheel and does not provide any other information then it seems to me that it is not a valid question.

But then again I have been wrong before.


:cool:

[Select Zone Five]

Thanks guys, I stopped short of the whole question because it was the first statement I didn't understand but here it is...

The fore and aft distance between the two main wheels and the single tail wheel of an aircraft is 8m. The unloaded weight of the aircraft is 20kN and its centre of gravity is 0.5m behind the main wheels. Calculate the force on the tail wheel and on each main wheel.

It sounds like a fairly simple forces question but I wasn't sure if I was reading the 'fore and aft' bit wrong??

Thanks again :)

[john_tullamarine]

I suggest that the question is testing your knowledge of statics and simple moment balances. Calculating the wheel loads is the reverse of doing the conventional calculation to determine the CG position.

Other than that, the question is a little pointless, not overly useful, and, I would have thought, not all that relevant to a pilot ... to the weights and balance man, sure ... but the typical pilot doesn't get to worry about this sort of thing.

The basic concepts, however, are quite important to a pilot.

[RadarContact]

The fore and aft distance between the two main wheels and the single tail wheel of an aircraft is 8m. The unloaded weight of the aircraft is 20kN and its centre of gravity is 0.5m behind the main wheels. Calculate the force on the tail wheel and on each main wheel.

CG is at 1/16th of the distance between main and tail wheel(s). So the main wheels together receive 15/16th of the resulting force or 9375 N each. The remaining 1/16th or 1250 N will be dealt to the tail wheel.

This will result in a momentum to CG from each side of 9375N*2*0,5m = 1250N*7,5m = 9375 Nm -> the A/C remains standing on the Apron.

Um..... did I spoil anything? :D

[Genghis the Engineer]

I agree with your answer Radarcontact, but your approach to solving the problem only works in simple cases and may not be the "approved" method. The way I'd treat it is:-

Taking moments about the mainwheel, there are two to consider. 20,000N at 0.5m and ftw N at 8m. Since the aircraft is not rotating, these sum to zero. Thus:-

0 = (-20,000 x 0.5) + (8ftw)

or

ftw = (20,000 x 0.5) / 8 = 1250 N, which is the force at the tailwheel.


Similarly, taking moments about the tailwheel, there are again two to consider. 20,000N at 7.5m, and fmw at 8m. Since the aircraft is still not rotating, these still sum to zero, hence:-

0 = (20,000 x 7.5) + (-8 x fmw)

or

fmw = 20,000 x 7.5 / 8 = 18,750


Assuming that the load on the mains is evenly distributed between them, divide by 2 to get 9,375 N per mainwheel.


And a crude check that you got it right is to add all three up and confirm that they add up to 20,000, which is the weight of the aeroplane.

G

[RadarContact]

I stand... well, not corrected, but improved :)

[Select Zone Five]

Thanks very much, I got the same answers.

I wondered if I was missing a vital piece of information in the "fore and aft distance" comment. Why not simply say "The perpendicular distance between the mail wheels and the tail wheel"?

Oh well, it's not the first badly worded question I've ever read!

Thanks again! :D

[Notso Fantastic]

Perpendicular? Horizontal! Put your books down and go make a cup of tea!

[Select Zone Five]

Thanks for the assistance everyone...and yes Notso my brain hurts and I didn't even start yet! :rolleyes: