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  #1  
Old 02-01-2008, 06:45 PM
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zkvii zkvii is offline
 
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Location: Auckland, New Zealand
Posts: 241
Question CofG theory

Hi,

Given the following weight and balance situation:

Code:
Pilot (kg)	85 kg	Baggage (kg)	0 kg
Passenger (kg)	0 kg	Fuel total (L)	60 L

		kg	lbs	arm ins	moment lbs ins	
Nose		152	334	39.11	13062.75	
Left Main	193	424	93.96	39875.67	
Right Main	191	421	93.96	39564.95	
					
Emtpy CofG	535	1179	78.43	92503.37	
Pilot		85	187	97.48	18267.06	
Passenger	0	0	97.48	0	
Baggage		0	126.78	0	0
Fuel TO (60L)	42	92	80.00	7354.62	
Fuel Landing	0	0	80.00	0	
Totals TO	662	1459	80.97	118125.06	
Totals Landing	620	1367	81.04	110770.44
How do I calcuate the Nose wheel weight from this data? Given that I have all the arms, weights and moments I'm sure it is possible - but I can't get my head round it at the moment.

Thanks,

Carl
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ZK-VII - RV 7A - New Zealand - NZNE - 700 hrs (2008 -2020)
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TMX OF-360 (FADEC), Dual GRT H1, Garmin 430 / SL30 / 327, TT VSVG

Last edited by zkvii : 02-01-2008 at 06:50 PM.
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  #2  
Old 02-01-2008, 08:06 PM
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AlexPeterson AlexPeterson is offline
 
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Location: Maple Grove, MN
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There may be some error in the following method, since W&B is done with the airplane level, while it may not be level sitting on the ground (the 7, 8 and 9A's seem to be nearly level, while the 6A is not).

Imagine looking directly at the plane from the left side. What you want to do is to sum moments (forces times distances) around the main gear, and set them equal to zero. So, there are two forces trying to "rotate" the plane around the mains. One is the total weight acting downwards (counter clockwise moment), the other is the nose gear force acting upwards (clockwise moment). The distance (from the mains) that the nose gear force acts through is known and is constant. In your plane, the distance is (93.96 - 39.11 = 54.85). The distance through which the total weight acts will vary depending on loading, but will be (93.96 - Cg).

So, if we define Fn as nose wheel force, Ft as total weight, and set the sum of the moments equal to zero, we get:

54.85*Fn = Ft *(93.96 - Cg)

and solving for Fn:

Fn = Ft *(93.96 - Cg)/(54.85)

So, take your example for TO weight:

Fn = 1459*(93.96-80.97)/54.85
= 345

Your LD example:

Fn = 1367*(93.96-81.04)/54.85
= 321

Make sense?
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  #3  
Old 02-01-2008, 08:34 PM
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zkvii zkvii is offline
 
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Hi Alex,

Thank you - I got a different (less elegant) formula (we worked it out just for the various delta) but same resultant answers which is encouraging.... (and yes I'm assuming aircraft is 'level enough')

Now I'll go and tie this in with the VANS Service Letter graph ( http://www.vansaircraft.com/pdf/Nose...ice_letter.pdf )

Thanks for the education,

Carl
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http://www.rvproject.gen.nz/
TMX OF-360 (FADEC), Dual GRT H1, Garmin 430 / SL30 / 327, TT VSVG
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  #4  
Old 02-02-2008, 09:49 AM
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AlexPeterson AlexPeterson is offline
 
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Glad I could help. Curious though, why you picked those fuel quantities... not very much, especially the landing one ...
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  #5  
Old 02-02-2008, 12:59 PM
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zkvii zkvii is offline
 
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Don't worry - I was using min / max for CofG limit planning / theory - 30L is my normal reserve....

My concern was with full tanks, pilot and no passenger or baggage I can get over the nose wheel weight limit. If I add 5 kg of baggage it is all within range again. The VANs graph is rather crude for getting a specific CofG line so I wanted to understand the maths.

Thanks again,

Carl
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http://www.rvproject.gen.nz/
TMX OF-360 (FADEC), Dual GRT H1, Garmin 430 / SL30 / 327, TT VSVG
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