

Sep 21, 2007, 08:28 AM  
Joined Nov 2005
20 Posts

Balance stands as described above work well when the C.G. of the aircraft is below the bottom of the wing. If it is above, then you can flip the airplane upside down and it will work. The tough ones are when the C.G. is located vertically within the airfoil section and no matter how you support it on a stand the C.G. is above the point you are trying to balance the airplane on. That is where the sling and plumbbob method comes to the rescue. Just a different way to skin the cat and the airplane can't fall off the stand and get dinged with hanger rash.

Sep 21, 2007, 05:14 PM  
Joined Sep 2007
5 Posts

If you're going to be serious about this, then you really do need to weigh the model and do the math (Courtesy of Mr Phil Clark, England):
1) The model has to be jacked up 'level'....by level, with the fuselage datum at '0' degrees 2) Hand a 'plumb bob' of somewhere convenient....in this case, the end of the crank shaft stocking through the prop (yellow arrow)......this determines our 'known position. 3) Measure the distance from this 'known position' to the centre (axles) if the main wheels. This is distance 'A' 4) Measure the distance from the 'known position' to the centre (axle) of the tail wheel. This is distance 'B' 5) Using a set of scales, place them under one main wheel and take a weight reading. When doing this, the other main wheel, and the tail wheel MUST be chocked up the same amount as the thickness of the scales so as to maintain the models 'level' 6) Swap the scales and chock to the other main wheel, and take a 2nd main wheel weight reading. Combine the 2 main wheel weight readings. This is 'Wt1' 7) Using the same method as above (with both main wheels chocked up) take a tail wheel weight reading. This is 'Wt2' We now have 2 distances (A and B), and 2 weights (Wt1 and Wt2). The 2 weight readings can now be combined to give us the 'total' weight of the model. The 4 values can then be plugged into the following formule, to determin where the CG is, in relation to the 'known point'. (A) (Wt1) + (B) (Wt2) = (Wt1 + Wt2) CG So.....here's the figures. (33.5" x 145.8lb) + (113.75" x 29.2lb) = (175) CG 4884.3 + 3321.5 = 175 x CG CG = 8205.8  175 CG = 46.89" back from the 'Known point' 
Sep 27, 2007, 12:59 AM  
Joined Jan 2007
2 Posts

Reply to #6:
Interesting and useful. This could be useful for models 10  50 lbs Generally we are given the C of G range. But where is it actually on your particular model with varying equipment placings? How about purchasing a kiddies seesaw (or make one)? Remove handles leaving flat plank and make sure it balances dead level Place on garage floor. Extend pivot point by painting a line on the garage floor on each side to make model alignment easier. Secure airplane to plank....foam cutouts etc.....with design CG on pivot point. Adjust equipment until plane balances level. 
Sep 28, 2007, 05:40 PM  
Joined Sep 2007
5 Posts

Quote:
You know the distance from the datum, and the weight of the equipment added and alter the existing figures to the following (in this case using 8lb's as the estimated nose weight requirement) (33.5" x 145.8lb) + (113.75" x 29.2lb) + (12" x 8lb) = 183lb x CG 4884.3 + 3321.5 + 96 = 183 x CG CG = 8301.8  183 CG will now be 45.36" rearward of our datum. Adding 8 lbs of weight on the nose of this model moves the CG 1.53" forward. The effect any object in the model placed in the model can be calculated in this manner, by replacing the '12" x 8lb' in the formula with the distance and weight of the object in question. 

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