ravill
Posts: 1915
Joined: 1/24/2003
From: Redwood,
CA, USA
Status: offline
|
Hi guys,
I think I may have found out why we balance inverted. I have come accross a very elegant solution using torque equations and simple fundamental mechanics. In short it seems Dragoonpvw is correct. Let me present to you my solution. I cannot, for the life of me upload an image that I made through apple works and saved as a jpeg. And, I left my digital camera at home. I usually bring it to work with me when I'm on call. Oh well, here goes.....
Assumptions:
1. The center of gravity(of the whole airplane) is "above" the wing. (not unreasonable since this is a low wing aircraft)
2. Masses that are used to balance an airplane are placed the same distance away from the CG point
explainations of assumptions:
1. The bulk of the airplane, including the engine, radio, batteries, etc..., will be located above the wing. If the center of gravity were on the other side of the plane made by the wing (which would be very unlikely unless the head of an inverted engine or drop tanks or missles were really, really heavy to pull the CG over) then my equations could be reversed.
2. It has been my experience that when I have to add weight to an airplane, I would likely use the same place wether I balance the airplane rightside up or rightside down. E.g, if I needed to add weight to the nose I might opt to use a weight just behind the firewall, this placement would not change wether I balanced the airplane rightside up or down. I would use the same spot.
I hope my assumptions are not to constricting and after some thought, seem reasonable to me. 
Since an airplane will tilt(or rotate) clockwise or counter clock wise when seen on side view, it makes the most sense to use fundamental torque equations, ie T= F X R. This expression is pronounced torque(T) equals force (F) cross the radius(R). Since this is in vector notation, it only makes sense qualitatively in this forum. For a quantitative approach, we need to use the absolute values. This means T= F x R x sin O. Read, torque equals the magnitude of force(F) times the magnitude of radius(R) times the sin of the angle between them. Imagine these two triangles(I will present balancing an airplane upright first) made by a tail heavy airplane
________________________
weight added to nose] _________________________
| [ measured center of gravity]
\/ |
gravity \/
_____wing top________________
______wing bottom____________
|
|
|
balance point as per manufactur, on bottom of wing
(this diagram comes out so well on the edit window, sorry it looks so horrible!)
Imagine a line, r, drawn from the weight to the balance point on the bottom of the wing, gravity going down, and the length(we'll call l) forming a perpindicular from the balance point straight over to the imaginary gravity line pointing down(the 90 degrees is between gravity line and this line l) Lets call the angle between line r and line gravity(lets call it Mu, u for up) O. Now imagine almost a mirror image of that triangle, this time formed by the line R from the measured CG to the balance point,
gravity(Fcg=force of gravity acting on the center of gravity) going straight down, perpindicular to another line L drawn from the balance point. This time call the angle between R and Fcg, A. I hope I haven't lost you. For an airplane to balance, both of the forces from the weight and the measured CG must equal each other or else the airplane would move/rotate!
then it follows from the torque eqns,
vector notation:
Mu X r = Torque produced from nose weight Mu= torque produced by the tail heavy CG= Fcg X R
then
Mu x r x sin O = Fcg x R x sin A (remeber Mu, r, O, Fcg, R and A can be ANY value since we have not put constraints on them!, magnitude Mu=mass x gravity, Fcg=mass of airplane x gravity)
remember from trigonometry
sin = opposite over hypotenuse, sin O= length l/ r, sin A= length L/R (remember l and L are arbitrary as well)
substituting
Mu x r x (l/r)= Fcg x R x (L/R)
simplifying
Mu x l = Fcg x L: Now we need more equations to help us solve this delima.
Lets look at the airplane being balanced inverted now, draw another picture as above but this time the balance point is ONTOP of the wing instead of the bottom. Two notable differences. 1. The imaginary lines drawn to the balance point WILL be shorter because they DO NOT traverse the width of the wing since balancing is now done from the TOP of the wing. 2. The angle made by the force of gravity acting on the nose weight,(the same on the center of gravity) and the line drawn from these two points to the balance point (moment arms), will be greater than 90 but less than 180. remember sinx=sin(180-X). We need to define some new terms, so we can identify the new triangles,
Md= mass of weight placed on nose to balance the airplane, at same spot as previously(assumption #2), note it does not necessarilly equal Mu.
rd= length of line from mass to new inverted point, it is shorter than r from above, this will not matter however.
B= angle between gravity of mass and rd, note it is greater than 90 but less than 180
l= length from balance point on top of wing to mass (note it is THE SAME as in the upright equations, assumption #2)
L= length from tail heavy center of gravity to balance point on top of wing (also SAME as upright equations, assumption#2)
Rd= length of line from tail heavy center of gravity to balance point on top
Fcgd= force of gravity working on tail heavy center of mass
D= angle between gravity of tail heavy center of mass and the line Rd
For balance, Torque from weight MUST equal Torque from tail heavy center of gravity, hence:
(1) Md X rd = Fcgd X Rd vector notation
(2) Md x rd x sin B = Fcgd x Rd x sin D
remeber sinx=sin(180-x),
(3)sin (180-B)=sin B= length l/ rd
(4)sin (180-D)=sin D=length L/Rd
substituting eqn (3) and (4) into equation (2)
Md x rd x (l/rd) = Fcgd x Rd x (L/Rd)
simplifying
Md x l = Fcgd x L
remember from our above calculations we obtained
Mu x l = Fcg x L
Notice that the lengths l, L are the SAME. This is from assumptiom #2. Notice also that Fcgd is just the force of gravity acting on the tail heavy center of mass, THIS DOES NOT CHANGE WETHER WE HAVE AN UPRIGHT OR INVERTED AIRPLANE!!!!
We now have
Md x l = Fcgd x L = Fcg x L = Mu x l
Md x l = Mu x l
l=l so:
Md = Mu QID 
Notice, 1. the differing moment arms R, Rd, r, rd although all different lengths did not matter!! This means that wing thickness cannot matter 2. the distance at which one chooses to place the balancing weight does not matter as long as you choose the same place wether on top or on bottom 3. Lastly, for an airplane to balance, regardless of whether upright or inverted, and using the same place to place weights, the weights NEED TO BE THE SAME!!!!!!!!
The only reason to balance an airplane inverted over upright is just because it is more stable as the balance point on top of the inverted wing will "catch" and stabilize the airplane as it tilts/rotates!!!!! Boy that was a long winded mathematical derivation. So rest assured all who balance. IT DOES NOT MATTER WHETHER YOU BALANCE AN AIRPLANE INVERTED OR UPRIGHT, YOUR AIRPLANE WILL FLY!!!!!!. Thank you Dragoonpvw!!!!!!
I'm sorry for all the math and physics, but if someone would be so kind to point out any mistakes, I think this would pretty much explain some reasoning behind some of the balancing wonders!
< Message edited by ravill -- 2/27/2004 12:52:10 AM >
_____________________________
RAVjets Demo Team. All We Do Is Fly.
|
Printable Version
All Forums >> RC Airplanes >> RC Warbirds and Warplanes >> RE: RE: World Models p-51(giant 80" WS) 
RE: RE: World Models p-51(giant 80" WS) - 
RE: RE: World Models p-51(giant 80" WS) - 
