Balance computation in %MAC

[Mad (Flt) Scientist]

if x is the reference point for the "leading edge" of the mac and "mac" is the length of the mac, then:

c=(a-x)/mac
d=b/mac

(which will give you cg as a fraction of mac; if you want it as a %age, then multiply then both by 100, i.e
c=100*(a-x)/mac
d=100*b/mac)

To prove this, consider that you already know:

h=a-(RA*b)/W

and if you know x and mac, then:

cg=(h-x)/mac

therefore

cg=( [a-(RA*b)/W] - x)/mac

= (a-x)/mac - [ (RA * b)/(W * mac) ]

= (a-x)/mac - [ (RA/W) * (b/mac) ]

so if cg = c - d RA/W, then c and d can be simply seen from the equation