Quote:
Originally Posted by Neko81795
if there are errors in my math, please correct me.
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Mostly typos:
Page 10 first equation has the denominators swapped
Page 16 what's the area calculation for? you can get H directly from sin(90-(theta+s))=H/Q
Page 19 should be sin(90-(theta+s))
Here's another way to do it:
Code:
I = 180-o
T = sqrt[R^2 + C^2 - 2*R*C*cos(I)] (Law of Cosines)
s = asin((R/T)*sin(I)) (Law of Sines)
w = 90-(theta+s) <-angle between T and Q
H = Q*sin(w)
B = Q*cos(w)
F = sqrt[H^2 + (T-B)^2]