# 1/2 factor in drag equation

Why the 1/2 factor in the equation $F_{R} = \frac{1}{2}\rho C_{d}A v^{2}$? Why not just divide every $C_{d}$ value found in tables by 2 and omit the 1/2 in the equation instead?

ShayanJ
Gold Member
The only reason I can think of is that the guy proposing this formula was thinking the force should be proportional to half of one of those things in it(I don't know why though!) and he wants to get sure people have a clue for that. I think the best candidate is the area.

Energy of compressed spring is ##\frac{1}{2}kx^2##. So according to you I will do some magic and replace all k value of spring in every table to k/2.
Now I have an equation ##F=kx##. This is the force exerted by a compressed /elongated spring. If I use the values given in the new table, I will get wrong answer. So I will have to multiply 2 every time I use it. Maybe ##C_d## has other uses and ##C_d## is more commonly used than ##C_d/2##.