I have a polar graph (on paper) with a curve similar to 7.8*cos(theta). Plotting the data estimate from a nice big print-out of the graph gives a fairly close fit, but I seems that the circle needs to made into an oval to fit better. In other words it looks like a scale factor in the "y direction" (if you'll forgive me, I know it's a polar plot.)
deg rad f(r) Real Data 0 0 7.8000 7.8 10 0.174 7.6815 7.7 20 0.349 7.3296 7.4 30 0.523 6.7550 6.9 40 0.698 5.9751 6.2 50 0.872 5.0137 5.3 60 1.047 3.9000 4.2 70 1.221 2.6678 3
(Beyond 70 deg a different curve applies in my problem domain so I'm not too worried about that part of the curve)
Is there another form of equation that I can use to modify r=7.8cos(th) to stretch the curve? The purpose is to use an equation to calculate the values at intermediate points, and to avoid a potentially complex lookup system.
You stretch the plot vertically by a scale factor $\lambda\gtrsim1$.
Your equation was
$$\rho=r\cos(\theta),$$ and in Cartesian coordinates
$$x=r\cos^2(\theta),\\ y=r\cos(\theta)\sin(\theta)).$$
It is now
$$x=r\cos^2(\theta),\\ y=\lambda r\cos(\theta)\sin(\theta),$$
which is no more simple in polar coordinates, as
$$\rho'=r\cos(\theta)\sqrt{\cos^2(\theta)+\lambda^2\sin^2(\theta)},\\ \tan(\theta')=\lambda\tan(\theta).$$
Eliminating $\theta$, this gives this little monster:
$$\rho'=r\cos\left(\arctan\left(\frac{\tan(\theta')}\lambda\right)\right)\sqrt{\cos^2\left(\arctan\left(\frac{\tan(\theta')}\lambda\right)\right)+\lambda^2\sin^2\left(\arctan\left(\frac{\tan(\theta')}\lambda\right)\right)}.$$
You'd probably better stick to Cartesian coordinates with the ellipse
$$\left(\frac{x-r}{r}\right)^2+\left(\frac y{\lambda r}\right)^2=1.$$