Am 09.04.2018 um 21:28 schrieb dick balaska:

> I'm trying to zoom in on my SpaceLoco as he flies away.  The angle
> option to the camera does exactly what I want.  I say "angle 20" and it
> is beautifully zoomed.  But I don't know why I'm saying angle 20.
> I want to know what the "default" angle is.  I'm stuck trying to solve
> for angle.  Doc says:
> 
> direction_length = 0.5 * right_length / tan(angle / 2)
> 
> So:
> 1 = 0.5 * right_length / tan(angle/2)
> 1 = 0.5 * 1.69 / tan(angle/2)
> 1 = 0.845 / tan(angle/2)
> tan(angle/2) = 0.845
>
> and then I'm stuck. I don't know how to extract that /2 so that I can
> take the arctan.

As both sides are equal, their /arcus tangens/ (written as arctan, atan
or tan^-1) should also be equal:

atan(tan(angle/2)) = atan(0.845)

Since the /arcus tangens/ is the inverse of the tangens (tan), the two
cancel out on the left side:

angle/2 = atan(0.845)
angle = atan(0.845)*2

I leave the rest up to you and a good pocket calculator ;)

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