Say I have this object that at 1 mile away, is 100 pixles wide. How many pixles is it at 2 mile wide?
I am probably asking a simple question, but I have no idea how size/distance is figured.
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Say I have this object that at 1 mile away, is 100 pixles wide. How many pixles is it at 2 mile wide?
I am probably asking a simple question, but I have no idea how size/distance is figured.
I imagine you can do this the same way cartographers plot cities on a map. Just put everything to scale. This is a proportion question. Do you remember how to solve those?
At 2 miles, it will be 50 pixels.
Callou collei we'll code the way
Of prime numbers and pings!
Unless my math is wrong, if you think about the "camera" being a single point, because of different ways to express the tangent of the angle,
w1/d1 = w0/d0
You ever try a pink golf ball, Wally? Why, the wind shear on a pink ball alone can take the head clean off a 90 pound midget at 300 yards.
You don't need to worry about tangents of angles -- because computer screens are flat, you want to think about how the rays of light are laid out as they pass through a plane in front of the camera.
w1*d1 = w0*d0 is the relationship to use.