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        riddles >> general problem-solving / chatting / whatever >> apparent area/position function
        
(Message started by: jarls on Jan 13^th, 2009, 2:39pm)

Title: apparent area/position function
Post by jarls on Jan 13^th, 2009, 2:39pm

An observers stands with eyes at a fixed location. An object moves away from the observer linearly with constant velocity and along the observer's line of sight/parallel to the observer's line of sight but not along it. What function designates the rate of change in the object's apparent area (imagine holding up tracing paper in front of the eyes and tracing the object) of the object for the observer with respect to time where velocity is some constant? If the trajectory of the object does not lie along the observer's line of sight what function designates at what rate it will appear to approach the line of sight of the observer?

Title: Re: apparent area/position function
Post by Grimbal on Jan 14^th, 2009, 5:22am

The size of the object would be proportional to 1/t where t is the time from the moment the object was closest to the observer. I assume a projection on a plane perpendicular to the line of sight.
The area would be ~1/t².

Title: Re: apparent area/position function
Post by jarls on Jan 14^th, 2009, 7:44pm

What would the function be?

Title: Re: apparent area/position function
Post by 1337b4k4 on Jan 14^th, 2009, 11:15pm

since the object is off to the side of you, if it gets really close the apparent area will go to 0 again. I don't know if there's a way to avoid some messyish trig.

Title: Re: apparent area/position function
Post by Grimbal on Jan 15^th, 2009, 5:18am

The position on the screen will be something like
(x,y) = (x0,y0)·(1/(t-t0)).

t0 is the time the object passes by the observer
(x0,y0) is the position on the screen at t=1.