Image scale and target surface?

Hello guys,

I know it must be a matter of simple mathematics, and so this is ultimately a dumb question, but I can't get it

How do I translate my image scale in area covered on the target?

I mean this: if I image an object at distance d at 1 arcsec/pixel, how many square meters on the object's surface does a pixel cover? Is there a simple formula that ties these quantities (image scale; distance; target surface)?

The area covered must be directly proportional to the distance d, but that's about as far as I can get. I also thought it could be approximated to calculating the base of an isosceles triangle with height equal to the target distance d and vertex angle equal to, in our example, 1 arcsec (the other 2 being then 1/2*(180°- known angle)). Unfortunately, I don't know how to calculate the length of a triangle's sides knowing its angles. If I could find the length of the long side, I could Pythagoras to calculate the short side and with it the area covered by one pixel and by my whole sensor.

I often wonder how much solar or lunar surface is pictured in my images and, with such a formula handy I would know at once

Sorry about the long post guys and thanks for any help you can give!

Luca

The size of your 1 arcsecond pixel at the target is
approximately d / 206000, where d is the distance to
the target.

It is basically a very long skinny isosceles triangle.
If the angle is 1 arcseconds, the ratio of the long
sides to the small side is about 20600.

An exact calculation using trigonometry is
target = 2 d (tan(angle/2))

For small angles with the angle measured in radians one can approximate
tan(angle/2)=angle/2

so you have
target = 2 d (angle/2) = d * angle = d / 20600

where one arcsecond is pi / (180 * 60 * 60) radians
or about 20600.