I would, but I don't feel the need to convert to standard, manageable units if you can't present the data in such a format. I'm also not going to bother trying to confirm your numerical manipulations, given the unnecessary complexity introduced by your choice of unit system.

The, you proved me wrong, but I'm just going to ignore it and change the subject, response.

It's simple. Here you go.

1. By knowing in your model that the sun is 32 miles across and 3000 miles away than we can calculate that the objects distance is 93.75 times it's diameter.

2. By comparing the diameters of the two objects, 32 miles for the sun, and 24,900 miles for the Earth, we can calculate that the Earth in your model is 778.125 times larger than your sun.

3. Now, to reduce the scale of your model to something more manageable, we use a common household item, a quarter, as your sun. The quarter measures .96875 inches across (you can measure that is you want. You'll get 31/32, I just turned it into a decimal.)

4. Now, knowing that your Earth is 778.125 times larger than your sun, we simply multiply the quarter's diameter, .96875 by 778.125 and we come up with 753.809 inches, or 62.817 ft.

5. Now to calculate the altitude of our sun. In your model, the altitude is 93.75 times the diameter of the object, so we can take the diameter of the quarter, .96875, and multiply by the 93.75, and we get 90.8203125 inches, or 7.57 ft.

6. Now you can build your scaled down model, if you're not too lazy.

7. Go somewhere flat, I recommend a large parking lot, and measure out a circle with a diameter of 62.817 ft..

8. Obtain a spotlight like your site claims the sun is. I'll even allow you to go bigger than a quarter. Hold your light source at a height of 7.57 ft.

9. Try to illuminate at least half of the circle you measured out.

10. Post your results.