Calculating Visibility of a Circular Structure on a Flat Earth Surface

[e-realmz]

I do not know if this would be the right thread for this forum but I have a few questions.

On a surface exactly equal in all properties on Earth but lacking in any properties which would create poor or limited visibility at any distance including trees, mountians or level differences, there is a circle structure exactly 4,500 ft. high. It is 66 miles in diameter from 0 to 3,200 ft. From 3,200 ft. to 4,500 ft., it evenly slopes until it reaches a center point at the top.

With that information, I need to know how far (in miles) would one have to be for the following *points of this structure to be exactly visible on the horizon.

*0+ (Structures foundation)
*3,200+ (Beginning of slope)
*4,500 (Structure no longer visible)

Could anyone answer that?

 

[Tide]

If I understand correctly what you are asking then you just need to find the distances to common horizon of both the observer and the observed point and add them. Pythagoras will accomplish that.

 

[dextercioby]

Well it's not differential geometry and nor tensor analysis. I'd say it's more like elementary geometry. As long as you assume Earth to be spherical.

Daniel.