Imagine there's an object floating 5 miles above the Earth's surface,and the Earth is perfectly smooth

Imagine there's an object floating 5 miles above the Earth's surface,and the Earth is perfectly smooth.
How does one calculate the size of the area that is in "sight" of the floating object?

I need the angle of the cone of sight.

With a really long tape measure.

you could just measure from one side to the other, earth is flat after all.

Do you mean distance? Worded kind of fucky

Depends on how big this Earth is and yeah the angle you're viewing the object at.

/thread

I'm a math major and I'm bored at work. Give me the radius of the earth in miles and I'll try to figure it out based on that

Easy. Just use this equation

8=D(0)^8

3959 miles

kek

Give me a minute

What are you looking for OP? I don't understand.

Are you looking for the area of the surface you can see from from that height?

leave it up to the math major, trips don't lie

Well focus on a numerical solution rn. I'm using geogebra, and a generalization can come in a little bit.

>Are you looking for the area of the surface you can see from from that height?
yes

You forgot something important to help calculate that OP hadn't specified.

You still need to know the field of view that the object can see. Not sure if I worded that right, but I'd imagine a FOV of 45 degrees would be a smaller area than a FOV of 80 degrees

bump

take it to /sci/ maybe?

Not that user, nor a math major (physics)
But hey, I dabble.
Ok, then OP, you mind me asking why are you looking for this? Let me attempt OP

I was just going with fov at the tangent lines

Pretty much what I thought of using, the tangent lines. Any who, so here's the plan OP.

If take the derivative of the function of a spherical cap and evaluate at where the tangent line meets I think we can find the total surface are visible.

Anyone in opposition?

>Ok, then OP, you mind me asking why are you looking for this? Let me attempt OP
Building a high altitude radio repeater.
I only have access to cheap latex balloons that can go only 5 miles up,so I just want an estimate of the line-of-sight area.
I am using spread-spectrum modules that can supposedly achieve ridiculously long ranges with very low power.

Bump

bump 3

/sci/ -->

>/sci/ -->
>>sci

>>/fag/

Take the equation of a sphere, with radius equal to that of the Earth, centered at the origin: x^2+y^2+z^2=R^2, where R is the radius.

Then take the equation of the bottom half of a cone, also centered at the origin, shifted upward by 5, and scaled appropriately enough by some factor k that aligns with the floating object's viewing angle (you can find k with basic trig from there).

Parameterize the spherical "cap" bounded by the cone. Best done in cylindrical or spherical coordinates. Then set up a surface integral and compute along the parameterized surface.

If I gave any more of a shit I'd be happy to show all the work, but alas.

Eqn. of cone would be z=5-k*sqrt(x^2+y^2)

Math major back. I computed it. Uploading photos of my work in a second, but I got 124532 mi^2

Shit! File size too large. Trying to figure out camera on mobile

...

Page 1. I just posted page 2 of my notes on it. I hope this clears it up. I couldn't get a nice big file to upload. I'll post the url's of the high def pics in a second

So I'm samefagging hard here but I didn't need to calculate theta, I just honestly haven't done a problem like this since maybe freshman year. Hope this explains everything OP. I'm sure someone else solved this too.

You dont, you know earths diameter