- #1
FelixLudi
- 26
- 1
(i.e the height you need to be at to look forward)
(This isn't in physics subforum because it's more of a geometry question, this is NOT a homework question and it's more of a random thought)
Basically, I've been thinking what height would you have to be at, to see 1° of Earth's curvature.
So this is my math;
Circumference of Earth = 40.075 km ~ 40 000 km
x = 40 000 km
Assuming that Earth is a perfect sphere, and that in 2D it's a perfect circle it's angle would be 360°.
Over 1° of Earth's curvature, the distance (d = x/360°)
d = 40000 km / 360°
d = 111,11* km
(In my school we put an asterisk at the end to define a number without end (e.g 2/3 = 0.66*))
Over that 1° of curvature, I assumed a shape of a triangle where the ground points are defined as dots A and B where the dot A has almost a sharp angle (90°) and the dot B has an angle of 1°.
In that triangle the cathetus between A and B is equal to d = 111,11* km, and the dot C which is elevated above dot A has the other cathetus of undefined height a, while the hypotenuse is c and is directly opposite of the dot A. In this triangle the hypotenuse and the cathetus d are near the same length, and I can not go further than this to calculate the height a needed to look over from dot A to dot B over the Earth's curvature (angle) of 1°. How do I adjust the equation of Pythagoras' theorem to this triangle?
Anyone that can help me?
Thanks!
(This isn't in physics subforum because it's more of a geometry question, this is NOT a homework question and it's more of a random thought)
Basically, I've been thinking what height would you have to be at, to see 1° of Earth's curvature.
So this is my math;
Circumference of Earth = 40.075 km ~ 40 000 km
x = 40 000 km
Assuming that Earth is a perfect sphere, and that in 2D it's a perfect circle it's angle would be 360°.
Over 1° of Earth's curvature, the distance (d = x/360°)
d = 40000 km / 360°
d = 111,11* km
(In my school we put an asterisk at the end to define a number without end (e.g 2/3 = 0.66*))
Over that 1° of curvature, I assumed a shape of a triangle where the ground points are defined as dots A and B where the dot A has almost a sharp angle (90°) and the dot B has an angle of 1°.
In that triangle the cathetus between A and B is equal to d = 111,11* km, and the dot C which is elevated above dot A has the other cathetus of undefined height a, while the hypotenuse is c and is directly opposite of the dot A. In this triangle the hypotenuse and the cathetus d are near the same length, and I can not go further than this to calculate the height a needed to look over from dot A to dot B over the Earth's curvature (angle) of 1°. How do I adjust the equation of Pythagoras' theorem to this triangle?
Anyone that can help me?
Thanks!