Find Earth's Radius Using a Sunset

Here I show how to calculate the radius of Earth to surprising accuracy using just a sunset.

I go into theory involving Earth's rotation and derive an equation to find the radius before reviewing my results from the experiment.

0:00 - Introduction
0:10 - Utility of Sunsets
1:10 - Theory Behind Method
3:46 - Deriving the Final Radius Equation
4:44 - Further Considerations
6:20 - Processing Results
7:16 - Review, Other Methods and Summary

[OVAstronomy]

Yes - there is a conversion error on screen when I show myself plugging in my height into the equation! A rookie mistake, I know!

[mrcopeland]

Your setup would be ideal if you were standing on the equator at either the spring or fall equinox. The reason your result is lower than the actual is due to the geometry that results from the tilt of the Earth at that time of year and of your latitude on Earth. If you do this experiment on the 1st day of spring or 1st day of fall, then you'd only need to adjust due to your latitude. If you divide by the cosine of your latitude (you sound British so lets say South England ~50 deg), you'd get 6222 km (4000 / cos(50)).

[zerobyte]

Was the observation made over water? That can cause superior/inferior mirages - and the amount of deflection of the Sun’s position will vary based on how close your view is to the water

[philoso]

Nice video and idea. Let me challenge our minds with this question.
What if we don’t have modern cameras or time keepers other than an evenly scaled rope and a good night sky? In other worlds how did ancient philosopher figure out radius of and distance to earth moon sun?

[michaelkinney]

I think the problem with your calculations is that you weren't on the equator when you took the measurements. Therefore your circle arc had a smaller radius, and the one from google. Think about how much smaller the latitude circles are on a globe through the uk compared to the equator of the globe.

[fug]

Wjay about light from the sun hitting the earth parallel?

[David_Lee]

Great video! I’m surprised you haven’t had droves of flat earthers screaming “nuh-uh!”

Walter Bisllin has an excellent discussion of refraction on his website, including a standard correction that can be applied for typical atmospheric conditions.

I was also wondering if you could reduce error by starting when you see the start of sunset on the beach, then run up to an elevated position like a beach hotel room or observation deck. As long as you can get a good estimate of the elevation, you would have a bigger “h” and a longer time.

[habtamumanaseb]

More hints, the space has 360 days as one linear year which is only 1hr for the light while only 24hrs day for time

[bernardtruchet]

c'est un exercice classique de trigonométrie qui permet effectivement de déterminer la distance de 2 arbres installés sur une machine, ce n'est pas valable en astronomie, ce calcul suppose un "rayon de soleil" matériel qui n'existe pas, il n'est pas prouvé que le soleil nous envoie "des rayons de soleil" visible comme des lignes droites, d'autre part la terre et le soleil se déplace à grande vitesse ???. Un raisonnement "similaire" a été appliqué pour déterminer la distance de la terre à la lune, le chiffre de 300 milles kilomètres est contestable pour ne pas dire faux

[christopherellis]

The minor differences in the metrics make little difference to the result, especially for someone who is only 179 cm tall.

[edwardlewis]

How can you work all this out knowing just one angle. Ie your 90 degree angle. You must know the radius in which case you know the circumference, or the distance to the Sun. Sure you know the time it takes but that does not give you the speed.

[DanPhysicsDoc]

Your calculation for angular speed uses a sidereal angle over a solar day.

[poneill]

Actually, from new learning we know the Earth to be banana shaped