I need to simulate the Earth’s curvature for some experiments. Is there a software that can handle extremely large objects while placing the camera on or at a specific altitude from the surface of these large objects with the ability to arrange other objects on the same surface?
If not, I thought about using the arc tool to create a small “strip” of the Earth’s surface, but I need to know the bulge distance.
I know the lengths of A, C, and, D but I need to know the distance from the midpoint of C to the highest point of the arc (B) for me to create the “strip” of earth.
Any mathematicians out there?
If A is 5 miles then D is 16.6667’.
The formula to find D is: D=(8"•A²)÷12
The “bulge distance” could possibly be found out since we know that it is a section of a circle that is 24,901.461 miles in circumference but I have no idea how to calculate my way to the correct answer.
Scale it down in sketchup, draw a circle in inches that is the miles radius of the earth and fit your triangle to it, then measure the bulge.
Very basic example, using 96 segment circle, the more segments the more accurate, to a point.
You can also calculate the Sagitta, here’s a link.
At that scale, the “A” would be 5 inches long and “D” 0.003156572 inches. I don’t know if “D” would be too short of a distance, but drawing the circle at 3,958.8 inches… there would need to be about 1,300,000 segments in the circle, which even with that amount would mean the flat surfaces of the earth would change angle once every 100 feet approximately, which translates to 0.018939 inches at the 1" to 1mile scale. The measurement taken using your method, even with this hypothetical circle, would not be accurate enough when “A” is just 5 miles.
However, 999 segments would make a completely useless measurement. It would be as if the earth was a bunch of perfectly flat fields that were 25 miles long.
Then there’s all the crazy navigation, clipping and move tool issues… just trying to interact with such a tiny triangle compared to the circle would present issues that… well… it’s frustrating, as I’m sure you know. SketchUp just isn’t made for objects the size of planets.
this may be my best option, though I wish it had more decimals for the calculator. I can do it myself with their formula. I’ll give that that shot!
If that means it would have the correct bulge according to the given radius then yes! I just wonder how accurate it will given be that I will be entering data in feet converted from miles. In this circumstance, the radius of the arc would be 20,902,464 feet.
The process would be, for a distance of 5 miles, to draw a line at 26400 feet, then from that end, draw a line vertically down at 16.6667 feet, start an arc from the start to end, and type “20,902,464r” making sure my units are set to decimal feet.
This is interesting, but it doesn’t seem like a 3D simulation, and since I am doing optical experiments, I need to basically be standing on 3D geometry.
If I may ask, what is the purpose of your experiments?
Are you planning on climbing a mast on a (sayling) yacht, to see if there are any large ships on the horizon?
To be exact, the earth is not a sphere but nearer an ellipsoid. The distance from the center to a pole is 21 kilometers (13 miles) shorter than to the equator.
This article would, I guess, add another layer of complexity:
He who would be a scientist shouldn’t be attempting to “prove” anything, but simply seeking the truth. When your bias is welded to your inquiry, your answer may be welded to self-deception.
Flatearth.ws is an interesting website that is dedicated to succinctly refuting the claims of people who believe the earth is flat one by one and I am trying to re-create some of those refutations. I’ve done this with other software before which have resulted in pretty conclusive evidence that the earth is round. Though, I’m not able to simulate atmospheric refraction.
