Quote from: Boobies00 on January 05, 2017, 06:51:16 PM

Quote from: notbatman on December 30, 2016, 12:38:12 PM

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Quote from: jam1nut0 on December 30, 2016, 11:57:42 AM

earth cant be flat, that's dumb

Educate yourself

I educated myself, and here is what I found:

Quote from: https://bitcointalk.org/index.php?topic=1009045.msg17342248#msg17342248

A transit with a powerful telescope, combined with observations 91 days apart, observations on the sun and on a horizon star, can show the 93 million mile distance to the sun (plus or minus 10%) by trigonometry. Nothing can refute this, since it can be repeated over and over to give a nearly accurate answer. Using a better telescope, and making similar measurements gives the 93 million miles rather accurately.

Now I'm happy. Cheesy

Fuck off, using a sextant and plane trigonometry the distance to the Sun is measured at about 3,000 miles.

A sextant is a doubly reflecting navigation instrument used to measure the angle between any two visible objects.

How do you get distance if you only have the angle? Can you show us the math behind this?

first off you can measure the distance based on angle, second of all the distance is a HELL of a lot farther then just 3,000 miles.
Here is the proof, math included.

What we actually measure is the distance from the Earth to some other body, such as Venus. Then we use what we know about the relations between interplanetary distances to scale that to the Earth-Sun distance. Since 1961, we have been able to use radar to measure interplanetary distances - we transmit a radar signal at another planet (or moon or asteroid) and measure how long it takes for the radar echo to return. Before radar, astronomers had to rely on other (less direct) geometric methods.

In more detail:

The first step in measuring the distance between the Earth and the Sun is to find the relative distances between Earth and other planets. (For instance, what is the ratio of the Jupiter-Sun distance to the Earth-Sun distance?) So, let us say that the distance between Earth and the Sun is "a". Now, consider the orbit of Venus. To a first approximation, the orbits of Earth and Venus are perfect circles around the Sun, and the orbits are in the same plane.

Take a look at the diagram below (not to scale). From the representation of the orbit of Venus, it is clear that there are two places where the Sun-Venus-Earth angle is 90 degrees. At these points, the line joining Earth and Venus will be a tangent to the orbit of Venus. These two points indicate the greatest elongation of Venus and are the farthest from the Sun that Venus can appear in the sky. (More formally, these are the two points at which the angular separation between Venus and the Sun, as seen from Earth, reaches its maximum possible value.)

Another way to understand this is to look at the motion of Venus in the sky relative to the Sun: as Venus orbits the Sun, it gets further away from the Sun in the sky, reaches a maximum apparent separation from the Sun (corresponding to the greatest elongation), and then starts going towards the Sun again. This, by the way, is the reason why Venus is never visible in the evening sky for more than about three hours after sunset or in the morning sky more than three hours before sunrise.

Diagram of Venus orbiting the Sun, as seen from Earth

Now, by making a series of observations of Venus in the sky, one can determine the point of greatest elongation. One can also measure the angle between the Sun and Venus in the sky at the point of greatest elongation. In the diagram, this angle will be the Sun-Earth-Venus angle marked as "e" in the right angled triangle. Now, using trigonometry, one can determine the distance between Earth and Venus in terms of the Earth-Sun distance:

(distance between Earth and Venus) = a × cos(e)

Similarly, with a little more trigonometry:

(distance between Venus and the Sun) = a × sin(e)

The greatest elongation of Venus is about 46 degrees, so by this reasoning, the Sun-Venus distance is about 72% of the Sun-Earth distance. Similar observations and calculations yield the relative distance between the Sun and Mercury. (However, Mars and the outer planets are more complicated.)

Historically, the first known person to use geometry to estimate the Earth-Sun distance was Aristarchus (c. 310-230 BC), in ancient Greece. He measured the angular separation of the Sun and the Moon when the Moon was half-illuminated to derive the distance between Earth and Sun in terms of the distance between the Earth and the Moon. His reasoning was correct, but his measurements were not. Aristarchus calculated that the Sun is about nineteen times farther than the Moon; it is actually about 390 times farther than the Moon.

Another ancient Greek astronomer, Eratosthenes (276-194 BC), estimated the distance between Earth and Sun to be either 4,080,000 stadia or 804,000,000 stadia. There is disagreement regarding the correct translation of Eratosthenes' value, and further disagreement over which length of a stadium was used by Eratosthenes. Various sources estimate that the length of a stadium (also called a stadion or stade), converted to modern units, is between 157 meters and 209 meters. Then 4,080,000 stades is less than 1% of the actual Earth-Sun distance, no matter which definition of a stade one chooses. However, 804,000,000 stadia is between 126 million and 168 million kilometers - a range which includes the actual Earth-Sun distance of (approximately) 150 million kilometers. So Eratosthenes may have found a fairly accurate value for the Earth-Sun distance (possibly with some luck), but we can't say for sure.

The first rigorous and accurate scientific measurement of the Earth-Sun distance was made by Cassini in 1672 by parallax measurements of Mars. He and another astronomer observed Mars from two places simultaneously. A century later, a series of observations of transits of Venus provided an even better estimate.

Since 1961, the distance to Venus can be determined directly, by radar measurements, where a series of radio waves is transmitted from Earth and is received after it bounces off Venus and comes back to Earth. By measuring the time taken for the radar echo to come back, the distance can be calculated, since radio waves travel at the speed of light. Once this Earth-Venus distance is known, the distance between Earth and the Sun can be calculated.

As you have indicated, once the distance between Earth and Sun is known, one can calculate all the other parameters. We know that the Sun, as seen from Earth, has an angular diameter of about 0.5 degrees. Again, using trigonometry, the radius or diameter of the Sun can be calculated from the distance between Earth and Sun, a, as 2×Rsun = tan(0.5 degrees) × a. Also, since we know the time taken by the Earth to go once around the Sun (P = 1 year), and the distance traveled by the Earth in this process (approximately 2πa, since Earth's orbit is nearly circular), we can calculate the average orbital speed of Earth as v = (2πa)/P.

Anyway, the relevant numbers are:

Earth-Sun distance, a = roughly 150 million km, defined as one Astronomical Unit (AU)
Radius of the Sun, Rsun = roughly 700,000 km
Orbital speed of Earth, v = roughly 30 km/s