Using parallax / triangulation to measure large distances in astronomy: from fizzics.org 

Thank you

That's good, I hope you enjoy your studies. Thank you for your comment.

I too have I love astronomy

Thank you

Good that it was useful

+Agasthya Rana When observations are made of very, very distant stars behind the star you are measuring, 6 months apart, when the Earth has completed half of its' orbit the change in the angle of observation will be negligible. It will be negligible because these stars are so distant compared to the 300 million kilometre shift of the earth around to the opposite side of its orbit. However a less distant star will seem to have shifted slightly against this background. It is that slight shift or change in the angle of observation that you measure. Hope that helps.

That's exactly everything I knew about parallax. I wanted to know about the exact method of measuring the angle as I am unable to think of it myself. An online journal might help! Great video though!

@@agasthyarana7750 its a little late but did you ever figure it out. The only thing I've found is that someone used a heliometer, but those don't seem to be very popular anymore

The distance of the moon is measured to a few millimetres using a laser and a reflector left on the moon in the moon landings.

Good, I'm glad it was useful. Thanks for the comment.

Good that it was useful

That's great, thank you

Hi no it does not have to be 90 degrees but the calculation is much easier if it is.

Many thanks.

Hi, to take your two questions in order. The earth is not quite spherical but the variation is small compared to the diameter. Compared to the diameter of our orbit the variation is infinitesimal. This method described here is not highly accurate for more distant stars in our galaxy because the angular differences we are measuring are tiny, maybe 1/100,000 of a degree or less and the percentage error possible in measuring this can be substantial. Measuring distances in astronomy is not a perfect science. Going on to your second question, we only use this method for stars within our own galaxy and these are not moving away from us. It is only other galaxies which are becoming further away. Hope that helps.

Good video and answer! A historic reason: Archimede thought that the Earth must have been a sphere. This was an error before Galileo. The Romans who computed the distance from the Earth to the Moon through parallex inherited this mistaken assumption.

How do you know the earth is a sphere?

Surely you would need a rough measurement to start with to know how far you could break a degree down?

I think that the identity of those responsible maybe lost in the mists of time, however the actual size of any unit we use for measurement is fairly unimportant, the key is that there is wide agreement and use.

Hi, I sympathise but it is difficult to get the level right. Those who know the maths accuse me of being condecending if I explain too much detail. Because there is a right angle (90) and we measured the other to be 83.5 then the smallest angle must be 6.5 (the total angles in a triangle is always 180 degrees. The tangent of an angle in a right angled triangle is a ratio of the length of the side opposite the angle to the nearest (adjacent) side.Looking at the diagram the tangent of the 6.5 degree angle is therefore 10m (we measured that) divided by the unknown side, which is the distance across the river. The value of the tangent can be looked up on any scientific calculator or online, it is 0.114. So if we rearrange the equation this unknown distance is 10 divided by the 0.114 which gives us 88m

Roger Linsell That still makes no sense.

Roger Linsell You're presenting us with circular reasoning. The distance across the river (88) is found by the tangent of the 6.5 angle, but to know the tangent of the 6.5 angle you have to know the distance across the river. We are not using a calculator to find tan of 6.5. We are or at least i am asking for the math that shows why .114 is the tan of 6.5.

lol. You need to have a knowledge in trigonometry and geometry in order to fully understand and appreciate the parallax effect and this lecture cos he can't teach trigonometry and geometry in a few minutes. Nice video btw

@@garyrolen8764 god you are so dumb

+Fred Caldwell Hi, I don't really understand your question. if you can take accurate measurements of angles on a reasonably long baseline at exactly the same moment (remembering the rotation of the earth causes the angles to change) then you should get an answer around 91 to 93 million depending on the time of the year and the point of the orbit. If the baseline is very long then the curvature of the earth might make a little difference in that the distance around the curve would be greater than a straight line.

Well, suppose the earth is flat for a minute. and some guy is in the middle of the Atlantic ocean, riht on the equator. He looks straight up and sees the sun at noon (his time) on March 22. At that same time, someone in Malapo Brazil (also along the equator) looks east at 10:33 AM (his time), and sees the sun rising at 60 degrees up in the sky, while someone in Equatorial Guinea, Africa looks up westward at 60 degrees from the horizon and sees the sun going down at 2:33 PM (his time). Would not the sun then have to be the same distance from the flat earth as the distance is between the two cities? Yes. And that calculation comes to less than 4000 miles away when completing a 180 degree equilateral triangle. If we try to do the same experiment on a ball earth and go up 60 degrees from each city, we know the total degrees from each city will be a lot more than just 60, because we have to add the degrees from an additional angle. The new angle must be calculated from an Imaginary Straight Line (ISL) through the earth between the two cities, back up through the earth to where each viewer sees the horizon.. So let's call the Malapo man's angle view: "A" (60 degrees), and the new unseen angle (from a straight line between the 2 cities): "B". Modern science tells us that the angles from our perspective here on earth for an equilateral triangle would have to equal somewhere in the neighborhood of 89.99 degrees from the two points on earth mentioned - and that (large but still acute angle) brings the equation to an approximate 93,000,000 miles. However, do the math yourself adding "A" plus "B" and you'll find you get a radical obtuse angle from each city that does not converge but diverge - throwing the accepted scientific answer out the window. I'll have to make a short video to give an visual example sometime. But try it yourself - if I've made any sense here!

Hi Not sure that I fully understand but remember that each observer measures the angle to their own horizontal and each of these horizontals, at a different places on the globe, will be far from parallel. In order to calculate these differences you would have to know the diameter of the earth.

That's fine. I'll never be accused of being a great communicator. But yes. The two important angles are where observers see the sun up at 60 degrees from the horizon at the same time. Keep in mind, these are not random locations but both along the equator. I used the accepted diameter of the earth and calculated the depth at earth's most curvature apex (to the ISL), from the man at sea in the Atlantic who's viewing the sun straight overhead at noon. (Its around 1000 miles from the apex of the curve, straight down to the Imaginary Straight Line between the two men. Each of those new angles have to be added to the 60 degrees already calculated, and when you do that, you come up with 107 or so degrees with an angle that diverges far away from the sun, instead of converging to the point of the sun. The only way there could be 89.99 degrees in each corner would be if we could not trust our own eyes seeing the sun at 60 degrees - and I don't think there could be that much light refraction in our atmosphere if the sun were only at 17 degrees up when both viewed from Malapo and Equatorial Guinea simultaneously. I'll try to make a video of it soon to show all the shapes / overlays.

Hi I have no means of checking or verifying your figures so I fear my answer is that I don't know.

You could, although since sin is opps/hypotenuse and we don't know or need the value of the hypotenuse (although it won't in reality be much different to the perpendicular distance of the "adjacent"). Using the tangent function is more direct.

What?

You don't understand it boi!

Nope.

In good conditions the atmospheric distortion would be minimal. To be thorough you could take measurements over several days to average out the effects.

It would give a rough estimate, and there are better ways to do it, and was done by the Greeks. Now there are even better ways, such as shooting a laser to the moon and watching the reflection.

I built a small one just because i was board and yes it works great.