Kriss3d

We know earth tilt 23 degrees. So how can the sun be directly above when at the equator ? Or is the equator only aproximatly where youd expect the sun directly above ( meaning that the line around earth would also tilt ) Looking at heatmaps of earth where you see most warm at equator. I would take it that its the mean temperature over a year ? Otherwise the "belt" of the suns path around earth being tilted would also have the belt tilt ? Or am i getting this wrong ?

Im working on a little problem for an idea im having. If you stand on earth and look out over the ocean. Your field of vision would be almost 180 degrees. But lets say 150 degrees for this example. Being 6 feet above sea level puts your distance to the horizon 3 miles out. This means that by trigonometry you can draw a triangle with 3 miles on two sides and the angle between them being 150 degrees. This would make the line between them be 5.8 miles. Ive been wondering. By math, how much of earth curvature would be seen on each side of the middle where the curvature from your point of view would be zero ? Earth will curve down to the sides in respect to the direct front of you. The wider this tangent line the more the curvature will be aparent. Obviously its not possible to register any curvature by looking at the ocean from this close. But math should be able to tell exactly how much it actually is. By my crude paint skills ive made an example. What id like to know is how to get the hight of the tangent to the surface of the curvature when i know that the tangent line is 5,8 miles long. I Just cant quite seem to find the right way to formulate this into the right formula.

Thanks. Thats what i had guessed as well but i wasnt sure.

Well not unless you mean that since earth is moving at 66,000 mph and you launch a rocket then it will already be going 66,000 mph ? Is that how it works ? Or why else would a rocket be able to travel in our solar system without having to also apply the force to obtain the speed just to keep up with the solar system moving ?

Ive been looking at some different claims about space travels. Correct me if im wrong here. But i was wondering how a rocket for something - say the moon or mars, is able to keep up with earth traveling at 66,000 mph through space ? I know how this works within the atmosphere where the air stands still in relation to earths rotating motion and its travel through space. Is there some kind of similar "atmosphere" that travels along with the entire solar system allowing rockets to not have to travel in an excess of 66,000 mph to even keep up with earth ? Or how does this work ?

Great. Well not that he's been sick. I hope he gets better soon. But that the project isnt dead. I still think the Alfred Wallace experiment had been much easier to perform and would have given a more conclusive result. But to actually have a modern proof that isnt relying on anything that only a select few are capable of - because lets me honest. Taking a rocket to space isnt something most of us will be able or permitted to launch from our backyard.. Its just a bit odd that there doesnt seem to have been any such experiments that I could find since the Rowbotham himself. But to minimize the refraction the next experiment should be done higher above the sea level right ? Another thing i thought of. Wouldnt there be a more accurate way to get an absolute straight line other than laser ? Wouldnt radio waves in theory be better as they wouldnt be subjected to refraction like light does ?

Is there anything new in this ?

Awesome. Thanks. Ive been putting up the dishes for quite a while in the past and we would just use the numbers on the mounting rack to aim for 26 degrees then ajust it untill we got the best signal strenght. I couldnt quite get why the hight i would get would be around 1800 miles above ground at equator. I just used the distance from some site from Denmark to equator as a line then the angle of 26 in a trigonometry to get me the lenght A. So I was way off. Thanks.

HI everyone. I dont know if this is the apropriate place to ask. Ive been having a discussion with Flat earth belivers ( yeah i know. ) One thing i couldnt quite get to match up with what i know. Ive in my past worked with satelite dishes and setting up these babies for people. Now here where i live in Denmark, we put the satelite at 26 degrees from horizon aiming them towards south. I searched for the hight of satelites such as the Thor 5 satelite which covers Denmark. Aparently the distance from Denmark to the equatorial line its just about 6700 miles. If i tap that into a trigonometry calculator with 6700 miles and 26 degrees, i get a hight of the satelite at 1800 miles. While that is within the geo stationary orbit hight, I read about the Thor 5 satelite that its 22.000 miles. Thats quite a difference. What have i missed in this ?