Recent content by aneikei

Thank you for your help btw. In the image, the angle is a static 55 degrees. At .7 inches there is less height than at 1 inch. At both, I want to know how many pixels that are 5177 nm in size can fit vertically at each position.

I'm trying to calculate If given 55 degrees and a run of .7 inches what's the number of (5177nm) pixels I can fit in that height?

Hello, I could some help. I've reached some sort of paradox. Ultimately I'm trying to calculate the number of pixels per inch, given a pixel size of 5177 nm and extended at a distance of .7inch (.0178 meters) (A) (60 arcsecs * 0.0178 meters)/206268 = 5177nm pixel per arcmin (B) 5177 * 60 =...

I apologize gentlemen or ladies. I had a previous post and when I needed an answer in a very particular way. I was browbeaten as to why would I ever do it that way. When I explained that although I understand the form I want the solution in isn't the "industry standard", but that I need it for...

I totally agree. The thrill and exhilaration is in the self-discovery.

Thanks, I already figured it out ##x=\sqrt {2y-y^2}## Everyone on this forum is mean. Instead of just kindly helping, you and those like you have to make sure to go out of their way to get their jabs in. Belittling anyone at any chance they can get. I take it to feel taller? How small you must...

using this equation ##1-\sqrt{1-x^2/c^2}## where c = 1 and x = 0.0 - 1.0 the speed of c for example ##1-\sqrt{1-.886^2/1^2}## = y = 0.5363147619 gives me the y values. How do I find the inverse? How do find for x inputting the values of y? Thank you.

Then I think there is a small bit of confusion as to what I'm looking for. If I'm standing on a neutron star and all I'm given is its radius and the amount that it dilates time at its surface. Which is 0.0910729244 Using the time dilation range 0.1 - 1.0 What is its mass? If it had a time...

Is this right? ## M = \frac{rc^2}{2G}## If that is the case there no term to plug in for my ##x = 1 - \frac{1}{\sqrt{1 - .50^2/c^2}} = 0.1339745962## Shouldn't it be something like this ## M = \frac{c^2\cdot r \cdot (1-1/x^2)}{2\cdot G} ##

Sorry about that, the time dilation here is 0.13 ##x = 1 - \frac{1}{\sqrt{1 - .50^2/c^2}} = 0.1339745962## So if given for x and the radius of the object. I'm looking for the mass of the object

Yes, that's exactly right. If gamma is 1 then it means I'm standing on a black hole which would have an infinite time dilation which is also 1. So if given that and if I know that the radius of the black hole I can calculate its mass. However, a gamma of .50 would give me a time dilation of...

Sort of. I'm looking for the equation to calculate the mass of an object if given its radius and the amount of the time dilation it causes using a range of 0.1 - 1.0. Where 1 would be infinite time dilation. To derive this scale we use the formula 1-(1/((1/(sqrt(1-(v^2)/(1^2)))))) where v is...

It for does to me.I'm working on a paper and need it converted this way Thank you for the help

"None of this has anything to do with the radius of the object. It has to do with the distance from the object's center." isn't that the very definition of the radius? And thank you. I know the formula to calculate the mass using the escape velocity. However, I'm looking the equation to...

Time dilation is a function of gravity. Which can be thought of as escape velocity from a gravitational field. So if you have to achieve .866 c to escape. Then the time dilation would be 1.99 years for every one year on earth. using 1/(sqrt(1-(.866^2)/(1^2))) Now that I have that and if given...