Recent content by neilparker62

We tend to take books for granted but it's a huge achievement to have actually written one. Congratulations on becoming a published author.

Mind blowing!

$$H=\frac{g}{2}\left(\frac{t_{0}}{2}+\frac{h}{gt_{0}}\right)^{2}$$ I just needed to see your formula.

Google Gemini. Will post all its workings generated after some "coaching" from me. I referred it to following desmos graph: https://www.desmos.com/calculator/mrmrkhoskf and told it to solve for Point A.

same thing I think. Just need to coach your AI to tidy up maybe.

Excellent diagram. Can solve by finding the area under the blue curve between it's x-intercept and point C.

https://archive.org/details/nicolaicopernici00cope_1/page/n7/mode/2up

Worth noting that if ##b=n^2##, then BM, BD and MD form a Platonic sequence for generation of Pythagorean triplets: $$\frac{n^2-1}{2}\;;\;n\;;\;\frac{n^2+1}{2}.$$It's not quite clear why this bears the name of Plato who was a philosopher rather than a a mathematician but then again even the...

Yes - in retrospect that should have been obvious. Even the highest resolution pic of our nearest galactic neighbour, Andromeda can only resolve stars as tiny yellow dots.

Generalized version of post #16: Construction of ##\sqrt{b}##

Alas - I will need to take responsibility for the above "solution" - the stated aim was to construct ##\sqrt{5}##. I don't necessarily get Gemini to find solutions - I just give it parameters for a diagram I want it to draw and it does that. Also on my instruction it drew and added in...

Well here's the construction of ##\sqrt{5}## anyway. Courtesy of Google Gemini.

Presumably you can always construct ##\sqrt{b}## by drawing a circle with diameter b+1 and placing a perpendicular at b to intersect the circle. Eg diameter 6 / radius 3 for ##\sqrt{5}##. The right triangle with h as an altitude has hypotenuse ##r=\frac{b+1}{2}## drawn from the centre and base...

Just a cellphone pic but managed to pick up 3 planets - Venus, Jupiter and Mercury. You might need to zoom in on the pic to see Mercury (or at least what I assume is Mercury based on Stellarium positioning) towards the top of the tree on the left.

Very tactfully put by Descartes! Nicolaus Copernicus in a header page of De Revolutionibus Orbium Coelestium writes somewhat more bluntly in Greek: "Let no one untrained in geometry enter here."