Recent content by bob012345

That is until someone falls into a black hole and gets spaghettified.

At the time I couldn’t figure out how to make Latex to do ##\approx##. Now I do and I edited the above post. Thanks. Regarding ##d=\sqrt{3}r##, the vector from the origin to the center of the small sphere has x,y,z components each equal to ##r##. Regarding the graph, it just visualizes the...

Looking at a projection of the 3D case we can see how the ball is smaller for the same ##R##.

Here is the next part; In the case of 4D, ##d=\sqrt{4}r=2r## then ##r=\frac{R}{3}##. In the general case ##d=\sqrt{n}r## and the ratio $$\frac{r}{R}=\frac{1}{\sqrt{n}+1}$$ The general equation for hypersphere volume is $$V_n(R) = \frac{\pi^{n/2}}{\Gamma\!\left(\frac{n}{2} + 1\right)} \...

Here is the first part; Let the center to center distance be ##d## and ##r## the smaller circle radius and ##R## the larger circle radius so ##d+r=R##. We know ##d^2=r^2+r^2## or ##d=\sqrt{2}r##. Solving for ##r## in terms of ##R## we get $$r=\frac{R}{\sqrt{2}+1}$$ therefore the fraction that...

Here is an interesting geometry problem given for general interest. In the figure, compute the fraction that is shaded. Then extend that to a sphere snugly packed in an octant of a larger sphere. That means the smaller sphere is tangent to each plane and to the larger sphere in the most compact...

I don’t know if this is simpler but it can be done as the intersection of a line and a circle. Note, I’m using ##x## as the standard coordinate. Referring to @kuruman ’s excellent graphic, If the origin is at A and letting D be along the x axis, the point C is the intersection of the line...

Thanks @Gavran for pointing out those cases. In my formulation, I computed the largest petal and the next largest petal which accounts for the overlap of the largest petals. Those overlaps don't exist for ##n=2,3##. The angle of the circular segment making the smaller petal goes as...

I worked up the general case for integer ##n## overlapping circles. Here is the Desmos page; The red circle is the generator of the ##n## overlapping circles.

My answer was actually <0.4 at about 0.3974 so I am investigating my math… EDIT: I did the calculation another way and agree with @Gavran.

If anyone is interested, here is the case of five identical circles with a common point as in this diagram. Notice that there are now two levels of petals, the larger and the smaller. In this case determine the fraction of the figure with any petals to the whole figure.

This is a little exercise for fun. Determine the area of the darker petals to the total area. The four circles are identical. Have fun!

As @Filip Larsen suggests, 2 clockwise.

At the Earth’s poles, the answer is indeterminate. East and West do not exist.