Recent content by bob012345

I see you just joined. Welcome to Physics Forums! One real benefit is that these kinds of problems help keep the mind sharp as one gets older. P.S. ##R_{(2,1)}##≈0.771

To answer part b, I used Foster’s Resistance Theorem. The average equivalent resistance is ##\frac{V-1}{E}## where V and E represent the number of vertices and edges which is 60 and 90 in this case. At any vertex there are three edges to average so we can write $$\frac{1}{3}(2R_{hex-pent}...

Thanks! There is a lot of interesting data in that paper. I liked the table of resistances between different vertices. You can get all the exact fractions with your simulation by multiplying the numerical answer by that factor 25080 and rounding to the nearest integer but that’s not the method I...

I started in a similar vein but a bit different. I drew the 2D projection of the TI modified to fit a flat x-y format. Then I drew it in LTspice also. It looks like this; The loops numbered 1-12 are the pentagons and the others are the hexagons. This gave me the numerical numbers that matched...

Here are my exact values. ##R_{hex-hex}=\frac{16778}{25080}## ##R_{hex-pent}=\frac{16273}{25080}##

Those numbers look numerically close but can you give me the exact values? How did you get these?

I worked on this problem during the World Cup. A truncated icosahedron (soccer ball) is made from 90 identical 1 Ω resistors, one on each edge. Part (a) Determine the exact equivalent resistance between two adjacent vertices across a hexagon–hexagon edge. Part (b)Without solving a second...

I saw something like that awhile back. I would see small voltages generated across thin film capacitors plates. Ultimately I noticed that when all the lab equipment was turned off the effect went away. I think things were charging from 60Hz noise from lab equipment. My dad used to light small...

Why do you think that?

You jest but since Lunar regolith is fine dust one could effectively use that. Use solar energy to move regolith to a high plateau, of which there are many, during the day and let it move down powering turbines during the Lunar night. Effectively a giant sand filled hourglass.

A mental model one can use is this, imagine you are standing on a smooth desert plain. It looks flat, very flat. But you are actually standing on a very large sphere, the Earth. The ground deviates from flat by about 8 cm per Km. If you walked a quarter way around the Earth, turned 90 degrees...

It reduces the launch mass enormously. The processing might be worked out and automated beforehand. Of course it probably wouldn’t be the first thing they build.

A good candidate for large Lunar batteries might be MIT professor Donald Sadoway’s dirt based liquid metal batteries because they could be manufactured on the Moon locally with majority lunar resources.

If ever there was a place for stable, consistent nuclear energy the Moon is it.

I believe batteries could work fine for Lunar bases as they work fine now for arctic bases. Obviously each situation has its own challenges but the article ignores basic strategies such as underground battery storage for thermal insulation and protection.