Recent content by Charles Link

I do think if the nail is a permanent magnet, that the tip will be of one polarity and the head of the other polarity. That would make it so that the tip gets centered on one region of the underlying magnet and the head on the other. The alternative is that the magnetism on the head is more...

I presume you are referring to any permanent magnetism, if there is any, appears to be minimal, but why not check for it, like @Baluncore suggested, with a compass.

From the response of the nail it is likely it is not permanently magnetized, if that is what you are referring to. It is responding how one might expect to induced magnetization. If the nail permanently had one pole on the top and the opposite pole on the tail, the response would be for each...

It is starting to make a little more sense, now that I have a clearer picture of it. I think the flat head is extensive enough that it is able to get magnetized in the x-y plane with a north pole and a south pole that will line up with the south and north poles respectively of the underlying...

You've got a very odd looking nail. I thought the head of the nail was the magnet. That is the head of the nail as I know it. You called it the flat end. Your nail has spiral grooves in it. I don't know if you can expect it to behave like an ordinary nail.

The pictures were unclear and the OP's explanations were lacking. I think that larger surface with 3 rings is the sensor, but I don't know.

Can you draw a diagram of where the poles are on this magnet? I had trouble following your explanations.

I want to repeat something from post 33 for those who might have missed it: See: @pasmith (post 11) basically has ## y_1'(x_1)=y_2'(x_2)=-(\frac{x_2-x_1}{y_2(x_2)-y_1(x_1)}) ##. where the slopes are set the same and set equal...

I think I basically did that solving the cubic by linear approximation to first order, where once I got a solution, I could get an improved approximation by taking the first solution and adding another delta to it, and solving it once again for the new delta to first order. This can be done...

Note: @Gavran 's method (post 15 above) needs a second power form to take the discriminant. @pasmith 's method works in the general case. The two methods are otherwise very similar. @Gavran sets both slopes the same and sets these to the slope of the normal line, just like @pasmith . @Gavran...

I did find from one article in a google that the expected speed at closest approach would be 3,139 m.p.h. I am lead to believe the website that showed no increase all day and a decrease to below 1,000 m.p.h. was inaccurate. Edit: I found a video on Facebook that may offer at least a partial...

Did the rocket speed up when moon's gravity became stronger than earth's? There was one website that a friend of mine found that showed speeds slowing down slightly all day from about 1200 m.p.h. to less than 1000 m.p.h. We kept looking for an increase up until about 5 pm, and saw none. The...

Yes, this way seems to get the result a little easier. I worked the ## y=4-x^2 ## and ## y=(2x-6)^2 ## problem by this method just now, and it does seem somewhat easier than other methods. i.e. Find the two points with same slopes that share the same normal line. @anuttarasammyak I see your...

Excellent. By @Gavran's method, I have ## x_2=\frac{a}{5}+\frac{12}{5} \approx 2.54 ##. :) (see also posts 24, 25, 29, and 30). Note: In the OP and the first several posts following it, we solved for the minimum distance between ## y=4-x^2 ## and ## y=(x-3)^2 ##. In these last couple of...

Please see the edits to my post 25. Thanks. <3 and I let Wolfram solve it: Three real roots. The smallest one is the one of interest and that is ## a \approx 0.71037 ##. I had no luck with Vieta's substitution on this one and looking at Wolfram's exact solutions, I can see why: They are...