Recent content by dbell5

A corollary to the above that I never thought of is that, even if the Earth were a perfect, homogeneous sphere, and ignoring the bulky clothes, I'd weigh nearly 5 pounds more at the poles than I do here at 37°N!

Huh! That would appear to be so! (Actually, 2G away, since the Earth's field is 1G inwards at the surface, vs. 0G inside a hollow shell.) OK, here's how I got there: a = v^2/r Circumference of Earth is about 41,000 km, or 41e6 meters Rotation is once in 86,400 seconds, for a tangential...

Funny that these same calculations result in a counter-proof to the hollow-Earthers (Koreshian cosmogony): If we did inhabit the inside surface of a hollow sphere the size of the Earth, the centripetal acceleration at the equator would be over 3G...

The *aapparent* artificial gravitational force has nothing to do with momentum of the objects inside thje station. It is actually centripetal force (what most people call centrifugal force) caused by the station's wall pushing up against your feet. If your motion was in a straight line (a...

Tanstaafl!

No, opening the switch of a flashlight adequately opens the circuit. Batteries lose charge due to internal leakage, as well as to external loads. Other than cold storage, there isn't anything you can do about that. As for increasing the chance of rupture and leakage, the only problem I see...

I was lazy and modeled it (B2SPICE). 0.415A, L to R. Dave

Assuming your segmented arms are rigidly attached - they don't rotate at the corner to the right - what you have is a new, shorter "virtual" lever arm, extending from the axis to the point where the weight is hung. You'd need to do the geometry to figure out its actual length, and what the...

... and I see where, now. Somehow, I kept reading µ0 as ~10^-17, not 10^-7. I'm going to blame my eyesight! I'll look for a considerably less sensitive device... Dave

I just re-read your note earlier: "iron has a relative permeability of about 200" I misread that as absolute permeability. 200 is a perfectly reasonable multiplier, but I'm still off somewhere, by a factor of 10^10 Dave

Getting back to this in the morning, I must be making some horrendous math error! Given \mu_0 = 4 \pi 10^-17, r = 0.01, h = 0.5, m = 42 I get B (in Teslas) = 3.36*10^-11, or 3.36*10^-7 Gauss. Also, the permeability of iron looks much smaller than 200...

Wow! Huge difference... 1 cm radius is probably close, and it has a solid (possibly soft iron, possibly ferrite) core. 2.7 Gs is pretty small, even with the most sensitive sensor of the family. But 3300 Gs is way up there! I'm thinking I will buy one on each end of the spectrum; they're...

Re-reading diazona's reply on the big screen (instead of iPhone), I see the formula. This related one was (also?) in the Wiki article: B(m, r, \lambda) = \frac{\mu_0}{4\pi} \frac{m}{r^3} \sqrt {1+3\sin^2\lambda} Since r is the distance from the center of the solenoid, this does make...

Yes, sensor would be very close, say within 1% of the solenoid's length. I think that's equivalent to half the field strength at the center of the coil. But is there any way to convert 42Am^2 to Gauss?

I have to spec a sensor to monitor the magnetic field of a DC solenoid element. The specification for the solenoid includes its length, DC resistance, current at the applied supply voltage, and most importantly, the expected Dipole Moment of the unit, as 42 Am^2. I plan to attach a solid...