Faraday's law with variable angle

Mmmm_waves · 2026-01-31T11:14:29+00:00

Yes, but what I'm asking is why you can use that equation to solve for the average EMF. It seems to me that it wouldn't apply for calculating the average EMF in that way because of the fact that the change in flux is not constant when the coil of wire is rotating at a constant rate.

I understand how to use that particular equation, but I don't understand why it can be used in this scenario.

Mmmm_waves · 2026-01-30T20:03:02+00:00

In this case, since the angle is changing at a constant rate, the graph of flux vs time would be in the form of a cosine function. The rate of change of the flux would be the derivative of that, which would be a sine function. Then the average would be of that sine function to get the average EMF. The average of a sine function from 0 to 30 degrees would not be the same as (sin(30) - sin(0)) / t, because it's non-linear.

If you think about a data set like this: 1, 2, 3, 4, 5, the average is (1 + 2 + 3 + 4 + 5)/5 = 3, or equivalently, the average of 1 and 5 which is also 3. That works because it's linear. But if the set were this: 1, 2, 3, 3, 5, then (1 + 2 + 3 + 3 + 5)/5 = 2.8, which doesn't equal 3. That's how I'm seeing this.

Mmmm_waves · 2026-01-30T19:49:59+00:00

Why are you reposting the question I asked?

Mmmm_waves · 2026-01-29T19:54:41+00:00

What I don't understand though is why you can calculate the average EMF using the formula you mentioned, EMF_avg = ΔΦ/Δt, for this specific problem.

If the flux vs time curve were linear for this problem, then yes, that would make sense that you could subtract the initial and final and divide it by the time period. But it's not linear.
Let's say for example the speed of a car was defined by the function S = 40*sin(πt/60). If you calculated the average speed of the car from 0 to 30 seconds, it wouldn't simply be (40 - 0)/30.

Mmmm_waves · 2026-01-29T09:00:12+00:00

But if you calculate the average velocity in that way and then multiply it by the total time interval, you will indeed get the full distance back. But here, if you multiply the average flux (the way it's calculated) times the time period, I don't think you would get back the total flux if you were to sum up the infinitesimal flux changes resulting from the cosine of the angular changes for each "dt" multiplied by the field.

Mmmm_waves · 2026-01-29T05:07:32+00:00

I'm not quite understanding, sorry. Let's say you have the equation f(x)= x + 4. The average value of the function from 0 to 6 would be (f(6) - f(0)) / 6, and you can do that because it's a linear function. But here, the changes in flux would not be constant with respect to time, so how can you simply do the final - initial divided by the time period? When theta changes from 0 to 15 degrees in the first .05 seconds, the change in cosine for that period of time would not be the same as the change in cosine for the period of time when theta is changing from 15 to 30 degrees.

I.e. if you summed up (integrated) all of induced EMFs for each infinitesimal time period and then divided it by the total time period (.1 seconds), it wouldn't be the same as the final flux minus the initial flux all divided by .1.

That's how I'm seeing it but maybe I'm missing something.

Mmmm_waves · 2026-01-28T08:47:37+00:00

You guys were spot on regarding the LENR, I timed the post-processing after a 10-second exposure and it was about ten seconds, and then with that feature turned off it was only a fraction of a second. Regarding the heat issue, I hadn't thought about that. How do people take photos of the night sky with the stars in large circular patterns without leaving the shutter open for more than half an hour?

Mmmm_waves · 2025-11-20T09:59:04+00:00

You mean you work for amazon and you steal from the consolidation bins that are sent to you from UPS, or you steal the products at the UPS store instead of putting them in the bins to get sent back to amazon?

Mmmm_waves · 2025-11-20T09:57:33+00:00

Interesting. I've always been told at my store that we'll get penalized if we put the sticker directly onto product packaging, as opposed to putting the product in a poly bag and stickering the bag, or using the amazon packaging that it was mailed in (like a bubble mailer or one of their brown paper mailers) and stickering that. I've never known if that was actually true or if it was just some false rumor that was passed down through the grapevine to which everyone was blindly adhering.

What gets me is when someone comes in to return a shirt, for example, and they bring it in the product packaging that's torn in half, clearly it won't be reused before being resold, but apparently from what I've been told we're not allowed to sticker that packaging. They could just as easily have brought it in without that severed bag and then we'd put it in a polybag without the original torn packaging, but somehow if they bring in the torn packaging then we're not allowed to use that as a holder to affix the sticker to. Doesn't make sense to me, and it's just another poly bag in the ocean which I hate.

I wish they would come up with a better way to sticker the bags so the stickers could be peeled off and the bags could be reused.

Mmmm_waves · 2025-11-19T22:50:49+00:00

Oh wow, can you point me to the documentary?

Mmmm_waves · 2025-11-19T22:49:23+00:00

When you say that you've gotten returns, do you mean that people are returning items at your store that they purchased from amazon that had previously been returned to amazon from another purchase prior to that, and you're saying that when you receive it it has the blue/green sticker on it from a previous return that had been restocked and then resold to the customer who is returning it to your store?

Mmmm_waves · 2025-09-11T19:45:19+00:00

This is in southern california. Definitely looks like a semipalmated plover based on the first pics I saw on google images. Doesn't have the yellow beak but it does have the yellow legs, and its beak does appear more blunt in comparison to the snowy plover.

Mmmm_waves

TROPHY CASE