kilotesla
kilotesla t1_ja0lh1v wrote
Yes, smaller loops of current do form. The current in a given loop is proportional to the rate of change of magnetic flux through it, and is inversely proportional to the resistance around the loop. With the slits as shown, the flux through each is much smaller than the total flux, and the resistance around the loop starts going up once the copper is divided into strips: The path length is roughly constant at double the length of a strip, and the conduction area gets smaller as it gets divided more. So the current per loop goes down as the square of the number of slits, and the effect become negligible.
kilotesla t1_j696fnt wrote
Reply to comment by agate_ in Why is wind energy generation greater during the night? by ZeroTheHero524
Thanks, that's a fantastic explanation of a really interesting phenomenon!
I wanted to see a graph of it and found this plot of wind speed by hour of the day at different heights above the ground, and it very clearly shows exactly what you're describing, with the neutral point between the two trends at about 150 m. Not all wind turbines are tall enough to be above that, but I'm not sure how representative that particular data is, and the height of a wind turbine tower is often augmented by putting it on a hill.
kilotesla t1_j4xb3jg wrote
Reply to Is it better to have warm or cool air for an external combustion engine? by Past-Loquat-4184
The answer for internal combustion engines is a little more nuanced. The simplest answer is that for a race car engine, where you want the maximum power output with the throttle wide open, you get that with cold air because the cold air has higher density (at the same pressure), and so you can get more air into the cylinder and can burn more fuel. And the efficiency goes up because the fixed losses are a smaller percentage of the overall power.
For regular internal combustion used for transportation, you rarely have the throttle all the way open and are often running with the throttle way back from wide open, creating substantial throttling losses. Lots of work on improved engine efficiency is finding ways to mitigate that problem. Variable valve timing is one example. But another is actually to deliberately heat the intake air. The lower air density will reduce the amount of mass flow of air into the cylinder, enabling scaling back the power towards the lower level that's actually needed with the throttle open more, or ideally even wide open.
So warm air can result in better efficiency given a specific engine size and a power output requirement well below its capability, because it effectively degrades the engine power without degrading its efficiency. However, if you are just operating the engine at its maximum efficiency point, with a wide open throttle, perhaps because you are using it for something other than driving a car on public roads, you'll get better efficiency with cold air.
A little more detail and links to papers with a lot more detail are on this Wikipedia page:
kilotesla t1_j4x8ruw wrote
Reply to comment by BobbyP27 in Is it better to have warm or cool air for an external combustion engine? by Past-Loquat-4184
This is specifically for the air supplying the combustion process. There's also a "cold side" of the heat engine, for example a steam condenser or a heat sink on a Stirling engine. If this is cooled by air, that air should be as cold as possible. (It might instead be cooled by water, in which case you want the water as cold as possible.)
kilotesla t1_j4gh09z wrote
Reply to comment by imeuru in Is there an upper limit on the size of a ship? by LilyFish-
You might also correct "almost 1/2 mile" to "almost 1/4 mile".
kilotesla t1_j4gguy4 wrote
Reply to comment by ColdFusion87 in Is there an upper limit on the size of a ship? by LilyFish-
Wikipedia has a list of the longest ships and the longest is an oil tanker that was 458.46 m (1,504 ft) long.
kilotesla t1_j46lu62 wrote
Reply to comment by die_kuestenwache in What does it mean to torque a screw to a specific torque? by alucemet
The third hazard of over tightening is stripping the threads, either the threads of the screw or the threads of the nut or threaded hole it is going into.
kilotesla t1_j46lmjd wrote
Reply to comment by abat6294 in What does it mean to torque a screw to a specific torque? by alucemet
>So if a screw spins with 5 in-lbs of torque, then you simply can't apply 7 in-lbs - it moves before 7 is reached.
Or if you did apply 7-in pounds, you would accelerate it rapidly and it would spend very fast, which isn't really possible with a manual tool although it could be possible with an electric tool, but spinning very fast would make it hard to avoid overshooting the final level of tightening that you want to achieve, and the whole thing becomes simpler when you go slow enough that inertial forces are negligible.
kilotesla t1_j3uolby wrote
Reply to comment by JamesTKierkegaard in Will water ice melt faster if allowed to drain, or remain in the meltwater? by terjeboe
>It's not glass compared to air that's the issue so much as compared to water or ice. The air is the environment in this situation so if the system is in a glass container the heat transfer to the air will be negligible compared to the surface area of the water and ice (which again depends on the shape of the container greatly).
In a series circuit with a very low resistance resistor, a medium value resistor, and a large resistor, fed by a voltage source, the voltage drop across the medium value resistor is affected a lot more by the large resistor than by the smallest resistor. If we have 1 ohm, 33 ohms, and 1000 ohms in series, the drop across the 33 ohm resistor is 3% of the source value, even if we drop the 1 ohm resistor to 0.1 ohms. We can't conclude that the 33 ohm resistor will have a lot of voltage drop because it is huge compared to 0.1 ohms. That doesn't work.
The glass outer surface will be very close to the same temperature as the water. The heat flow per unit area is determined by the temperature difference between the water and ambient. If the outer glass surface were at 1 C instead of zero, the temperature difference with respect to ambient would not change significantly. And the surface temperature wouldn't even be that high.
>Realistically, radiation is going to be a meager source of heat loss, even hot water radiators to heat houses only supply about 5% of their heat contribution through actual radiation, and that's at higher temperatures.
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5% is way too low. Modern "radiators" have fins which enhance convection but not radiation, so convention is typically larger, but radiation is still about 25%, even just counting the outward facing surface.
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In the range of temperatures we are talking about, radiation is reasonably approximated by a linear function of temperature difference. Yes I know, that's counterintuitive with that fourth power, but it's T1^4 - T2^4 , not (T1-T2)^4. On the other hand, natural convection is nonlinear enough that it drops as a fraction of overall heat transfer when the temperature difference gets smaller.
kilotesla t1_j3tzhd4 wrote
Reply to comment by januarytwentysecond in Will water ice melt faster if allowed to drain, or remain in the meltwater? by terjeboe
Those sound like valid results to me. One thing that I'd want to be a little bit careful of is the initial temperature of the bowl. If you had a bowl that started out at room temperature, and it was thick and heavy, it's thermal mass could contribute to the faster melting.
One other question is whether the ice was in the form of cubes, such that air could flow in between them in the strainer case, or whether it was perhaps frozen in the bowls so it was one solid hunk of ice
kilotesla t1_j3tz1rk wrote
Reply to comment by JamesTKierkegaard in Will water ice melt faster if allowed to drain, or remain in the meltwater? by terjeboe
That's good advice that the specifics matter. Since there will be heat transfer by radiation as well as convection, the emissivity of the surface of the container also matters, and if it's polished metal, the low emissivity would retard melting. Also, if the thermal mass of the container is significant, its starting temperature would matter.
When considered in comparison to air, glass is not a poor conductor of heat. It's conductivity is 33 times higher than that of air, and it's likely pretty thin, such that the heat transfer through the air will be the dominant thermal resistance. Air of course has the advantage that it is moving and so carrying heat by convection, not just conduction but it's still going to be the dominant thermal resistance.
kilotesla t1_j3txubj wrote
Reply to comment by bisnotyourarmy in Will water ice melt faster if allowed to drain, or remain in the meltwater? by terjeboe
That's possible but not certain. If you had a pile of ice cubes on a grate, such that water could drain away, and warm air could filter through accessing that full surface area, that might result in higher heat transfer than the same ice in a bucket full of melt water where the surface area accessible to the air is the surface of the bucket and the surface of the water, and is smaller than the total surface area of all of the cubes.
But if you froze the ice in the bucket to get a roughly cylindrical block of ice, and put that on the grate with the water draining away, the surface area would start out similar to that of the bucket, but would gradually decrease as the ice melted, whereas if you kept it in the bucket with the melt water, the surface area would stay constant.
Topics left as an exercise for the reader are consideration of heat transfer by radiation and the possibility of putting it in a bucket with a hole in it such that the water drains out. And then the challenge of how to fix a bucket with a hole in it without access to a working bucket, which might be important for getting water necessary to sharpen tools for carrying out the repair operation.
kilotesla t1_j3twpjl wrote
Reply to comment by vacri in Will water ice melt faster if allowed to drain, or remain in the meltwater? by terjeboe
I believe that the 33° was supposed to be fahrenheit, for about 0.5° C. Perhaps you knew that and were making a joke, but I'm giving you the benefit of the doubt.
kilotesla t1_j34bt89 wrote
Reply to comment by [deleted] in How is stereo information encoded into a vinyl record? by caedin8
Are you imagining the system that used L/R and up/down as the two channels being a rectangular groove? If it was still a V-groove, you wouldn't have the problem described in your first paragraph, but I agree that the system that is actually used is better, for the reason described in your second paragraph.
kilotesla t1_j334mm8 wrote
Reply to comment by Weed_O_Whirler in How fast would a body fall to earth if there was no atmosphere to stop it from accelerating past a terminal velocity? by straubzilla
In case people want to put some numbers to that and are too lazy to look them up, that's 11.2 km/s or 40,000 km/h. For americans, that's 25,000 mph.
kilotesla t1_j2xlq9l wrote
Reply to comment by hmartin430 in How close does one need to bring two coloured lights together to perceive a compound colour effect? by romxza
>low amplitude is only gonna excite the rods? High amplitude will allow the excitation of cones and at that point frequency will determine which cones are excited?
Yes, and yes, if that helps. Of course, the rods' response is not independent of frequency, but since there's only one type, you have no way to distinguish colors using them.
kilotesla t1_j2u6omm wrote
Reply to How close does one need to bring two coloured lights together to perceive a compound colour effect? by romxza
The underlying physics of this is really the underlying physiology. The different cells in the retina are sensitive to different colors of light, and when we see a color it is because of the comparison between the responses of those different cells in the retina. So a simple way to describe that is that the two sources of light have to be close enough together that they look like one point of light to us, and then the color we see is the color of that light. In other words, it's determined by the resolution of our vision.
The resolution of our vision is partly determined by the density of receptors in the retina, but it's also determined by how well we can focus. If you have red and green LEDs right next to each other, some meters away, and you can still see them as two separate leds you could make them merge into one yellow light by either taking off your glasses if you need glasses to see them, or putting on glasses that are wrong for you, to make the view blurry. Or you could simply walk further away from them until you only see one point of light, and that you will see it as yellow.
A common place to see this effect is on a large, but not very high resolution television screen. If you get up very close to it you can see the individual pixels, and for many display technologies, you will see them as individual red, green, and blue pixels, rather than as a full range of colors including yellow. If you don't have an appropriate display where you can see that, you can use a magnifying glass with a higher resolution display, even a phone.
kilotesla t1_j2mri32 wrote
Reply to comment by RandomThrowaway410 in Does cold temperature make vistas more 'crisp' looking? by colorado_hick
I put this in another comment as well, but to provide a direct answer to your request, here's a link to the EPA's guidebook on air quality impacts on visibility, written with an emphasis on views in national parks. https://www.epa.gov/sites/default/files/2016-07/documents/introvis.pdf
kilotesla t1_j2mg39m wrote
Reply to comment by uber_snotling in Does cold temperature make vistas more 'crisp' looking? by colorado_hick
The references both to colder air holding less water vapor, and relative humidity, might be confusing. The water vapor, when it stays in the form of water vapor, has minimal impacts on visibility. A larger impact, captured in the equations above, occurs when the water vapor combines with pollution in the air, specifically ammonium nitrate and ammonium sulfate in those equations. Those are hygroscopic salts that will become hydrated if the relative humidity is high enough, forming larger particulates that will scatter light more, especially if they start to clump together as they can do. Whether they become hydrated is controlled by the relative humidity, not the absolute humidity.
For more on this, see the EPA's guidebook on air quality impacts on visibility, written with an emphasis on views in national parks. https://www.epa.gov/sites/default/files/2016-07/documents/introvis.pdf Section 4 is the most directly relevant, particularly starting on p. 22 (p. 31 of the pdf).
The relative humidity in Colorado tends to be low in both summer and winter, and also varies greatly with time of day, so I'm not sure to what extent humidity explains what OP is seeing.
Edit: the monthly average RH in Limon CO (on the plains, east of the mountains) is a little bit higher in the winter than in the summer, but the swing in relative humidity over the course of a single day is often much larger than that seasonal variation.
kilotesla t1_j2mcqmg wrote
Reply to comment by LawOfSmallerNumbers in Does cold temperature make vistas more 'crisp' looking? by colorado_hick
It sounds like you're talking about a parameter that would describe the visibility of stars, etc., which you have to look through the whole atmosphere to see, whereas OP is looking horizontally, just through the lower atmosphere. The humidity would matter for both, but specifically counting the total in a vertical column would be less relevant for the horizontal view.
kilotesla t1_j1j4nf8 wrote
Reply to comment by man_o_brass in Can the Doppler effect make sounds inaudible by shifting frequencies out of human hearing range? Or vice versa? by IonnoFry
Adding to this, human hearing doesn't abruptly stop at a specific frequency, but becomes less and less sensitive as you go further to the extremes.
kilotesla t1_j1fxjsd wrote
Reply to comment by Phat_Potatoes in which is more efficient, using the sun to boil water and generate electricity from team or using solar panals? by Phat_Potatoes
There was a lot of interest and excitement about it a while ago, maybe 15 years ago. What made people generally turn away from it was simply that silicon solar cells and panels got less expensive really fast, so even though when solar thermal electric power projects looked like they'd be competitive, by the time they were finished, the silicon solar cells were so much cheaper that they were no longer competitive.
However, the need for storage is leaving to renewed interest: storing heat can be cheaper than storing electricity.
kilotesla t1_j1da7rf wrote
Reply to comment by samskiter in Why do we use phase change refrigerants? by samskiter
It could within some range. It would be best as a heat pump for heating from moderately warm temperatures up to high temperatures, perhaps 40° C up to 100° C, for example.
Note that that's also how a conventional heat pump works, with an HFC refrigerant. Let's say it's operating between 10° C and 40° C. The evaporator pressure will be set up so that the boiling point of the HFC is around 10° C, and the condenser pressure will be high enough to make the boiling point there 40° C.
kilotesla t1_j1aic57 wrote
Reply to comment by BigWiggly1 in Why do we use phase change refrigerants? by samskiter
To expand on this point:
> Water vapor is useless for space cooling, because it condenses at 100 C.
That's at atmospheric pressure. Part of the concept of how a refrigeration cycle works is to change the pressure so as to change the boiling point. So then the question becomes, why not lower the pressure an lower the boiling point, and use water as the working fluid for an air conditioner? A problem with that is the low pressure needed (circa 0.01 atmosphere) would mean the gas would be very low density, and you'd need to flow and compress a very large volume of it to move significant heat. (The other problem is that you couldn't go below 0 C without the water freezing.)
kilotesla t1_ja0lrto wrote
Reply to comment by Movpasd in Would the magnitude of the electric field at a point away from the center of a sphere depend on the radius of the sphere? by cozymikey
This is a good answer. Like the other answers, it assumes that the sphere of metal is far away from other charged or polarizable objects. If the two spheres being discussed were close to each other, the charge would no longer be uniformly distributed on either of them and the analysis isn't as simple.