7 ms·
Quote: the metal casing helps spread that heat around It does, to a degree. But that is absolutely not what we are seeing in that IR picture. Shiny bare m
by kees99 7y ago
Quote:
the metal casing helps
spread that heat around
It does, to a degree. But that is absolutely not what we are seeing in that IR picture. Shiny bare metal is a mirror at the wavelengths that thermal camera uses. So what you see on thermal image where wifi-module shield and CPU are is reflection of ambient (room) where this picture was taken. Try waving your hand around, you'll see it reflecting there too.
To measure real metal surface temperature by IR, you have to paint that metal (ideally, with black matte paint), or apply similarly-textured sticker.
- gadgetoid 7y agoThere's a better image here- https://medium.com/@ghalfacree/benchmarking-the-raspberry-pi-4-73e5afbcd54b https://medium.com/@ghalfacree/benchmarking-the-raspberry-pi...
- nixgeek 7y agoHow was that accomplished? Extremely nice image!
- JorgeGT 7y agoMany IR cameras have actually two cameras, a normal optical one with high resolution (~10^3 x 10^3), and the IR one (microbolometer) which is low resolution (~10^2 x 10^2). They ensemble the two pictures later and produce these results.
- geerlingguy 7y agoI have the cheaper Seek imager, which just has a low res IR camera. Can’t afford the fancy tools where not needed! I mostly use it to check the house for air leaks and thermal issues.
- book-mind 7y agoThat image is still highly misleading, you can see from the overlayed optical image that the CPU is not coated. Ideally you'd want to coat any metal surface with black epoxy, and set the IR camera Emissivity Coefficient to 0.9 (the coefficient for black epoxy).
- delinka 7y agoWhy would the overlay need to be with the board that had a coating? Buy two boards, cover the parts on one board that are silver, shoot IR. Use second board for visible wavelength photo, combine for a beautiful overlay result.
- godelski 7y agoWhy would you assume that they are different boards?
- stan_rogers 7y agoIt's not an assumption, it's a suggestion for making an thermal/visible light overlay photo where the IR photo shows only emitted light.
- outworlder 7y agoWouldn't that change the heat dissipation characteristics and skew the results? Is a metal surface coated with epoxy exactly the same (from transmission, convection and radiation point of views) as a naked metal surface?
- steve76 7y agoCan't you just put it on a thumper? I dunno Atmos or something... ... ... Phonons and dat shiznit! - DAMNED ENGINEERS :)
- geerlingguy 7y agoTrue, I worded it a bit wrong. But the metal case is a huge improvement over the plastic die package on the Pi 3 B and earlier—a heat sink helps a little but you don't yet _need_ one if you use a fan, since the metal is much better at dispersing the heat.
- kees99 7y agoIndeed - there is enough aluminum/copper in the SoC's metal lid and PCB's inner (ground/power) layers, that the whole thing is effectively a 2-3 square-inch heat-sink. Eben himself has been talking about this technique around the time RPi3+ was released. And your point here is completely understandable too - there is a kinda-heatsink there already. Just add a fan, and we're good. My point is - seeing a large black hole where SoC goes (color corresponding to +20-ish degrees C by the chart) is utterly confusing, since SoC is the hottest point actually.
- deleted 7y ago[deleted]
- hinkley 7y agoTo what degree does scanning it in a cold room help?
- JorgeGT 7y agoYep, a lot of people nowadays is using IR cameras without any calibration or understanding of the physics behind the measurement. The truth is, the camera requires an emissivity coefficient to associate the received radiation with the emitter body temperature. There are tables for different types of materials, but the best is to calibrate against a surface thermocouple or similar. A corollary of this is, when you see an IR picture of a product with very different surfaces like rubber, shiny metal, matte paint, plastic, glass, etc. you know that almost for sure the measurement is unreliable because at most they calibrated the camera for the emissivity coefficient of one of the surfaces. And they vary quite a lot...
- DINKDINK 7y ago>people nowadays is using IR cameras without any calibration or understanding of the physics behind the measurement The four factors are Emissivity (ε), Absorptivity (α), reflectivity (ρ) and transmissivity (t). Emissivity is only one facet. Ideal is high Emissivity and low reflectivity. https://en.wikipedia.org/wiki/Radiation_properties https://en.wikipedia.org/wiki/Radiation_properties
- JorgeGT 7y agoCertainly, in most cameras the factors are in fact accounted for by setting the emissivity factor, a reflected apparent temperature, the distance to the object and the relative humidity. Thermography is certainly not an straightforward "point and click" process...
- CliffStoll 7y agoAs a once-infrared-astronomer, I can certainly confirm your statement: Thermography is certainly not a straightforward "point-and-click" process. Six years of grad school...
- sizzle 7y agoSounds like a cool profession, what are you up to nowadays?
- deleted 7y ago[deleted]
- burnte 7y agoHe's measuring the temp of the board next to the CPU, not the shiny IHS on the CPU.
- tagrun 7y agoPhysicist here, and I disagree. Where else could that radiation with ~10 micron wavelength at that intensity with that localized spatial profile from that particular direction be coming from? Yes environment typically will have some residual "noise" at those wavelengths, which you can check its intensity and spatial profile by taking a "dark frame" if you're in a strange environment and are really suspicious, but it's hardly going to alter what you're seeing in any qualitative way. Assuming someone isn't sending a focused beam of exactly that size at exactly that spot at an exactly correct angle at that particular wavelength.
- klickverbot 7y agoPhysicist here too. What exactly do you disagree with? The parent comment is sound – thermal imaging cameras typically under-read on shiny metal surfaces. Their emissivity/absorptivity at relevant wavelengths is low, and reflectivity is high. Thus, their own Planck spectrum is (approximately) scaled down by their emissivity, and consequently the radiation in the measured MIR band is mostly what is reflected, which tends to come from the room-temperature environment. A polished piece of metal makes a shitty black body. This is also why shiny metal (foil) is used to curb unwanted radiated heat transfer everywhere from thermos flasks and cryostats to space probes. (The lower emissivity further improves the efficiency of multi-layer insulation.)
- tagrun 7y agoLet me try again. Assistant Professor of Physics here (not a grad student). Yes, reflectance of room temperature aluminum at those wavelengths is pretty good (not true for all metals BTW). Yes, this usually makes it hard to distinguish thermal radiation and reflected radiation with metals. What are you trying to say though? That whatever comes off from a metal must always be a reflection coming from somewhere else? > Thus, their own Planck spectrum is (approximately) scaled down by their emissivity, and consequently the radiation in the measured MIR band is mostly what is reflected, which tends to come from the room-temperature environment. I don't know what you mean by "Planck spectrum is (approximately) scaled down" (as "Planck spectrum" only refers to thermal radiation and is generated in a separate process from reflected photons [one is governed by the conduction band whereas the other is governed by everything up to Fermi level] and you can't hope to suppress thermal radiation by simply shining random environmental light on a metal --there is no such thing as "scaling down" of thermal radiation unless you engineer such property), but there is just no way that 10 micron photons at that intensity could be coming from a room-temperature environment. So your blanket statements about metals aside, the hot area in that picture is due to a very specific signal which can't be due to something that's reflected from the environment. No significant fraction of those 10 micron photons coming off from that localized the area around the CPU could have originated from the environment --assuming that those pictures aren't taken in a hot oven and someone focused the thermal radiation on to the heatsink to get that amount of intensity. And as I mentioned, that's pretty trivial to test. If those 10 micron photons were coming from the environment as you or the parent comment suggest, the thermal camera would report ~60C even when you look at Pi 4 when it is cooled (again, this is something can use as "dark frame" and subtract off from all readings if you're trying to be more accurate). This is clearly not the case, though, as you can see in the video on the blog post.
- jeffalyanak 7y agoUnderstanding the reflectivity and emissivity of the material you are measuring is something that almost nobody accounts for outside of critical, professional applications. Not that it makes the measurements in question any less inaccurate, but it's very common to see these problems in temperature measurements.
- pankajdoharey 7y agoCheck the Ice Tower cooler for raspberry pi https://youtu.be/RyUXC3886Ic https://youtu.be/RyUXC3886Ic
- logicallee 7y agohow about touching a thermometer to it? (granted that is one spot, but I'd think it's most precise of all.)