Understanding thermal imaging fundamentals - Page 1
Forum section: electronics / general

author: WittyUsername
author_rank: Newbie
timestamp: December 30, 2018, 04:55:32 am
content: Hi team,I've been watching of a bunch of youtube videos and reading explanation websites for understanding how thermal imaging works (On a fairly basic level), however there are quite a few a few gaps missing that I would like to clarify.Q1: Does a particular wavelength correspond to a particular temperature? That is, do the wavelengths of 1um correspond to temperature of x° Kelvin and the other end of the spectrum having approx wavelength of 1000um, correspond to a temperature of y° Kelvin. This would mean temperature is analogous to the colour of visible light.Q2: If Q1 is no, is the temperature reading analogous to brightness of visible light? i.e. intensity.Q3: If Q2 is a yes, then why can thermal imaging cameras work when they seem to use only one section of the spectrum, i.e. using only the short wavelength IR, medium wavelength IR or long wavelength IR part of the spectrum. To get brightness of visible light for example, we must get information across all wavelengths of visible light.Q4: What governs your choice between getting a SWIR, MWIR or LWIR based camera? Is it the medium which you wish to look through? Again using an analogy to visible light, if you wanted to detect visible light penetration through water, you would consider detecting only blue or green wavelengths as they penetrate the farthest in the medium of water.If detecting visible light in a vacuum then you would just detect any type of light (Or whatever is the cheapest to detect) because they all penetrate equally?If these questions don't make sense then I'm obviously way out in understanding basics of thermal imaging.Thanks for any clarifications/corrections.I'll be sure to do more research.

author: Rerouter
author_rank: Super Contributor
timestamp: December 30, 2018, 05:20:15 am
content: Q1. its not in the sense your talking about, the distribution of wavelengths shifts towards shorter wavelengths as things get hotter, but its not like saying 404.2nm light is 485C, after all a photon is just a packet of energy that matched the gap between the electron charge states when it was released. As more energy is building up in the material, higher wavelengths begin to appear as the materials electrons are charge to higher states before they emit a photon.Q2. Its not intensity per say, rather the amount of energy captured in a given time frame, e.g. a photo of deep infrared contains less energy than a photon of blue light. so by capturing how much energy is presented instead of counting photons you can measure the amount of radiated energy, and thus temperature. (blackbody Radiation equations and corrections for different materials, its not a straight answer, and needs to be converted)If you take any common webcam, unscrew the lense and remove the little rectangular filter from the bottom, and replace it with something like developed film negative you can make yourself a Infrared intensity camera, the filter normally blocks infrared and ultraviolet light, with it replaced the film mostly only lets infrared through, for the most part it shows you what is reflecting or attenuating infrared, its also quite good at letting you see what plants are well watered, as they reflect the most IR light.To clarify this is not a flir style camera, just lets you see through an infrared light camera. you get no temperature, only intensity like your thinking. e.g. that bonfire is bright in infrared, but there is no easy way to tell its relative temperature to anything without atleast 2 different selective filters so you can try and map back to the blackbody equations. (e.g. long and short IR like the handheld thermometer guns work)Q3. To get brightness of visible light you don't really need to get the whole spectrum, for us humans you only need 3 peaks, and essentially count photons that have been limited by filters such as the bayer pattern, otherwise why would it stop at the visible, if you wanted to take a photograph filtered to 1 specific wavelength, it can be done, just be aware it looks a bit weirdQ4. This really comes down to what your intentions are, the longer the wavelength the more obscure, seeing as they have so much less energy per photon you need to catch many times more for a similar photo, generally resulting in much larger sensors to keep the noise down which increases there price, they are definitely used in some farming applications to spot plants that are sick or lacking water (leaves are effectively covered in steam on a hot day which makes them shiny in IR)

author: WittyUsername
author_rank: Newbie
timestamp: December 30, 2018, 10:13:17 am
content: Ok excellent, I think that has helped a lot.I've now done a bit of reading on the black body curve and emissivity.So, hopefully I've got this right. Instead of the RGB detectors present in every pixel of a visible light camera, a thermal imaging camera (Such as flir) has a SWIR and LWIR detector per pixel (And perhaps a MWIR too?). The amount of energy received in each of these detectors is then used in an equation to calculate the temperature of the body.I had not realised you need to select the material (To get the emissivity value) of which you are photographing with a thermal imaging camera in order to get an accurate temperature reading. I assumed it was a point and shoot situation and no user input was required.

author: Rerouter
author_rank: Super Contributor
timestamp: December 30, 2018, 10:32:32 am
content: off by a little again, they call these camera cores a "microbolometer array"Now this shoots of on a tangent, but what is the best way to measure the amount of power presented by an RF signal? using a power detector, and how did they measure it for signals too high frequency to capture any other way? measuring the resistance change of a resistor as the power from the RF signal heated it up.each cell you can think of as a resistor with a deliberately poor but linear temperature coefficient, a small change in temperature creates a decent change in resistance. there coated so that only the wavelengths they care about are absorbed and heat this resistance, reflecting anything higher and lower, to make a coating that reflects all visible and ultraviolet light while not reflecting long infrared and with a decently even attenuation is costly, so most units just hit the ranges that are easy, SW and MW,If you where keen enough you could likely rig up a proof of concept one using say 16 x 01005 PTC resistors under a webcam lens or a pinhole camera with a visible light filter. would just need to mount them on an insulating substance. and neurosurgery level soldering skills to wire it up.

author: Berni
author_rank: Super Contributor
timestamp: December 30, 2018, 10:40:37 am
content: Yes thermal imaging is fundamentally different than visible light imaging.The thermal imaging sensor is essentially measuring the energy of the radiation that is hitting its pixel. The temperatures we commonly tend to find in the environment around us tend to be in the range of -50 to 500°C. These things radiate light in the longer infrared wavelengths so the optics of a thermal imaging camera are designed to work for those wavelengths.Colder than that and the radiated energy gets hard to detect so you need a special camera optimized for that. Hotter than that and the radiated energy starts to creep into the visible spectrum so you can see it with your own eyes without the need for a fancy thermal camera.Thermal camera sensors in the current generation of cameras (Microbolometer) are actually arrays of tiny thermometers on tiny pixel shaped plates with a IR lens infront. This uses the principle that when two black body objects are allowed to radiate towards each other they start exchanging energy. If you hold your hands parallel about 1cm apart you can feel the heat coming off the other hand (This is radiation not convection), if you move your hand away it disappears. When your hands are near each other they are radiating IR photons towards each other, but because they are at the same temperature means that as much energy one hand radiates out it receives back in in the form of photons from the other hand. But when you remove one hand its still radiating IR photons out into the room but because the room is colder its getting less IR energy radiated back so it starts loosing energy to the environment and you feel that as cold. This is essentially how a microbolometer thermal imaging chip works. The temperature at the surface of the chip changes trough radiation according to what is infront of it. If its pointed at something hot more radiation comes in and warms the surface, if its pointed at something cold then the sensor is radiating more IR energy out of itself and cooling itself down. Trough calibration we can figure out what temperature on the sensors surface correlates to what temperature at the target. Final piece of the puzzle is the lens that focuses the environment in front of it on to individual pixels to get a sharp image rather than the average eviorment temperature.There are some other methods for doing thermal imaging but the one described above is what your modern FLIRs and those little phone based thermal cameras use.EDIT: Okay the hand example probably doesn't work in Australia that well right now given that its summer there. But it should work as long as there is a cold object behind one of those hands.