I imagine that engineers have to decide where an inertial measurement unit (or just an accelerometer, gyro, etc.) is placed in a rocket or spacecraft. I'd like to know how are these positions decided? Do they put them in zones of least vibrations for the smoothest signal, etc.? Particularly for sensor fusion, how would one decide on sensor location for the best attitude estimation of a rocket? Thanks for your input!
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2$\begingroup$ Short answer: it's a tradeoff. $\endgroup$– Organic MarbleNov 19, 2015 at 14:28
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$\begingroup$ @OrganicMarble Sorry but that's very vague. If you're knowledgeable in the field, at least point me to other sources where I can learn about what I am asking. $\endgroup$– space_voyagerNov 19, 2015 at 15:35
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3 Answers
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For spacecraft it's "where they fit and can be oriented properly". Harnessing constraints are somewhat considered (how do you get wires to it) and thermal constraints might be pretty important (ir telescopes need the detector to be very very cold so anything that produces heat is as far away as possible), but in general for satellites all motion is pretty slow so vibration isn't much of an issue. I can't talk about rocket design.
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$\begingroup$ For launch vehicles vibration can matter, The shuttle ground vibe tests ended up changing something about the RGAs. I can't remember if they added a filter or moved the packages (I think the former). It's mentioned on page 329 here without any details ocw.mit.edu/courses/aeronautics-and-astronautics/… $\endgroup$ Nov 20, 2015 at 0:23
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$\begingroup$ @OrganicMarble that seems like something that should be put as s comment to the main post or as an answer, not as a comment to my answer. $\endgroup$– SamNov 20, 2015 at 0:27
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See NASA Technical Note D-5869. This document includes a discussion about choosing the location of the navigation instruments of the Saturn V rocket. See the figures on pages 11 and 13 and nearby text (page 12).
Several considerations were mentioned, including the effects of bending modes, and other environmental factors of possible locations, length of cabling needed, and so on.
The bending mode shapes show the magnitude of the amplification of lateral motion. You would prefer not to place motion sensors at a position that bends (a lot) at a similar frequency to control motions of the rocket, as the bending could feed back into the control loop and lead to instability.
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It depends upon the mission. Ideally you try to put them as close to the center of mass as you can so that you don't have that offset between the IMU and the center of mass to worry about, but these days with everything modeled up to the millimeter in CAD, even that isn't that important any more. In practice, unless there is a mission driver, you put it where it makes sense for practical reasons, like to balance the mass distribution.
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$\begingroup$ @Dave Why would that offset matter? In principle, a rigid body has the same angular velocity, and hence position and acceleration, at every point. So in that sense mounting the accelerometer anywhere should give the same reading, no? Other than vibrations... $\endgroup$ Nov 19, 2015 at 21:37
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1$\begingroup$ @space_voyager, circular motion requires acceleration, whether it is an orbit or whether it is an off-axis point on a rigid, rotating body. Points farther from the axis must undergo more acceleration for the same angular velocity. $a_c = r \omega ^2$ $\endgroup$ Nov 19, 2015 at 22:42
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$\begingroup$ Sometimes you want large radius. STS had rate gyro packages on the SRBs. science.ksc.nasa.gov/shuttle/technology/sts-newsref/… $\endgroup$ Nov 20, 2015 at 0:17
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$\begingroup$ @space_voyager One example of where it matters is if you have a camera mounted on a platform, but your IMU is mounted somewhere else, say your camera is in the tail of an airplane, and the IMU is in the nose. The IMU will tell you the rotations it experiences; imagine rotations about the center of the IMU. If your plane is sitting in some X-Y-Z coordinate system, the IMU only rotated and hasn't changed it's coordinate position, but the camera at the tail of the plane has, and the amount of motion is has undergone goes as the length between them (the lever arm). $\endgroup$– DaveNov 20, 2015 at 1:00