I Does a rotating accelerometer moving linearly measure linear velocity?

[FGD]

If an accelerometer is rotating as it moves forward through the air, what force is acting on the accelerometer? Say we remove gravity and just concentrate on the horizontal axis.

I know there is centripetal and tangential acceleration that shows up as an offset in the sin wave, but what causes the sinusoidal pattern?

For example if the accelrometer was in place and rotating at a constant speed, all you would see is the centripetal force as an offset. Tangential acceleration would be non existent since the speed is constant.

As soon as it moves forward though, you see a sinusoidal pattern emerge.

The only thing I can think of is velocity of the accelerometer is somehow effecting the readings as it spins. Is this what is happening?

If so, is there math/physics to back this up? Like could you reverse engineer the values to get the velocity?

Say:
we remove the offset so that the centripetal acceleration is not a factor.
ω = ~70 rad/s
period = ~0.09 seconds
amplitude = ~20 m/s^2
we’ll just say phase is 0 to make things simpler.

What would be the math to figure this out? Any help would be very appreciated.

I’ve attached an image of the 2d plane if that helps. There is no acceleration when the accelerometer is 90 degrees to the velocity direction. And negative values from 90 to 270 degrees.

 

[FGD]

I may have said that wrong now that I look at it closer. Maybe the larger values are on the sides closer to the 90 degrees. I would still love help with this. Been thinking about it quite awhile now. :)

 

[Nugatory]

If an accelerometer is rotating as it moves forward through the air, what force is acting on the accelerometer?

Moving at a constant speed in a straight line?

 

[FGD]

Moving at a constant speed in a straight line?

Yes, in order to make is simple, I'm just looking at the constant speed in a straight line.
Accelerometer measures change in velocity, so there must be a way to determin the actual velocity from this. I just haven't been able to track down or figure the formula yet. Thinking about it more: from the radial/circular diagram above it looks like the biggest change in the y direction (the axis that aligns with velocity) is at the -90 and +90 positions, but then the accelerometer only acts along it's axis so that change in acceleration is probably close to 0 at those points. It's got me stumped at the moment. Any suggestions?

 

[berkeman]

What do you mean by "rotating an accelerometer"? If you just spin an accelerometer, it will measure zero acceleration (depending on the construction of the accelerometer). It needs to be off-center from the axis of rotation in order to sense the centripetal acceleration.

Think of one of the simplest accelerometers -- a mass on a spring. If you spin that accelerometer about the center of that mass, there is no force on the mass to generate a displacement, so the accelerometer reads zero...

 

[Nugatory]

Accelerometer measures change in velocity

Actually an accelerometer measures something called "proper acceleration", loosely the difference between the path an object follows and the straightline path it would follow if there were no forces acting on it.

Say we place the accelerometer at a point on the rim of the disk. The only force acting on that point is the centripetal force that is stopping that point from moving in a straight line tangent to the rim. That force has constant magnitude and is directed toward the center of the disk, so of course the accelerometer also shows constant acceleration towards the center.

The speed of the point on the rim in the direction of motion does change. That is explained by calculating the component of the centripetal force vector in the direction of motion (this is easier in cartesian coordinates than in the polar coordinates that we more often use for rotating objects). You will find that all along the trailing side of the disk that component is positive so the point is accelerating ("change of velocity" acceleration here) in the direction of motion; all along the leading side of the disk that component is negative so the point is decelerating; and of course the maximum and minimum speeds will be reached at the inflection points where the acceleration switches between positive and negative.

It may help to visualize wheel rolling without slipping: The contact point of the wheel is at rest relative to the ground, the top edge of the wheel is moving forward at twice the speed of the hub, and all it takes to make that happen is the centripetal force that holds the wheel together.

 

[Lnewqban]

If an accelerometer is rotating as it moves forward through the air, what force is acting on the accelerometer? Say we remove gravity and just concentrate on the horizontal axis.

Like in this case?:
The accelerometer is set to sense only horizontal accelerations of the geometric center of a flying Frisbee disc?

 

[FGD]

Like in this case?:
The accelerometer is set to sense only horizontal accelerations of the geometric center of a flying Frisbee disc?

Yes, that is a good analogy. The accelerometer is aligned with the radial axis. (AKA: aligned with the centripetal axis). Sorry if I am not explaining this clearly enough. Here is a chart with real world data. I took the data and removed the offset (Which I believed was the centripetal acceleration. (I don't know what else would cause a constant offset.)).

Then I fitted a sin wave to the data. The accelerometer is almost center, but as berkeman mentioned it is not perfectly center. Using the equation ac=ω^2⋅r, I get that R = 0.001103m (So the accelerometer is about 1mm from the center of rotation). In my new diagram I have the velocity vector pointing in the direction I believe Nugatory described where the negative values are aligned with the velocity vector. Once the Centripetal acceleration is removed isn't what is left related to linear velocity?

BTW, Thank you berkeman, Nugatory, and Lnewqban for taking the time to help me with this. I appreciate you!

 

[Lnewqban]

Since the values of accelerations seem to be perfectly symmetrical about the y axis, the radius is only 1.1 mm, and the rpm are only 700, I initially thought about partially blaming the cyclic readings to the accelerometer sensing device not being perfectly coplanar with the disc.

But, after a second, I realized that the readings are respect to the disc rather than to an inertial frame.

Then, I contradicted myself remembering that a small gyroscope fights back any attempt by an exterior torque to change its spatial plane of rotation.

 

[FGD]

Since the values of accelerations seem to be perfectly symmetrical about the y axis, the radius is only 1.1 mm, and the rpm are only 700, I initially thought about partially blaming the cyclic readings to the accelerometer sensing device not being perfectly coplanar with the disc.

But, after a second, I realized that the readings are respect to the disc rather than to an inertial frame.

Then, I contradicted myself remembering that a small gyroscope fights back any attempt by an exterior torque to change its spatial plane of rotation.

Love the brain storming! I actually attached a 3 axis accelerometer, but want to keep things simple and just think about the linear velocity. However, just for info sake, the z axis pointing down actualy has roughly twice as fast of a rotation. X and Y are similar since they are about 90 degress out of phase.

 

[FGD]

Actually an accelerometer measures something called "proper acceleration", loosely the difference between the path an object follows and the straightline path it would follow if there were no forces acting on it.

I visualize the accelerometer like a mass with a spring, like the diagram below. Is this correct? That is why I thought it might be possible to figure out the linear velocity. (Sorry the masses look like W's. I origionally drew it upside down. :) )

 

[Nugatory]

I actually attached a 3 axis accelerometer....

The three axes of your three-axis accelerometer are of course the cartesian x,y,z axes and it s reporting the components of the proper acceleration vector in cartesian coordinates. Calculate the components of the centripetal force vector and you will see that you are getting exactly the expected acceleration in the x and y directions. The z oscillation will not be present in the idealized system you described in the original post - likely you are seeing flutter or some other artifact of imperfect balancing of the disk.

 

[FGD]

The three axes of your three-axis accelerometer are of course the cartesian x,y,z axes and it s reporting the components of the proper acceleration vector in cartesian coordinates. Calculate the components of the centripetal force vector and you will see that you are getting exactly the expected acceleration in the x and y directions. The z oscillation will not be present in the idealized system you described in the original post - likely you are seeing flutter or some other artifact of imperfect balancing of the disk.

I am not totally sure the R value to calculate the centripetal force. I only derived it from the calculation based on thinking that the offset was the centripetal acceleration.

If I understand you correctly your saying the whole thing is centripetal acceleration and the linear velocity has no effect on the measurement? Is the answer to the question no you can not calculate linear velocity from a rotating accelerometer?

I took the magnitudes of the x and y and got the following plot. It looks like the z axis shows up in the magnitude. (This is the raw data, with the offsets removed).

 

[FGD]

Here is another sample for comparison. Different amplitude. ω = 62.8318530717959

 

[vela]

Love the brain storming! I actually attached a 3 axis accelerometer, but want to keep things simple and just think about the linear velocity. However, just for info sake, the z axis pointing down actualy has roughly twice as fast of a rotation. X and Y are similar since they are about 90 degress out of phase.

I wonder if the factor of two has to do with what Feynman mentioned about a wobbling plate in one of his books. Here's a video illustrating the phenomenon.

 

[Nugatory]

If I understand you correctly you’re saying the whole thing is centripetal acceleration and the linear velocity has no effect on the measurement?

Yes. Consider that the disk moving in some direction relative to us is the same situation as the disk at rest while we are moving in the opposite direction; surely the accelerometer readings must be the same in both cases. But if they are same in both cases they cannot be used to distinguish them.

(This is an example of Galilean relativity. If you are not familiar with that idea, Google will bring up some good references)

Is the answer to the question no you cannot calculate linear velocity from a rotating accelerometer?

In general, we cannot. There are some special cases: for example if we know that the accelerometer is attached to a wheel that is rolling without slipping then the rotation rate, which the accelerometer will give us, allows us to calculate the speed of the wheel relative to the road.

 

[Orodruin]

Is the answer to the question no you can not calculate linear velocity from a rotating accelerometer?

No, you cannot. This would violate Galilean relativity. There is no such thing as an absolute velocity and you should get the same result regardless of whether the accelerometer is moving relative to you or just spinning in place.

 

[Lnewqban]

... just for info sake, the z axis pointing down actually has roughly twice as fast of a rotation. X and Y are similar since they are about 90 degrees out of phase.
View attachment 349044

 

[A.T.]

Is the answer to the question no you can not calculate linear velocity from a rotating accelerometer?

If you know the initial linear velocity and orientation, and keep track of the linear acceleration (accelerometer) and angular velocity (gyroscope), then you can calculate linear velocity by integration. It's simple in theory, but tricky in practice, because small initial errors add up over time (drift).

https://en.wikipedia.org/wiki/Inertial_navigation_system

 

[FGD]

No, you cannot. This would violate Galilean relativity. There is no such thing as an absolute velocity and you should get the same result regardless of whether the accelerometer is moving relative to you or just spinning in place.

I took a day to learn about Galilean relativity. (Then got caught up in special relativity. lol, lost a day). Thank you all again for your insight. I looked for experements of acceleration that included both rotational and translational components. However, every one I found only showed either translational or rotational acceleration. Does anyone know of experements that show data for both translational and rotational acceleration at the same time? Just the fact that there is a sin wave that is produced says to me there is something more going on here. If you watch the following video, it looks like the rotational acceleration shows up only as an offset, with very little gravitational sin wave showing for the table not being perfecly orthoganal to gravity. It does not look like the data I collected that included both rotational and translational components. (I guess we need to move away from constant speed.)

 

[FGD]

Yes. Consider that the disk moving in some direction relative to us is the same situation as the disk at rest while we are moving in the opposite direction; surely the accelerometer readings must be the same in both cases. But if they are same in both cases they cannot be used to distinguish them.

I am trying to grasp this in my little brain. I took a day to study Galilean relativity as you suggested. The problem I am struggling with is the sin wave that is generated. What is causing this? In the video above you see the accelrometer reading the centripetal acceleration as a constant offset.

Here is my thought at the moment. (Just brain storming) Could this sin wave be due to the deceleration of the disc caused by drag? If you knew the coefficient of drag, could you figure the velocity?
(Sorry, I think my attempt so simplify with constant speed made things more complicated. :/)

 

[A.T.]

Here is my thought at the moment. (Just brain storming) Could this sin wave be due to the deceleration of the disc caused by drag? If you knew the drag, could you figure the velocity?

Are we talking about a thrown object, like a frisbee? The linear proper acceleration that an accelerometer at the center of mass measures, is solely due to the total aerodynamic force (lift and drag) acting on the flying object.

 

[FGD]

Are we talking about a thrown object, like a frisbee? The linear proper acceleration that an accelerometer at the center of mass measures, is solely due to the total aerodynamic force (lift and drag) acting on the flying object.

Yes like a frisbee. But the accelerometer is not exaclty in the center. It's offset more like this. The accelerometer could be placed anywhere on the radius circle. So X and Y are not neccessarily parallel to the radial of the plate/disc/frisbee. It is hard to know where the accelreometer is inside the manufactured casing exactly.

 

[Nugatory]

Does anyone know of experements that show data for both translational and rotational acceleration at the same time?

Very strongly suggest that you not take on the problem of translational acceleration until you've resolved the original problem you posed: constant translational velocity, therefore zero translational acceleration.

 

[FGD]

Very strongly suggest that you not take on the problem of translational acceleration until you've resolved the original problem you posed: constant translational velocity, therefore zero translational acceleration.

I believe the original was solved with the answer it is not possible because of Galilean relativity. Should I start a new thread for the translational acceleration problem?

 

[Nugatory]

Could this sin wave be due to the deceleration of the disc caused by drag? If you knew the coefficient of drag, could you figure the velocity?

The sine wave has nothing to do with drag, or anything else except the centripetal acceleration produced by the centripetal force. You may be confusing yourself by thinking that the centripetal acceleration is constant - it is not. Its magnitude is constant, but being a vector it has both magnitude and direction, and the direction is not constant.

The sine wave is just how all circular motion works. Consider an accelerometer moving counterclockwise in a circle of radius $R$ with angular velocity $\omega=\frac{v}{R}$. Using ordinary Cartesian coordinates with origin at the center of the circle (an arbitrary choice that makes the math easier) and starting our clock as the object crosses the positive $x$ axis (another arbitrary choice that simplifies the math) the $y$ position of the particle at time $t$ is $y(t)=\sin\omega{t}$. We differentiate twice to get the acceleration along the $y$ axis $\ddot{y}(t)=-\omega^2\sin\omega{t}$ and there is the sine wave that your accelerometer is reporting, caused by the centripetal force that is keeping the accelerometer on its circular trajectory. (We can of course do a similar calculation with the $x$ position to find another sine wave, this one ninety degrees out of phase).

Now consider the whole rotating disk moving in the positive $y$ direction with constant speed $V$.
We use the Galilean transformation to calculate the $y$ position, and we get the rather unsurprising result $y(t)=Vt+\sin\omega{t}$. Differentiate twice to get the acceleration in the $y$ direction and we get... the exact same $\ddot{y}(t)=-\omega^2\sin\omega{t}$ because the only acceleration involved is that produced by the centripetal force.

(Sorry, I think my attempt so simplify with constant speed made things more complicated. :/)

No, it is less complicated The variable speed problem is worked exactly as above but with the additional complication that $V$ is not a constant but instead some more or less messy function of $t$.

As I said in the previous post, you do not want to take on this more complicated problem until you clearly understand how the logic of the Galilean transform applies in the simpler constant velocity case.

 

[Nugatory]

I believe the original was solved with the answer it is not possible because of Galilean relativity. Should I start a new thread for the translational acceleration problem?

Sure, but Galilean relativity will still apply (the math just gets messier without providing any additional physical insights) so if you understand why the answer to the constant velocity case is what it is, you won't need the new thread - you'll already know the answer.

 

[Orodruin]

The sine wave has nothing to do with drag, or anything else except the centripetal acceleration produced by the centripetal force. You may be confusing yourself by thinking that the centripetal acceleration is constant - it is not. Its magnitude is constant, but being a vector it has both magnitude and direction, and the direction is not constant.

The sine wave is just how all circular motion works. Consider an accelerometer moving counterclockwise in a circle of radius $R$ with angular velocity $\omega=\frac{v}{R}$. Using ordinary Cartesian coordinates with origin at the center of the circle (an arbitrary choice that makes the math easier) and starting our clock as the object crosses the positive $x$ axis (another arbitrary choice that simplifies the math) the $y$ position of the particle at time $t$ is $y(t)=\sin\omega{t}$. We differentiate twice to get the acceleration along the $y$ axis $\ddot{y}(t)=-\sin\omega{t}$ and there is the sine wave that your accelerometer is reporting, caused by the centripetal force that is keeping the accelerometer on its circular trajectory. (We can of course do a similar calculation with the $x$ position to find another sine wave, this one ninety degrees out of phase).

Now consider the whole rotating disk moving in the positive $y$ direction with constant speed $V$.
We use the Galilean transformation to calculate the $y$ position, and we get the rather unsurprising result $y(t)=Vt+\sin\omega{t}$. Differentiate twice to get the acceleration in the $y$ direction and we get... the exact same $\ddot{y}(t)=-\sin\omega{t}$ because the only acceleration involved is that produced by the centripetal force.
No, it is less complicated The variable speed problem is worked exactly as above but with the additional complication that $V$ is not a constant but instead some more or less messy function of $t$.

As I said in the previous post, you do not want to take on this more complicated problem until you clearly understand how the logic of the Galilean transform applies in the simpler constant velocity case.

This explanation requires the accelerometer to also be a gyroscope such that its axes remain fixed. This is not the case for most accelerometers. Instead, they are usually fixed to an object (like a smart phone), meaning the reported axes co-rotate with the object. In the ideal case with the accelerometer performing perfect uniform circular motion, it would therefore report a constant acceleration in the radial direction.

 

[FGD]

This explanation requires the accelerometer to also be a gyroscope such that its axes remain fixed. This is not the case for most accelerometers. Instead, they are usually fixed to an object (like a smart phone), meaning the reported axes co-rotate with the object. In the ideal case with the accelerometer performing perfect uniform circular motion, it would therefore report a constant acceleration in the radial direction.

I plotted the 2 graphs as I undertand it. Here are the plots according to how I understood Nugatory's explanation.

And the plot according to how I understand Orodruin's explanation if the accelerometer is aligned with the radial axis.

Correct me if I am misunderstanding. (Learning lots! Love all the help!)

 

[Nugatory]

I plotted the 2 graphs as I undertand it.

You are understanding them right.
The plots from your posts #13 and #14 show the sinusoidal pattern, so we know you are using an accelerometer that includes the gyroscope @Orodruin describes in his post. Without that gyroscope the accelerometer would have no way of knowing that it is pointing in different directions at different times (rather like “down” on the other side of the earth from us is pointing in the direction that we call “up”) and you would have seen the straight-line plot of your second graph.

 

[Orodruin]

we know you are using an accelerometer that includes the gyroscope @Orodruin describes in his post

Yes and no. What I described was an accelerometer performing perfect uniform circular motion. This will generally not be the case if the accelerometer is attached to a rigid body with anisotropic moment of inertia. Freely rotating rigid bodies can do all kinds of interesting stuff. The Dzhanibekov effect is a great example of this:

 

[A.T.]

In the ideal case with the accelerometer performing perfect uniform circular motion, it would therefore report a constant acceleration in the radial direction.

Which might be the constant offset of the sinusoidal acceleration, while its variable part is due to the external aerodynamic force, which rotates in the body fixed system.

 

[FGD]

The plots from your posts #13 and #14 show the sinusoidal pattern, so we know you are using an accelerometer that includes the gyroscope @Orodruin describes in his post.

This is the accelerometer I am using. https://www.adafruit.com/product/4464. The accelerometer is fixed 3 axis attached to the plate. It has a separate gyroscope that has no effect on the accelerometer from what I see. So the sin wave has to be from another source correct? Also the simulation I ran earlier has acceleration values much higher than my measurement.

 

[FGD]

The Dzhanibekov effect is a great example

Way to make things more complicated. lol
It's a relief the earth won't flip on it's axis though. (Had me worried for a sec) :)

 

[FGD]

Which might be the constant offset of the sinusoidal acceleration, while its variable part is due to the external aerodynamic force, which rotates in the body fixed system.

That is what I was wondering earlier. If it measures deceleration then I believe you can determine instantaneous velocity. (Haven't seen info on this though, so maybe I'm missing something.) I love all the different ways of looking at this problem so far. Very interesting.

 

[Nugatory]

The accelerometer is fixed 3 axis attached to the plate. It has a separate gyroscope that has no effect on the accelerometer from what I see.

The separate gyroscope is what allows the accelerometer to calculate and report accelerations along fixed x and y axes, the ones that make sense to us watching this whirling device. Without it would be reporting constant acceleration in the direction of the centripetal force, which does not change relative to the orientation of the accelerometer.

 

[A.T.]

That is what I was wondering earlier. If it measures deceleration then I believe you can determine instantaneous velocity.

This won't be very accurate though. If your object is disk-like, you will have lift and drag. And part of the aerodynamic force will be along the z-axis (spin axis), where it is the constant component, while the double frequency sinusoidal variation is likely the wobble mentioned in post #15.

But at least you can get the current flight direction in local space, which might help you to correct drift errors, when using integration of the gyro data.

 

[A.T.]

The separate gyroscope is what allows the accelerometer to calculate and report accelerations along fixed x and y axes,

I doubt the device is actually doing this, but @FGD should check the documentation, just in case.

Without it would be reporting constant acceleration in the direction of the centripetal force, which does not change relative to the orientation of the accelerometer.

The data does have a constant offset and a sinusoidal variation on top of that.

 

[Nugatory]

I doubt the device is actually doing this, but @FGD should check the documentation, just in case.

These little (and ridiculously inexpensive) accelerometer devices are used in drones, an application which really cares about orientation in space and accelerations relative to the earth's surface so I wouldn't be that surprised. But yes, the documentation will be authoritative and I'm speculating.

 

[FGD]

This won't be very accurate though. If your object is disk-like, you will have lift and drag. And part of the aerodynamic force will be along the z-axis (spin axis), where it is the constant component, while the double frequency sinusoidal variation is likely the wobble mentioned in post #15.

But at least you can get the current flight direction in local space, which might help you to correct drift errors, when using integration of the gyro data.

Yeah, I will have to take into account lift, orientation, etc. But then it should be better than what kalman filters can do alone atm. Just need to make sure it is actually deceleration causing the sin wave.

 

[FGD]

These little (and ridiculously inexpensive) accelerometer devices are used in drones, an application which really cares about orientation in space and accelerations relative to the earth's surface so I wouldn't be that surprised. But yes, the documentation will be authoritative and I'm speculating.

Yeah, it is a fairly cheap acclerometer. The z gyro saturates past 4000dps. The x and y gyros actually give a fairly similar sin wave to the accelerometer data. Here is an image of the accelerometer and gyro data plotted together. So, after seeing this would you conclude the sin wave is from decelleration, wobble, or something else? I actually don't know why I am seeing the sin wave in the gyroscope.

 

[A.T.]

So, after seeing this would you conclude the sin wave is from decelleration, wobble, or something else? I actually don't know why I am seeing the sin wave in the gyroscope.
View attachment 349178

It could be from the wobble. If you look at the video below you notice that the angular momentum vector (which is parallel to the angular velocity vector) is fixed in the inertial frame, while the vertical body axis revolves around it. In the body frame it is the other way around and the angular velocity vector revolves around the vertical body axis, so its X and Y components are sinusoidal with quarter period phase offset.

It might seem weird that the XY-gyro frequency (related to the wobble) is the same as the XY-accel frequency (spin_frequency), and not twice as much like Z-accel (wobble_frequency). But this is because when you transform the wobble_frequency (= 2 * spin_frequency) from the inertial frame to the rotating body frame, you have to substract the frame rotation (spin_frequency), so the wobble_frequency in the body frame ends up the same as the spin_frequency.

The Z-accel doesn't transform like that. because it is linear acceleration along the frame spin axis, so it shows the same wobble_frequency, as in the inertial frame. If your object is disk-like and wobbling, the variation in the Z-accel is likely due to the changing angle of attack, which can lead to big changes of the aerodynamic force direction (lift/drag ratio), especially at small angles of attack.

How does the Z component of the gyro look like?

 

[FGD]

That makes sense. I am going to try and model this but need to make sense of the data. It looks like from the model in the video that the foot of the bear (logo) is pointing down on the right side and then pointing up on the right side so there must be a rotation around the axis. (Like the video says.) The orientation of the sensor according to the datasheet is as the following diagram shows.

I guess when the x axis (gyro) rotates it would effect the linear value of the y axis (accelerometer) which would be 90 degrees out of phase. And the Y (gyro) effects the X (accelerometer).
So I guess the linear accelerometer would just get a centripetal force acting on it from the rotation? But offset also somehow needs to be accounted for.....

 

[FGD]

Ok, so I've been working with the data and the wobble does not seem to contribute a lot to the accelerometer values. I used a fairly large value for the z offset from the center of rotation. (A value of 13mm). I take the gyro value (in rad/s) and calculate centripetal acceleration using this formula.

Where 'w' is the gyro reading and R is the 13mm. This gives these values for the centripetal acceleration. (Which is only 1.4 at the highest and 0.8 at the low end.) Am I missing something?

Ignore the units in the chart, thay are in m/s^2.
Here is the origional data plotted against the data with the centripetal acceleration from wobble removed. (Very little actual difference)

 

[A.T.]

Ok, so I've been working with the data and the wobble does not seem to contribute a lot to the accelerometer values.

The Z-accel-component variation, with the double frequency, was supposed to be due to the wobble. You are plotting only X & Y now. What is your point here?

I used a fairly large value for the z offset from the center of rotation. (A value of 13mm).

What is Z here? And how did you determine that value?

I take the gyro value (in rad/s) and calculate centripetal acceleration using this formula. View attachment 349808 Where 'w' is the gyro reading and R is the 13mm. This gives these values for the centripetal acceleration. (Which is only 1.4 at the highest and 0.8 at the low end.) Am I missing something?

If the XY-accel are in the body fixed frame, then centripetal acceleration is the constant offset, while the sinusoidal phase shifted variation is the external aerodynamic force.

Here is the origional data plotted against the data with the centripetal acceleration from wobble removed.

What is "centripetal acceleration from wobble"? Centripetal acceleration comes from the spin, while the wobble modulates the external aerodynamic force.

 

[FGD]

The Z-accel-component variation, with the double frequency, was supposed to be due to the wobble. You are plotting only X & Y now. What is your point here?

Yes, this is plotting the X, Y axis. All of the plots in this post are mostly from the same data so as to keep things consistant. As Z wobbles, so does the rotation around the X, Y axis. I am testing the conclusion that the sin wave in the accelerometer X, Y data is due to the wobble. In post #41 it shows the gyro and the accelerometer data. I'll repost so you can see the Z value for the accelerometer as well. (The Z gyro is saturated, so the value is useless and not shown in the plot. But the Z accelerometer is good and shown in green below.)

What is Z here? And how did you determine that value?

I measured it with a ruler. (The offset should be a little less than 13mm.)

If the XY-accel are in the body fixed frame, then centripetal acceleration is the constant offset.

Yes, I can easilly get the centripetal acceleration from the offset. (The offset has been removed from the data above.)

while the sinusoidal phase shifted variation is the external aerodynamic force.

This is exactly what I want to calculate. The aerodynamic force. Do you know how to do this from the data? I've been scouring the net. There is lots of info on stationary rotating objects but not much on accelerometer attached to a spinning flying object.

What is "centripetal acceleration from wobble"? Centripetal acceleration comes from the spin, while the wobble modulates the external aerodynamic force.

Wobble is a short spin that reverses isn't it?

Thanks again for your reply.
I appreciate you!

 

[A.T.]

Here is the origional data plotted against the data with the centripetal acceleration from wobble removed. (Very little actual difference)
View attachment 349812

Just to make sure I understand: You used the (small) XY-gyro values (from wobble) to calculate the centripetal XY-accel due to wobble only, and substracted that from the XY-accel data. That makes sense. And it also makes sense that it's just a minor difference, because the XY-gyro are small.

It's a pity that Z-gyro cannot be used, because that would be the best way to compute and remove all the centripetal linear acceleration, from spin and wobble. But you can remove the constant offset of the XY-accel so it is centered around zero, as you already did. But note that it is not really constant, as the spin slows down over time due to drag. So you should do it step-wise.

This is exactly what I want to calculate. The aerodynamic force. Do you know how to do this from the data?

Once you have removed all the linear acceleration from spinning and wobbling, all is left is the linear acceleration from the external aerodynamic force. Which is of course revolving in the rotating body frame.

But to transform this into the inertial frame you would need to know the orientation, for which you would have know the starting orientation and then integrate the gyro data, mainly the Z-gyro, which you don't have. But you might estimate the Z-angular velocity from the XY-accel, based on the centripetal offset you had to remove and/or the frequency of the XY-accel sinusoidal component.

 

[FGD]

Ok, thank you very much.
I think this has been put to bed. The concensus is that you can not measure linear velocity from a rotating accelerometer. Velocity needs to be integrated from a known state. (like rest).
I also thank all the ones who gave their insights.
I learned a lot in this thread. I hope it helps others who maybe had the same questions.
This forum has come a long way.
The mentors and advisors are great.
Thanks again!