Does a rotating accelerometer moving linearly measure linear velocity?

  • #36
FGD said:
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.
 
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  • #37
FGD said:
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.
 
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  • #38
Nugatory said:
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.
Nugatory said:
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.
 
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  • #39
A.T. said:
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.
 
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  • #40
A.T. said:
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.
 
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  • #41
Nugatory said:
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.
1722379207833.png
 
  • #42
FGD said:
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?

 
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  • #43
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.
1722633557029.png

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.....
 
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  • #44
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.
1723498600665.png
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?
1723498678353.png

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)
1723499801401.png
 
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  • #45
FGD said:
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?

FGD said:
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?

FGD said:
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.

FGD said:
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.
 
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  • #46
A.T. said:
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.)
1723505698407.png

A.T. said:
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.)
A.T. said:
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.)
A.T. said:
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.
A.T. said:
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!
 
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  • #47
FGD said:
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.

FGD said:
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.
 
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  • #48
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!
 
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