Acceleration forces acting on an elliptical mirror

In summary: I divided the tangential acceleration component by 9.81 to get the number of g's. I thought a decent way to double-check might be to analyse this as a simple harmonic motion problem also, but I'm not too sure about that.
  • #1
Doc
47
4
Hi all,

I need to bond the back surface of an elliptical mirror (75mm major diameter and 37.5mm minor diameter) onto a voice-coil actuated mirror mount that is going to be operating at 500 Hz and has an adjustment range of +-1.5 degrees. I need to determine the g-force likely to be acting on the mirror to determine the amount of adhesive that I need to use.

To do this calculation I am examining a point on the outermost 'edge' of the mirror as shown in Fig 01 attached. I am assuming curvilinear motion in that the point is rotating around a central point (essentially a see-saw action). I am only interested in the range of motion from the mount's total range of motion, so from the very bottom of its travel to the very top: 3 degrees, as shown in Fig 02. The time period for one total oscillation is 2ms, but I'm only interested in half of that as shown in the sketch, 1ms.

From here I simply calculate the angular acceleration, and then the tangential component of that, see Fig 03. I divide the tangential acceleration component by 9.81 to get the number of g's.

I'd just like some comments about whether this is the correct approach or not. It's been quite a while since I've had to do anything theoretical.

I thought a decent way to double-check might be to analyse this as a simple harmonic motion problem also, but I'm not too sure about that.

Thanks,
Doc
 

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  • #2
The maximum acceleration of the mirror is ##a_{max}=(2\pi f)^2 A## where ##f=500 Hz## and ##A## is the amplitude of oscillation in meters. You can divide that by 9.81 m/s2 to get the number of g's. It is not clear to me whether this mirror moves back and forth in linear motion with amplitude ##A## or rocks in angular motion about its center with angular amplitude 3o.
 
  • #3
Hi kuruman,

It is not clear to me whether this mirror moves back and forth in linear motion with amplitude A A or rocks in angular motion about its center with angular amplitude 3o.

It will likely do both, whichever calculation yields the larger g-force is the value I will use to specify the glue amount.

With respect to moving back and forth in linear motion with amplitude A (2mm) I get something like 1970g: I'm not sure that that seems reasonable?
 
  • #4
Doc said:
With respect to moving back and forth in linear motion with amplitude A (2mm) I get something like 1970g: I'm not sure that that seems reasonable?
Why not? The period is ##T=\frac{1}{500}= 2 ms##. A quarter-period is ##t=0.5 ms##. The maximum speed is ##v_{max}=\omega A=2 \pi f A=6.3~m/s =14 ~mph##. What kind of acceleration do you think is reasonable for a mass to go from 0 to 14 mph in half a millisecond? Assuming that the acceleration is constant (which it isn't in this case), we can find the average acceleration ##a_{avg.} =\frac{v_{max}}{T/4}=12600~m/s^2=1280g.## That's an average and should be smaller than the maximum, however it is of the same order of magnitude.
 
  • #5
kuruman said:
Why not? The period is ##T=\frac{1}{500}= 2 ms##. A quarter-period is ##t=0.5 ms##. The maximum speed is ##v_{max}=\omega A=2 \pi f A=6.3~m/s =14 ~mph##. What kind of acceleration do you think is reasonable for a mass to go from 0 to 14 mph in half a millisecond? Assuming that the acceleration is constant (which it isn't in this case), we can find the average acceleration ##a_{avg.} =\frac{v_{max}}{T/4}=12600~m/s^2=1280g.## That's an average and should be smaller than the maximum, however it is of the same order of magnitude.

What you say does sound reasonable. I've done the calculations on the amount of glue that I need and it seems reasonable given the large accelerations.

Any insight on why there is a factor of ten difference between the calculations?
 
  • #6
Doc said:
Any insight on why there is a factor of ten difference between the calculations?
Where is the factor of ten difference? In post #3 you got 1970g for the maximum acceleration. In post #4 I got 1280g with my calculation for the average acceleration which I did crudely to convince you that the numbers are reasonable. The crude average is 65% of the maximum. The exact calculation of the average (rms value) yields ##a_{avg.}=\frac{a_{max}}{\sqrt{2}}=71##% of the maximum.
 
  • #7
kuruman said:
Where is the factor of ten difference? In post #3 you got 1970g for the maximum acceleration. In post #4 I got 1280g with my calculation for the average acceleration which I did crudely to convince you that the numbers are reasonable. The crude average is 65% of the maximum. The exact calculation of the average (rms value) yields ##a_{avg.}=\frac{a_{max}}{\sqrt{2}}=71##% of the maximum.

I attached photographed hand-written calculations in post #1. I calculated the angular acceleration and converted to a tangential acceleration for a point on the edge of the mirror's major diameter. This calculation yielded about 200 g's.
 
  • #8
OK, let's go back to basics. If there is angular acceleration, there must be a torque acting on the mirror. To have a torque, you need a force applied at some non-zero lever arm from the axis of rotation. What is the force in this case and where is it applied? It seems to me that the external force is applied at the center of mass of the mirror and the torque about the center of mass that might cause rocking is zero. Another way to say this is, what reason does the mirror have to rock "right to left" as opposed to "front and back", or in between? There must be some asymmetry in the driving force with respect to the center of the mirror in order to introduce rocking motion and angular acceleration. I think what you have to worry about is the maximum linear acceleration and whether the mirror will peel off if it reaches maximum distance from the coil and is about to move backwards.
 
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  • #9
kuruman said:
the maximum linear acceleration and whether the mirror will peel off if it reaches maximum distance from the coil and is about to move backwards.
That works for a drive with Square, Triangular, Sawtooth waveforms. For a Sine wave the maximum acceleration is at the zero crossing.
What is the driving waveform?
 
  • #10
Tom.G said:
For a Sine wave the maximum acceleration is at the zero crossing.
Zero crossing of the displacement, or the velocity curve...
That should be specified, otherwise we will have some confusion arising.
 
  • #11
kuruman said:
I think what you have to worry about is the maximum linear acceleration and whether the mirror will peel off if it reaches maximum distance from the coil and is about to move backwards.
Tom.G said:
That works for a drive with Square, Triangular, Sawtooth waveforms. For a Sine wave the maximum acceleration is at the zero crossing.
What is the driving waveform?
Are you saying that for a sine wave the (magnitude of) maximum acceleration does not occur at maximum displacement? If the displacement is described by a sine, so is its second derivative, the acceleration. Am I missing something?
 
  • #12
kuruman said:
If the displacement is described by a sine, so is its second derivative, the acceleration.
Uhmm, First derivative?

Sounds like we are talking about two slightly different conditions.
I was referencing to the displacement of the mirror.
You may have been referencing to the driving voltage.

If frequency of the driving voltage is below the mechanical resonance, the drive and displacement are nearly in phase.
If frequency of the driving voltage is above the mechanical resonance, there will be a significant phase shift between drive and displacement.

Cheers,
Tom
 
  • #13
We may be talking about two different things. Here is what I am talking about:
Whatever the driving mechanism of the mirror may be, if it undergoes harmonic motion
a. The displacement of the mirror is described by ##x(t)=A \sin(\omega t +\varphi)##
b. The velocity of the mirror is described by ##v(t)=\frac{dx}{dt}=\omega A \cos(\omega t +\varphi)##
c. The acceleration of the mirror is described by ##a(t)=\frac{d^2x}{dt^2}=\frac{dv}{dt}=-\omega^2 A \sin(\omega t +\varphi)##
Thus, ##a(t)=-\omega^2 x(t)##. Therefore, the magnitude of the acceleration is maximum when the magnitude of the displacement is maximum.
 
  • #14
You are correct for a mass & spring harmonic oscillator.

However we are dealing with a driven mass here, or a forced oscillator... which I believe changes the analysis.

Anyone else care to jump in and help clear our confusion here?
 
  • #15

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Related to Acceleration forces acting on an elliptical mirror

1. What is an elliptical mirror?

An elliptical mirror is a type of curved mirror that has an elliptical shape. It is commonly used in optics and can be found in telescopes, microscopes, and other optical devices.

2. How do acceleration forces affect an elliptical mirror?

Acceleration forces can cause an elliptical mirror to deform and change its shape. This can lead to distortions in the reflected image and can affect the performance of the mirror.

3. What is the significance of acceleration forces on an elliptical mirror?

Acceleration forces can impact the accuracy and precision of the reflected image from an elliptical mirror. They can also affect the stability and durability of the mirror over time.

4. How are acceleration forces measured on an elliptical mirror?

Acceleration forces on an elliptical mirror can be measured using specialized equipment, such as force gauges or accelerometers. These devices can detect and quantify the amount of force acting on the mirror.

5. What are some common techniques for minimizing acceleration forces on an elliptical mirror?

Some techniques for reducing acceleration forces on an elliptical mirror include using materials with high stiffness and strength, optimizing the design and shape of the mirror, and using supporting structures to distribute the forces evenly.

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