Re: Need simple equations for vibration motion

On 11 Jan 2006 09:55:57 -0800, "Robert A. Macy" <macy@xxxxxxxxxxxxxx>
wrote:

>Brian,
>
>Thank you again, but the material is only 170 pounds per cubic foot,
>not steel.
>But with only 2.8 psi, 5 mils is perfect. right?
>How do you go from 2.8psi causing 5 mil deflection to 1 Watt of sound
>will only deflect 0.5 mils, not my estimate of 5 mils?
>Perhaps, I should state that the sound power is not transferring from
>air to the diaphragm, but the diaphragm is being driven by a piezo
>transducer.
>2.8 psi across the diaphragm is a relatively small number, yes? I mean
>I've worked with poisonous gas cannisters that were 2,000 psi. Don't
>airless paint sprayers put out a whopping 400-800 psi, at least enough
>to inject into flesh? Water pressure at my home was 63 psi. So, 2.8
>psi is small.
>These answers are EXTREMELY helpful. So far I'm within a magnitude of
>accuracy, would like to get closer to more like 10% though. Greatly
>appreciate the help.
>
> - Robert -

This will probably be my last offering on the topic, but it should get
you close enough. But first, a word about the engineers who did NOT
respond. There are several people who monitor this group who would
avoid answering the question you posed, although they are well
equipped to do so This is the land of the ambulance-chaser.
It is the question of liability. In contrast I have been willing to
answer because this is not my field of expertise, so you are at your
own risk.
The question I asked myself was, if I pushed a diaphragm of the kind
you describe, how far would it deflect for a given force?
The reason for this question is because there is a particularly simple
relation between the product of force and displacement in the
direction of the force - this defines work done - with no fooling
around with scale factors if expressed in SI units.

Now it happens there is a rather cheap but effective code for getting
deflections etc for perforated and solid disks. It is called WinPlate.
It is sold by Archon Engineering, and they offer a 30 day free trial
- no strings - on a usable download. You could do that.
It needs the diameter, thickness, and edge fixing method and above all
the youngs modulus for the material - which essentially describes the
material's stiffness.
You did not give this, so I deduced it from the specific gravity that
you provided, namely 2.724 (which you expressed as pounds per cubic
foot.) This SG fits aluminum, glass, quartz etc. These materials
happen to have comparable stiffness of 71 MPa or 1E7 psi - a third
that of steel.

I could now centrally load this disk and see how the force and
deflection compared. I used a central circle of 0.2 inch diam to apply
a 19.1 lb force initially. I finally realised that if I gradually
increase the force from 0 to 19.1 lb the deflection increases from 0
to 0.005 inches. This takes only half the effort to displace a
surface with a constant force.

I mentioned previously that if you expend 1 watt at 43kHz, each cycle
takes 1/43000 of that power per second. So one cycle can expend
up to 1/43000 joule. This can be expended by displacing the surface
by a certain amount twice, with a certain force.
The displacement and force on the diaphragm scale pretty linearly down
from the large initial value I tried (Remember? 0.005 in using 19.1
lb)
When I scale these values to newtons and meters, I can deduce the
value for deflection and force provided by the energy I mentioned
above.
Or to put it another way 0.005 X 19.1 X scale factors to give
newtons X meters times a factor to fit this energy to the actual
energy available which is 1/43000 joule.

Without troubling you with the details, it turns out that a central
pressure of 3.8 psi on a circle of diameter 0.2 inches gives a central
force of 0.12 lb or 1.91 oz and this provides a deflection of 0.00023
inches.

Now the real world issues - If the energy is not dissipated by this
deflection, it is stored, and the deflection increases resonantly
until the surface DOES dissipate the cycle's worth of energy, each
cycle. So though the deflection figure is pretty respectable, it
could be much bigger if the coupling to the external fluid is low.

That's as far as I go.

I enjoyed addressing the question
Thanks

Brian Whatcott Altus OK
.

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