Re: Question on "air bearings"

J

Jim Y 20 years ago

I am modifying a manual tread mill to roll automotive tires.....

>
> I want to load these tires up to 1000 pounds.....maybe a bit more.... >
> Obviously, the nylon/teflon/whatever pad under the belt will withstand the
> momentary, moving footprint of an adult of sizeable weight, but I suspect a
> stationary tire footprint of 1000 pounds will be something much different. >
> Rather than get involved with some sort of messy lubrication scheme, I
> thought an air bearing just might do the trick.
>
> I can drill the support pad with the right-sized holes, and feed the right
> air pressure from my shop compressor.
>
> Question is, "How does one calculate the size of the holes and the air flow > needed?"
>
> I've seen automotive machinery that allow you to slide massive truck engine
> components around on a cushion of air with one hand, and I can pretty much
> pinpoint the tire footprint - which should allow me to concentrate on the
> affected area only.
>

I suggest you get a copy of Schaum's Outline Series, Engineering Mechanics, McLean and Nelson. In the Second Edition, Chapter 19, page 298, problem 24, describes a water jet. The formulae are there, just change the mass from that of water to air. Watch your units!

Jim Y

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M

Michael 20 years ago

Unless you're planning on a pretty short test, I have my doubts about trying to make a treadmill work--you're well in excess of the design load of the machine, and the belt friction is only one area of concern.

There are SAE papers available that publish a "standard" design for test machinery of this sort... a trip to a decent engineering library would be time well spent.

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J

Jim Y 20 years ago

McLean and Nelson. In

The formulae are

Here is the problem from the both the second and fourth edition, chapter 20 Impulse and Momentum, page 417:

See several example problems dealing with this situation in the chapter.

Values are based on Standard Temperature and Pressure (STP)

water = 62.428 lb.cu ft

gravity = 32.174 ft/s/s

diameter of stream of water = 2 inches

velocity of water = 80 ft/sec

delts t = time interval of particles of water (Note: delta t cancels out, but required in formula)

mass = (area of stream in sq ft) (velocity of stream in ft/s) ( density of water in slug/(cu ft)

mass = (.25 x PI x (2/12) x (2/12)) x (80 x (delta t)) x (62.428 / 32.174)

mass = 3.3865(delta t) slugs Note: delta t cancels out below

---------------

Sum of the forces = mass(v" - v')

P(delta t) = 3.3865(delta t) x ( 0 - 80) = -270.92 lb

P = -271 lb

Force of water on the plate is +271 lb to the right.

The density of air at standard conditions (STP) is 0.074 lb/cu. ft. (Marks Handbook)

Using the above with the density of air, the mass = 0.004(delta t) slugs and P =

0.32 lb.

And 1000 lb / .32 lb = 3125 nozzles. You may modify your velocity and nozzle diameter..

Jim Y

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Re: Question on "air bearings"

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