Re: Question on "air bearings"

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

Reply to
Jim Y
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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.

Reply to
Michael

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

Reply to
Jim Y

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