Re: Question on "air bearings"
- From: "Jim Y" <j.s.yablonsky@xxxxxxxxxxxxxx>
- Date: Wed, 18 Jan 2006 18:57:53 GMT
"Jim Y" <j.s.yablonsky@xxxxxxxxxxxxxx> wrote in message
news:Bvuzf.271788$qk4.105628@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>
> "*" <nospam@xxxxxxxxxxxxx> wrote in message news:01c61c35$f99258c0$f9a4c3d8@xxxxxxx
>>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
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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