Need somebody smart to do a calculation for me...

[RazorRacer]

The HP loss is the same at 10 mph as it is at 120 mph.

[Zeke]

There are AL threaded valve stems available. I think I saw them at Jeg's or Summit. By reducing the weight of the stem itself (I mean how much do those steelies weigh anyway?), one could assume that one could add back a little weight. Certainly something perched on the top of a stem at any appreciable angle, and given the forces you guys have come up with, the bending over of the valve stem has to be considered. The rigid AL stem would do that job.

The balancing guru that I listen to, Nate Jones of Nate Jones Tires, and balancer extraordinaire for Bonneville cars and bikes, explained this to me some time ago. He didn't run the calcs for me, just told me that a quarter once at speed weighs POUNDS.
His thesis has always been on truing tires before any balancing. As he explained, the weight of the encentric portion of the tire tends to exacerbate the problem as the tire grows even more out of round at spped. Of course, he started doing this 40 years ago when tires were not as round as they are today. Nevertheless, at 300 MPH, it has to be a very critical item on the car (or even bike nowadays).

These little tire pressure monitors need to be some how tapped into the side of a special valve stem enabling better placement, IMHO. They would be hidden, to some extent, as well.

[randywebb]

I'd just get the strap on things that mount to the whel circumerence inside the tires. They have radio transmitters that a receiver in the car picks up.

[masraum]

I have to tell you that there's no way I'd put something that big and heavy on my valvestem.

If I want a monitor I'm getting one of the in tire models.

[A Quiet Boom]

Interesting calculations, you guys are scaring me about my big lathe though. The 4-jaw check is 18" diameter and each jaw weighs about 10lbs. It spins as fast as 1800rpm. I don't even want to think what the dynamic weight of those jaws is.

Anyway, I'd recommend against the all threaded valve stems. Those threads create a stress point that would eventually snap under repeated loads listed above. In fact I'd recommend against steel stems period as they all have some threads exposed I believe. I ran them on a 4x4 until I snapped one out in the woods in a deep mud pit. Now if you ran those little stem supports with rubber stems and rebalanced the wheel I think it would be fine.

Oh an the calculations missed one thing, the stems aren't at the radius of the wheel, they can be as much as 3/4" closer to the center depending on the wheel.

[cstreit]

bringing the stems in 1" only reduces the force by 100 G's or so...

[A Quiet Boom]

Originally posted by BK911
I think we are close enough Rick. I used 18" as where the stem would actually be, for r = 9", and used 20" as the rolling diameter to determine W. I used a different formula though, I used:

F = m w^2 r

If I may ask what is "W"

[rick-l]

Originally posted by A Quiet Boom
If I may ask what is "W"

Small omega the angular velocity.

I think we used the same formula since w*r = velocity, w = v/r

What units did you use for the mass?

[oldE]

There is another factor involved.

All calculations so far apply to a spinning wheel which is not travelling along a surface. In effect, a point on the circumference of a wheel has no forward velocity while it is in the plane between the axle centre and the surface upon which the wheel is travelling. It is then accelerated to twice vehicle speed as it passes through the plane above the axle then decelerates to a "stop" again. The effect is reduced as you return to the centerline of the wheel.

I suppose this effect is an order of magnitude less than the rotational forces, but they could certainly add up at 150 mph, as the valve stem would accelerate to 300 mph and back to 0 several times/sec.

(Kind of makes you glad you weren't born a valve stem, doesn't it.)

Les

[Bill Verburg]

Oh an the calculations missed one thing

See my post, 19" wheel,

valve cap is 9" from the axis orf rotation

[cstreit]

Just realized that I screwed up and used the diameter rather than the radius!!! (My math teachers would be aghast.)

LETS TRY again using 9" radius :

((176 * 176) / 32.2) / 0.75 = 1282 G's... er....

Well anyone figure that out? Now I've confused myself. It's really the old F=ma calculation... Perhaps it's

((176^2) / 32.2) * 0.75 = 721 ?!?!

Ahh,.. Now that I've sufficiently confused the issue, Chris OUT!

[Zeke]

Originally posted by oldE
There is another factor involved.

All calculations so far apply to a spinning wheel which is not travelling along a surface. In effect, a point on the circumference of a wheel has no forward velocity while it is in the plane between the axle centre and the surface upon which the wheel is travelling. It is then accelerated to twice vehicle speed as it passes through the plane above the axle then decelerates to a "stop" again. The effect is reduced as you return to the centerline of the wheel.

I suppose this effect is an order of magnitude less than the rotational forces, but they could certainly add up at 150 mph, as the valve stem would accelerate to 300 mph and back to 0 several times/sec.

(Kind of makes you glad you weren't born a valve stem, doesn't it.)

Les

I'm far from an engineer and even further than that from being a mathematician (I guess that's why I'm not an engineer ), but somehow I don't think the valve stem knows about the road and therefore, the acceleration and deceleration you allude to. Now, there may be some of that effect when the vehicle is braking or accelaerating (such as wheel spin or the awesome acceleration of a drragster), but I question this at a constant speed. I'm not picking nits here, I'm learning by questioning. Comments?

[hcoles]

The load of an object on the valve stem attached to the valve stem is cyclic. Remember that the tire rotates about the ground not the axle.
E.g. the top of the tire is moving at 240mph... so there may be some "radius" adjustment needed in the calculations.

[rick-l]

Originally posted by cstreit
LETS TRY again using 9" radius :

((176 * 176) / 32.2) / 0.75 = 1282 G's... er....

Well anyone figure that out? Now I've confused myself. It's

You have V^2 * weight/accel / radius or V^2 * m / r which is what I came up with.

Your equation is normalized to 1 pound weight.

If it weighed .35 oz or had a mass of .00068 slugs it would have 28 pounds of force on it.

[rick-l]

I think BK911 has the correct answer. You have to get the angular velocity from the total diameter of the wheel and tire (DOH) and from that find the centripetal acceleration at the radius of the rim. The tangent of the rim is not going 120 mph, the tangent of the tire is. Oops

>> 26"od = 6.8' circumference
>> 9"r = .75'
>> 120mph = 162.4 rad/sec
>> .35ozs = .022 lbs
>>
>> F = m w^2 r / g
>>
>> = .022 0.75 162.4^2 / 32.2
>>
>> = 13.4 lbs

[cstreit]

So I was close originally then...?

[Bill Verburg]

angular velocity of 235/35x19 Michelin PS2 26" OD 799rev/mi @120mph is

 revs/sec       radians/sec

26.63333333	167.3421687

[Formerly Steve Wilkinson]

And I thought this would be easy.

[2.7RACER]

Where F is lbs
F=M(V*V)/R*32.2
M lbs
V ft/sec.
R radius in ft.

Problem: What is the G force or Weight of a 0.35 oz. Object
on a 18" wheel on a car going 120 mph.
Object is 9" from the center of the wheel
Tire:Michelin Pilot Sport 275/35 ZR18
Tire dia. 25.7" Revs per mile 807
Revs @ 120 mph 1614 revs/min

V @ 120 mph 9" radius on a 25.7" tire
Radius x 2 x PI x 1614
.75 ft x 2 x PI x 1614= 7605.7596 ft/min. 7605.7696/60= 126.7632 ft/sec
V= 126.7632
M= .35 oz.
M= 0.021875
R= .75 ft.
F= .021875 x (126.7632 x 126.7632)/.75 x 32.2
F= 351.5077/24.16= 14.55 lbs @ 120 mph
G= 14.55/.021875= 665.38 G's @ 120 mph

[jester911]

You guys are too funny. If anyone needs any problem analyzed to the point of distraction just post it here. All these engineers and there is really no one that can drive the train.