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

A company sent me a set of "air alert tire-valve caps" to try out and ideally write about. They're big and heavy, because they each contain a battery and a red LED that blinks if the tire pressure falls by four pounds. Very nice, but they weigh 0.35 ounce apiece, and I wouldn't dream of putting them on a fast car for fear of snapping the vale stem off at speed.

Can somebody good with numbers tell me what their effective weight would be at, say, 120 mph (a speed I'm sure a lot of us have done at one time or another...maybe more) on a 19-inch wheel? I want to be able to sound as though I know what I'm talking about when I tell the company that makes them that they're dangerous--which I think they are.

I think you'd be in the 300 RPM range or so at 120mph... Ran a quick calculation I think it's like 600 G's or some ridiculous thing.. in other words, it weighs a LOT.

((feetpersec * feetpersec) / 32.2) / radiusinfeet


feetpersec= 120mph*5280ft / 60min/60sec = 176

((176 * 176) / 32.2) / 1.58 = 608 G's...

Which means each valve cap effectively weighs 13.3 pounds at that speed, right?

Question for anybody, does that sound like weight that would do damage to a tire stem?

I don't think I'd put it on my rubber valve stems... :o

No I dont think that would be much of a problem. The tire monkeys have long levers to remove the stems. I think somewhere over 20 pounds and it never tears them no matter how old they are. Certainly this should be installed on shorty stems and not the long truck stems. Furthermore it would require a rebalance of the wheel. The steel stems would probably be best.

Porsche saw fit to put stem supports on even basic rubber valve stems,... I won't even run them on the racecar ... Seems risky to me.

608 G's! Wow.

.35 ounces X 608 G = 212.8 ounces = about 13.3 pounds.

Mind you, the tip of the valve stem is significantly inboard of the outer radius of the tire so the G- forces wouldn't be quite as high as indicated above (this was based on the rolling radius of the tire), but it is still probably on the order of 1 pound. If the weight acting on the stem is not perfectly centered, centrifugal force will tend to force the valve stem out toward the edge of the tire, possibly leading to valve stem failure where it attaches to the rim. However, if it's one of the bolted-in types, that one pound or so of force probably would be insignificant.

This is just a guess on my part, btw.
(edited for a stupid math error)

I am an EE so I need practice. Somebody check

weight-mass
0.35 ounce = 0.021874997 lbs mass = 0.000679898 slugs

velocity
120 mi/hr = 176 ft/sec

radius
19 in dia = 9.5 rad in = 0.791666667 ft


F= m v^2 / r 26.60277889 pounds

I did a real quick calc assuming an 18” rim with a 20” rolling diameter and came up a centrifugal force of about 22#’s each at 120mph.

It’s been awhile since I’ve done math so….

We have 3 answers now, is there a prize?

A 19 inch wheel would have an 9.5 inch radius right?

Lets ask Wayne. Those MSMEs are supposed to know this crap.

The problem I would see with this is every time you went 120 mph the stem would flex over against the rim and then return again when you slowed down. How many times would a valve stem do this?

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

Interestingly, I queried the company about this, and they told me hundreds of thousands of these units have been sold in Germany and they have never had a valve-stem failure even in Autobahn use.

I came up w/ an answer close to CS's


here are my assumptions
235/35x19 Michelin PS2 26" OD 799rev/mi @120mph
valve cap is 9" from the axis orf rotation

answer
~653g

The cap will weigh ~653x it's static weight

Just stick them on the racing steel valve stems, and you'll never need to worry about them....

I'm sure they'd be fine on racing steel valve stems, but how many ordinary consumers who will buy them at AutoZone even know what a racing steel valve stem is?

Ok, this bothered me so I used Bill's numbers in:

F = m w^2 r

and came up with 13.44 lbs, which is what CS said an hour ago. I like his method better. :D

If they were sold in Europe/Germany, don't they have to pass TUV testing?

TUV: yes, that's what I assumed, and I told the guy that if they were good enough for the TUV, they were good enough for me...

You know, wheel weights are typically in 0.25 oz increments. The caps weigh 40% more than that!

I can see it now. Joe Average buys the caps at Pep-Zone and installs them. Now his Ford Taurus has a shimmy at 80. He takes it in to the shop. They take the wheel off the car, take the cap off, check the air pressure, and spin the wheel. Hmmm.... it's in balance.

Eventually it dawns on someone (probably Joe) that the caps are the problem. After all, replacing shocks, ball joints and tie-rods didn't work. After 30 minutes, the tire jockey decides to humor Joe and balance the wheel with the caps on. Viola! The problem is resolved.

On the ride home, Joe looses one of the caps...

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

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.

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.

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.

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. :eek:

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.

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

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?

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

See my post, 19" wheel,

valve cap is 9" from the axis orf rotation

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!

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?

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.

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.

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

So I was close originally then...?

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

<pre> revs/sec radians/sec
26.63333333 167.3421687</pre>

And I thought this would be easy.

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

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.