The direction of the damper retaining bolt certainly makes a case for the R/H or L/H thread argument, as does the braking theory.

I don't follow you. Please explain. I am a little slow today.

David
:):):)

I wonder what the millions of Small Block Chev owners would think of the above crank damper pics given that most of the 'plain jane' SBC dont use a bolt or washer to retain the damper/timing sprocket on the crank snout, relying totally on the light press fit at assembly. Ford Australia did the same with most of the inline 70's 6cyl stuff as well, this in spite of the crank snout already being drilled/tapped for a retainer bolt.

I asked the Aussie V8 Supercar guys which thread direction they use on the centerlock nut/hub assy's- apparently they originally made both LHT/RHT nuts/hubs available, but most teams now use RHT only- the KISS principle at work. With all thread direction the same you dont need rattleguns/crew specific to each side of the car for pit stops.

As mentioned above, there isnt a problem until the nut comes loose- working out what is the cause of the nut coming loose appears to be the problem.

Incidently,I remember trying to degree a cam on a Boss 302 years ago while the entire valve train was in place, had removed the single bolt holding the CAMSHAFT sprocket while making the necessary adjustments and while turning the crank sheared the dowel in the cam nose simply for the lack of that bolt .

I think I may understand what you mean. Please correct me if I do not explain things clearly.

I can see where the harmonic balancer illustration could be a little confusing--but it is actually an excellent example of my point. Let me explain.

The harmonic balancer in the above pictures is of an Ford FE. The Ford FE has a clockwise rotation (as you look at the harmonic balancer from the front of the engine). Therefore, as the pistons push down on the crank to rotate the crank, the crank turns clockwise. The harmonic balancer is designed such that it RESISTS this clockwise motion--or dampens the clockwise motion of the crank. It does this by means of a rubber inertia ring (or a viscous fluid depending upon the style of dampener) that is between the hub of the dampener and the heavy outer ring of the dampener. The outer ring of the dampener can't instantaneously "catch up" with the hub/crank with each of the piston pulses because there is a "slip" motion from the rubber or the viscous fluid between the inner hub and the outer ring. The outer ring is heavy and has quite a bit of inertia so it doesn't "want" to catch up with the crank/hub--in other words, it lags behind. That is how it dampens the vibration in the engine. As it lags behind, it is effectively running counter-clockwise, or "unscrewing" the harmonic balancer bolt.

As you can see from the above the direction of the thread (right-hand threads on a Ford FE harmonic balancer bolt) didn't "work" like it was supposed to. In other words, once you loose the preload on the harmonic balancer bolt, hub wing nut, spinner, or whatever other bolt you have, you are doomed to failure.

Another way to think about it (without threads at all) is to think of a bearing that is pressed onto a shaft. There is considerable torque applied to the bearing (there is drag on the bearing) yet the bearing doesn't spin (hopefully). Why? Because it is "clamped" to the shaft by the interference fit of the race to the shaft. When you tighten a bolt, you create the same interference fit with the threads as the press fit does on a shaft.

I hope this explains it.

David
:):):)

True enough...

Unfortunately, the Ford FE doesn't have a press fit...as you can readily see.

David
:CRY::CRY::CRY:

Right hand thread on the damper retaining bolt, although the crankshaft spins in the same direction disputes the argument for R/H threads on drivers side (L/H on pass. side) knock offs, as does the brake-torque which is opposite of the drive-torque. Get my drift? It means your theory trumps the rest.

Thanks, just a little s l o w today! Spending WAY too much time on the billet chassis car!

David
:LOL::LOL::LOL:

David,
Thank you for your insight and explanation of clamping load. I totally agree that it is the clamping load applied by the threads that keeps the wheel on.

I agree with your statement, “If the nut is properly tightened, then it is absolutely IMPOSSIBLE for the wheel and the nut to move at different angular velocities”. But, what if they are not properly tightened? What if the knockoff becomes loose? Wouldn’t the “self-tightening” element of RHT or LHT play a part in keeping the knockoff on?

Now I'm out of salt......:p

As for "self-tightening" playing a part...

Would that be under braking or under acceleration that you would want to keep your nuts on? :)

See above harmonic balancer...

See Murphy's wheel...

See ancient Aztec civilization dude asking Senor Torture Chamber Inquisition Thug (all in the interest of saving his soul, you know) which screw is best for his...

Bottom line: The only thing that is going to help you if you find your nuts loose is Tide or a condom. But, at that point...it is probably to late for either one to help too much.

David
:):):)

I guess that's where the safety wire comes into play.:D

I stand corrected; I should add safety wire to the above list of "helping devices."

David
:):):)

Not having the technical background and expertise. might some one clarify:

I had a Austin Healey 100-6 and then a XK-E, both with factory spinners/knockoffs that were not factory safety wired. I do not recall any recommendations to safety wire.

A tool was placed over the knock-offs for use with the hammer rather than hitting the knockoff directly. Recommendation was to retighten after a few miles and the knockoffs were marked right and left.

My understanding, the knockoffs originated to facilitate speedy tire replacements for race cars prior to powered devices to remove the single or multiple nuts. When replaced, the knock-offs were not safety wired during the race,

With a very quick check of the Dayton Wheel site, does not appear to mention safety wires for their spinners/knockoffs.

What would be the difference in the knock off wheels mentioned on this Thread and the Austin Healey and Jaguar wheels ?

About 40+ years of lawyers, judges & juries.

David,

This is such an interesting subject I had to do some homework. I received this information from a computational aerodynamicist on one of the F1 teams.

"While the PEAK value of longitudinal acceleration on a Cobra is approximately the same as the PEAK value of its deceleration, maximum acceleration tails-off as speed increases, whereas maximum deceleration remains nearly constant. Add to this the fact that on an F-1 car, in particular, the peak acceleration is around 1.5G's, whereas the peak deceleration is close to 6.0G's, and it’s clear that deceleration generally is the significantly more severe condition. This conclusion applies to high-performance cars of every type and from every era."

"All other things being equal, one should choose the thread handedness such that inertial forces tighten the center-lock nuts under braking. The wheels on the right-hand (i.e., relative to the seated driver) side of the car rotate clockwise. Hence, under braking, the inertial loads on right-hand-side wing nuts are clockwise. For clockwise loads to tighten, the thread on the right-hand-side wing nuts must be right-handed. The same logic dictates left-hand threads on the left-hand-side wing-nuts."

It appears that traditional thinking has been 'reversed' all these years :3DSMILE:

Your XKE & Austin Healey wirewheels would have been of the 'Rudge' design and spline drive as opposed to pin drive.
In the Rudge design the 'knock off or nut' is a concave design where it contacts the wheel center and as such should the nut ever loosen slightly the car weight bears on the top part of the nut. The alloy Halibrands etc have a convex knockoff/nut and if these loosen the car weight bears on the lower part of the nut.
When the epicyclic effect takes place these two different designs tend to rotate the nut in opposite directions when viewed/fitted to the same wheel on a car. note-- this only happens when the 'nut' is in a loose state and this principle was founded back in the days of relatively narrow rims etc. With all the 'extra' factors we have now of better tyres, wider rims,better brakes, more power, huge load offsets, and lateral cornering loads etc-- these all have to be taken into account with each application .

I found a link where I think the author must have been sitting at the end of the tequila line as he wrote it. Perhaps a version of this (or perversion I should say) is where people got to thinking about wheel nuts spinning "faster" than the wheel. If your wheel nut is tight it is impossible they can move at different angular velocities. How on earth anyone could have ever thought (or posted such a thing) I can not begin to understand. Those who are not in the business are, of course, forgiven. It is all too easy to believe the "experts." (Most politicians immediately come to mind here.) Also, just ask Jamo if you should accept MY legal advise! :LOL:

Make sure you have a bottle of tequila beside you when you read this...I don't think there is any other way to understand it. It is good for laughs, if nothing else. You should NEVER believe hook, line, and sinker, ANYTHING that is written to the web by ANYONE...ME INCLUDED! (Come to think of it ESPECIALLY ME! :LOL: ) Seriously, you should run your OWN thought experiment on EVERYTHING ANYONE posts or advises you (religion comes to mind here and so do lawyers--no offense Jamo, or my other esteemed lawyer buddies out there).

Here it is: I think you better sit down in a chair that is not too easy to fall out of while reading this one.

http://www.mgaguru.com/mgtech/wheels/wl102.htm

Now you know why I never trust Wikipedia. My brother always tells me it is a "distillation of the entire world's stupidity."

A-Snake,

As I mentioned in my above post, "Remember, an F1 car can decelerate at some 4 g's and they certainly can't accelerate at 4 g's so for the "Jamo empty bottle argument" F1 should make make the threads REVERSE of what they are on our cars... Alas, they don't."

I guess F1 cars now decel at 6 g's! That is AMAZING to me!

As for the rotational inertia of the spinners...the rotational energy of a rotating wing nut at 2000 rpms is basically nothing in comparison to a 22 ton (probably MORE, but I haven't had time to run the numbers--see there, don't believe ME either!) clamp load exerted by the threads on a properly tightened wing nut. Rotational inertia is so small in comparison at that point to the clamp load it is simply lost in the noise of the system.

Remember CLAMP holds the wheel on...not the thread direction.

David
:):):)

Not to change the subject...but I have a little problem with your "expert" here. (Granted, I may misunderstand what you have written as well.)

I must take issue with his "whereas maximum deceleration remains nearly constant" comment.

NO it does NOT!

Drag INCREASES with the square of the velocity whether (Jamo :p ) you are driving a Cobra, an F1, or a Pinto!

seen NASA website:

http://www.grc.nasa.gov/WWW/K-12/airplane/drageq.html

Again, RUN THE THOUGHT EXPERIMENT IN YOUR OWN MIND...DO NOT TRUST ME TO POUR KNOWLEDGE IN YOUR HEAD. If you are driving down the road at 10 miles and hour and you lift off the gas (assume car in in neutral and you are on a flat road) you will barely be able to feel the car decelerate from wind resistance. (You will most likely only feel tire resistance and bearing/drive train resistance slowing you down). Now, drive down the road at 150 mph (legally of course on a race track :D ) and let off the gas again...notice the IMMEDIATE braking effect of the wind as your anti-submarine belt tightens up on your family jewels :eek:

Back to our F1 cars. The F1 car can only accelerate at 1.5 g's because that is all the grip the tires can achieve (assuming low speeds where aerodynamic drag and down force is not significant to screw up the numbers). Now, the tires have basically the same grip accelerating, decelerating, or cornering (that is known as the traction circle). Therefore, an F1 car can only brake at 1.5 g's before the tires say no more! No more? But wait, why can an F1 pilot brake at 4, 5, or even 6 g's? Aerodynamic drag! An F1 can NOT brake at 6 g's when he is going 20 mph--the tires simply don't have that much grip. But he can certainly brake at 6 g's (I am taking your word for it here on the 6 g part) when he is bombing down the straight at 200+ mph and looking at a concrete barrier coming up to say hello because aerodynamic drag is "helping" him slow down.

Back to nuts.

David
:):):)

(edit) ps. I have a clarification comment below. As I re-read this, I can see I didn't explain it very well.

Back to our F1 cars. The F1 car can only accelerate at 1.5 g's because that is all the grip the tires can achieve (assuming low speeds where aerodynamic drag and down force is not significant to screw up the numbers). Now, the tires have basically the same grip accelerating, decelerating, or cornering (that is known as the traction circle). Therefore, an F1 car can only brake at 1.5 g's before the tires say no more! No more? But wait, why can an F1 pilot brake at 4, 5, or even 6 g's? Aerodynamic drag! An F1 can NOT brake at 6 g's when he is going 20 mph--the tires simply don't have that much grip. But he can certainly brake at 6 g's (I am taking your word for it here on the 6 g part) when he is bombing down the straight at 200+ mph and looking at a concrete barrier coming up to say hello because aerodynamic drag is "helping" him slow down.

Back to nuts.

David
:):):)[/quote]

Your 'aerodynamic drag' is a factor at high speed, but more important is the aerodynamic downforce added to the cars static weight which increases the grip available from the tyres when operating at higher speeds. After all 'they' say that at around 200mph most modern single seaters develop enough downforce to enable them to be driven on an upside down surface ( and I dont mean any track downunder in S.A. Aussie, or NZ.):). To accomplish this there has to be around 2g + of generated downforce.

"While the PEAK value of longitudinal acceleration on a Cobra is approximately the same as the PEAK value of its deceleration, maximum acceleration tails-off as speed increases, whereas maximum deceleration remains nearly constant. Add to this the fact that on an F-1 car, in particular, the peak acceleration is around 1.5G's, whereas the peak deceleration is close to 6.0G's, and it’s clear that deceleration generally is the significantly more severe condition. This conclusion applies to high-performance cars of every type and from every era."

I interpret the above as follows;
The peak acceleration of 1.5G’s of an F1 would be achieved at a relatively low speed. It’s not going to achieve 1.5G’s when the car reaches 200MPH. Nothing to do with tires here. The maximum acceleration simply tails-off.
If he is braking hard from 200MPH to 40MPH wouldn’t the G’s remain much more constant then when the car was accelerating from 40 to 200MPH? :confused: