Tuesday, February 24, 2009

Quick Release Force

Have you ever wondered about  quick-release clamp force? (that which holds your wheel in the frame) ?   Way back, this came up in a forum, and it was mentioned that Bicycling Magazine actually tested/measured it! (imagine the time wasted!) At the time,  I commented that, that was pretty silly because it is easily calculated. Which I didn't bother to do until now and yes it is easy calculation!   

From BLOG Pictures

It is a statically indeterminate calculation, meaning it is not solvable with force equations alone, deflection/stiffness of components  is also considered.  Since Metal (steel ti or alum in most cases for bicycles), under elastic conditions, behaves linearly (simply meaning that if the force doubles, so does the deflection, and so on), this is not an issue.  The components can be thought of as Springs, where Force=SpringConstant*Deflection  or  (F=k*x).

In the case of a bicycle hub, the main components are the quick release rod elongating, and the hub's axles, compressing.  Also involved are the nuts, and frame dropouts. 

For items under axial loading the equation to calculate the spring rate is ElasticModulus*CrossSectArea / Length    or   (k=EA/L).

The amount of deflection introduced to the system is the "throw" of the quick release lever, which is typically reccomended to be 90 degrees (so we will use that).  With most quick releases this equates to 1 mm of deflection.   

In the table below (done with MS Excel) the measurements came from a typical shimano rear hub setup.  Everything is assumed to be steel (thus E=30e6psi).   Most of the measurement are not exact and for these purposes really don't need to be, with maybe exception of the quick release rod.  Assumptions also have to be made as to where in the threads does the force path change from nut to bolt. (good assumption is 3 threads in...)
ItemModulusLengthOuter DiaInner DiaOD of clamped volX-Sect AreaSpring Rateamount of deflection
ELdodid3AAk (EA/L)
Axle Nut30E638179.5156.10.2421910150.00091
Axle Nut30E638179.5156.10.2421910150.00091
QR Nut30E66154164.10.25412721510.00014
QR Nut30E65160201.10.31218698800.00009
QR Rod30E61565019.60.03058530.02958
Overall Spring Rate [lbf/in]4398
amount of throw [in]0.039
Amount of Clamping Force: [lbf]173
Amount of Force if consider only QR rod:  [lbf]230

Only 170 lb !  Not much.  Mostly dependent on the Quick Release rod, as expected since it is considerably more flexible.  Other notes:
-Compare to a clamp force of bolting: ~8000 lbs vs 170 lbs.  Thus the need for locknuts that "bite" into and deform dropouts.  Thus  vertical dropouts, Lawyer Lips, disc brakes causing pullout, etc. 
-If you going to use a Ti-rod, be wary.  With a modulus of about half of steel, you will only get 1/2 the clamp force (<100lbf!)>
-The calculation is a lot easier and can be almost as accurate if you consider only the qr rod.  It =   E*A/L*AmtThrow.  If the axle is titanium or aluminum with small x-sect area then you will want to include that as well.  
-Obviously if you are able to apply more Throw/deflection into the quick release, the amount of force goes up linearly.  Ie in this particular case, 2mm would give 340 lbf.  
-If "things" plastically deform during the clamping, then all bets are off on clamp force. 
-How about quick releases with plastic (actual plastic) in the force path ??   Sucks to be you, Good luck !!    No actually the plastic is so thin, ie the "L" is small thus the spring rate is still high. But yeah, personally I do not use them.  
-Note the approx 7 mil of axle compression (for a typical 10mmx1 steel axle..).  Thus the reason to have a little play in the bearings under no-load. 
-Note the nice mix of American and Metric Units.  Ametrican!  Like a lot of other American Engineers, we tend to use what we are use to.  Another quirk,  7 mil = 7 thou (thousandths) = 0.007 inch (mil is not millimeters, but rather is short for a milli-inch, crazy huh?)

Modifying table-saw tilt-screw.

Bought an old craftsman saw for $75.  My initial interest was because it has its own vacuum with overhead and cabinet suction.   But also I know see this is a very solid saw that will be worth future updating.  Had a lot of rust but cleaned up nicely.  


One of my first improvements:  I didn't like the tilt screw, basically a ball running on stamped sheet metal...  a lot of friction.  

So I updated it with bearing on a pivot.  This turned out quite well mainly cause it didn't take a lot of effort or time to make out of stuff I had laying around.  Used a 6002 bearing, and just happened to have some 4130 tube that it slip fit in (saved some lathe boring time!).  Put 180 deg apart holes, chamfered, in the tube.  Had some square tube, drilled and tapped for 3/8 bolts that were 60 degree point turned on the lathe.  Took the "ball" off, brazed a new sleeve on that I turned to 15mm for the 6002 bearing.  Added a groove for a clip.  Welded some "wings" on the square tube that will used to attach to table saw cabinet. 

Friday, February 20, 2009

Bolting a Bicycle Wheel

In general this applies to Single Speed or Fixed Gear bicycles with a horizontal slot dropout but could also apply to Tandems or to any bicycle where you want to insure a secure wheel. How to insure your rear wheel doesn't slip (with subsequent chain throw).

Although this is common sense, most of the hub manufacturers don't have it figured out (using small diamter serrated locknuts means you don't have it figured out). Not even the expensive "Boutique" hubs. (yet another re-inforcement of my opinion that most of these expensive hubs are yes beautifully machined but poorly engineered).

The proper solution is simple: proper clamping surface area is required. Just like snowshoes are used to prevent "post-holing", so should large enough lock nut and track nut washer diameters be used to prevent permanent deformation of your frame .

Serrated nuts and small-clamp-area are fine for derailleured bikes where digging into the frame doesn't matter. But it doesn't work that well for single speed and track bikes

Serious indentions shown in pic

Fine adjustment of chain tension becomes difficult if not impossible depending on severity of indentations and the nuts will tend to slip "downhill" into the deformity and preload will be lost and wheel will slip and maybe the chain too.
[BTW, an easy chain adjustment method is described by Sheldon Brown: Basically walk the wheel back in the slots by tightening and loosening one nut or the other and pushing sideways on the tire. http://www.sheldonbrown.com/fixed.html#wheel . ]

Another solution is to have track fork slots be made of high strength steel. Something with high yield strength, or maybe even case hardened. But If you don't have that already, it would entail a frame or dropout replacement which is of course absurd.

One could also use “tug nuts” but they are huge hassle.

How much clamping surface area is needed to prevent frame deformation and to properly hold a wheel in the frame? That is, how big of a diameter should the "lock nut" and "track nut" be? It depends on:

-Strength of frame track slot or horizontal dropout. The contact stress should not exceed the yield strength. If we assume an inexpensive stamped steel dropout, the yield might be as low as about 35,000 psi.

-Amount of Clamping Force. A 10mm bolt class 8.8 can easily be tensioned to 7000-8000 lbf.
[compare to quick release force..]Most likely this much will not be used. Assume a torque of 20 ft-lbf (which is reasonable, 6 inch wrench, 40 lbf). This will generate approx 3000 to 4000 lbs of clamping force (calculated using torque vs clamp force equation...) . To be conservative assume 5000 lbf.

bc defg
2ID mmOD mmClamped Area mm*mmClamped Area in*inForce lbfContact Stress psi

[ fyi, from ms excel, the complicated equation above is finding the clamped area : D3=C3^2*(0.25*PI()-0.5*ASIN(B3/C3))-0.5*B3*C3*(1-B3^2/C3^2)^0.5 ]

Only OD is changed in the table above. At 16 mm outer diameter, which are most lock nuts, the contact stress would be way too high. Thus why many have issues with chain adjustment or throwing chains. Higher quality hubs usually run with 18mm luck nuts, which is still not adequate. I would want at least 20mm. But even larger diameter would be smart if you want to play it safe on not deforming your frame or if clamping on aluminum

Far left is overkill at 24mm diameter. Far right is 18mm.

A related aside, how much holding force is really needed to keep a wheel from slipping forward in the frame? Being conservative, at the most 300-400 lbs could be put into a crank (if you are really strong). This would translate into 800 to 1000 lbs of chain force. (well if much more the chain is going to break).

Assume someone only clamps with 3000 lbf , assume friction of 0.15 of steel on steel (which is about as low as it should ever be), equals only 450 lbs of friction force capacity (compared to 800-1000 lbf in chain !). Oops, not enough !? Nope, because there are actually four faces that the clamping action is on. (so this is the reason for multi-plate clutches...).

Too be conservative, of course only two of those, on the drive side, should be considered, giving 900 lbf of capacity before the wheel would slip. But also this could be more than doubled with more wrench torque (which you are going to do if you if able to generate 1000lbf chain force). So if your wheel slips, its either not clamped with enough force, or too little clamp area, plastic flow occurs (frame indentations). When things go plastic, force isn't maintained.

What can a person do to get a large diameter locknut? (I made the one pictured below, on a metal lathe). Do the best you can. I did find a locknut in my spart nuts bin, that was a little over 20 mm OD. I think it came from a Surly hub. Yah smart people in Mn.

Yeah this works. 24mm dia and some rust (magic friction increaser).