Chain cleaners

On Saturday, December 7, 2013 6:08:06 AM UTC-5, Graham wrote:
"Jeff Liebermann" wrote in message ...

On Fri, 06 Dec 2013 21:15:51 -0800, Joe Riel wrote:



Jeff Liebermann writes:



On Fri, 06 Dec 2013 19:59:15 -0800, Joe Riel wrote:



Completely wrong. Each scale reads 1000 lbs.



You're right. I screwed up big time. I just HATE it when that

happens:

http://www.physics.byu.edu/faculty/christensen/Physics%20121/Sample%20Problems/Oscillatory%20Motion/Two%20Springs%20in%20Series.htm

My apologies for the faulty physics and thanks for the correction.





Pin contact area:

Pin dia = 3.30 mm

Sleeve width = 2.54 mm

1 link = 2 pins

Half the cylinder surface area = contact area =

1/2 * Pi*radius^2 * height =

0.5 * 3.14 * (3.30mm/2)^2 * 2.54mm = 10.9 sq-mm



Cylinder surface area is 2*Pi*r*height.



Maybe I should have given up while I was only one major screwup

behind. Circumference of the cylinder top times the height. I should

have noticed that the units didn't work out.

Thanks again (grumble...) Third try:



Pin contact area:

Pin dia = 3.30 mm

Sleeve width = 2.54 mm

1 link = 2 pins

Half the cylinder surface area = contact area =

1/2 * 2 * Pi * radius * height =

0.5 * 2 * 3.14 * 3.30mm/2 * 2.54mm = 13.2 sq-mm



Therefore, the pressure on each pin is:

454 kgf/pin / 13.2 sq-mm = 34.4 kgf/sq-mm

or = 49,000 PSI

or = 337 MPas

http://www.convertcenter.com/pressure (for units conversion).



49,000 PSI still seems too high.



I'm depressed... maybe an ice cream will help.



Try this:



Power transmitted by a chain is HP=(tensionxspeed)/33,000 where tension is in lbs and speed ft/min



At cruising speed you are delivering about 0.2HP via the chain to the back wheel.



On a 52 tooth big ring with a 90rpm cadence and a 1/2" pitch chain your chain speed is (52x90x0.5)/12=195 ft/min so your chain tension will be: (33,000x0.2)/195=33lbs.



Now very approximately the pins in a 9 speed chain are 1/8" diamenter and their width is again 1/8" so the bearing area is 0.125x3.142x0.5x0.125=0..025sq ins



That gives a bearing pressure of 1344 lbs/sq in.



I think you will find this sits quite well with the design spec of chains for low speed operation.



If you want to continue with your sums then typical rolling friction coefficients are 0.21 dry and 0.14 lubricated.



Graham.

On a quick glance, the calculations look OK, except that bearing pressure (pins, shafts, etc.) is always calculated based on projected area. For any cylindrical surface like that, the projected area is what's seen in the side view, so to speak: a rectangle, whose area is simply diameter times length. No Pi term. For a pin 1/8" diameter and 1/8" long (in the region it's supporting bearing stress) that area is 1/8" x 1/8" = 0.016 in^2.

Bearing pressure is thus 1.6 times what's been calculated, or about 2200 psi.

- Frank Krygowski

On Saturday, December 7, 2013 7:11:08 AM UTC-5, John B. wrote:

Cylinder surface area is 2*Pi*r*height.

I thought it was (Pi x (R squared)) x height ?

Nope. You want projected area for bearing calculations. Area = Diameter x height.

- Frank Krygowski

On Saturday, December 7, 2013 1:04:33 PM UTC-5, Graham wrote:

OK max practical case. Take someone plus their bike around 230lbs climbing a 30% grade at 2mph in with a 26 tooth granny ring at 30rpm. Power requirement is double the normal case at 0.4HP and the chain speed drops by a factor of 6 to 32.5ft/min increasing the bearing pressure by a factor of 12 i.e. 16,128 lb/sq in. Now this getting high for a chain but for how long can anyone keep this up?

Well, now we're getting into chemistry. What's in the rider's blood?

- Frank Krygowski

On Sunday, December 1, 2013 9:37:46 AM UTC-5, Lou Holtman wrote:
On 2013-12-01 14:21:51 +0000, sms said:



On 11/30/2013 5:53 AM, somebody wrote:

I like the convenience of clamp-on chain cleaners but haven't run into

one that does a good job. Most leave a lot of grunge on the side

plates, especially the innermost one.



I've tried the Park tool, one originally sold under the Vetta brand

and later other names, and a generic cheapie.



Have any of you tried the cheap Chinese versions?

http://www.focalprice.com/HL0697L/Mi...aner_Blue.html

http://www.focalprice.com/HLF77L/Pro...aner_Blue.html

http://dx.com/s/chain+cleaner



They all work pretty much the same.



As Sheldon Brown wrote "The on-the-bike system has the advantage that

the cleaning machine flexes the links and spins the rollers. This

scrubbing action may do a better job of cleaning the innards."



The key thing is that you have to change the solvent multiple times

during the cleaning process. Eventually the solvent will run clean and

you're done. They can get messy if you're not careful (running it

through too fast) so you want to have rags covering the chainstay. Of

course even if you're cleaning the chain off the bike you want to

change the solvent until it runs clean.



As I said (even before Sheldon!), the big advantage of the chain

cleaning machines is that the links pivot as they go through and you

end up getting the chain a lot cleaner than if you remove the chain and

put it into a bottle of solvent and shake it.



What's more difficult to do on-bike is proper lubrication. Since you

need to get lubricant actually inside the links, not just on the

surface of them, you need to either submerse the chain or use a foaming

spray on lubricant that penetrates the lubricant into the links. You

can have second chain cleaning machine that you fill with oil to do the

lubrication.



If you have a chain with a link that opens for chain removal it's not

that much of an advantage to do on-bike cleaning but if you need to use

a chain-breaking tool then on-bike cleaning is a much better option.



Every time I go to Interbike I'm amazed at the number of ridiculous

chain cleaning and lubricating chemicals that are being marketed (if I

added up the retail value of all the free samples I get then I'm

actually making money on the trip since it costs me only about $100 in

transportation costs to fly to Vegas and get to the convention center).

I really don't need coffee-scented chain lube

http://cf2.thefancy.com/default/276484325_06be9387d878.jpg and this

is a real product. The reality is that "expensive" has little, if any,

advantage over "cheap." Clean the chain with a petroleum-based solvent

(never use water-based cleaners) then lubricate it with a petroleum

based lubricant (dipping in SAE30 oil or spraying foaming motorcycle

chain lubricant for non-O Ring chains). Avoid hot wax at all costs.



Given the fact that a chain that is cleaned by running it through a

rough rag before lubrication after every wet ride give about the same

lifetime this is a questionable, messy and unnesessary troublesome

regime.

--


Lou

I have to agree with you Lou. By the time you go your first mile after doing one of those obsessive chain cleanings your chain is probably not much cleaner than my chain is after I give it my daily cleaning (see below).

Say that you clean your chain like this once a week and you ride 200 miles a week and you get 5 or even 10 miles of real benefit from having a pristine chain before the chain is a little dirty. If you do the math here you will find there is not much benefit in meticulous chain cleaning unless you want to carry the cleaning kit with you and stop every ten miles or so to clean your chain.

I typically get around 5000 miles out of a chain and have cassettes and chain rings that have close to 20,000 miles on them. For reference, I am a big guy and I ride through winter here in Indiana. When it comes to keeping a drive train working well year after year, frequency and methodology count for more than getting the inner side plates squeaky clean.

Here are the lessons I have learned:
1) Clean after every 30 miles or every ride if you are in rain/snow/sand
2) Wipe your chain down with the rag you used after applying lube last time
3) Clean the chainwheels, even a simple wiping with a rag goes a long way
3) Wipe the crud off your cogs and idler wheels too
4) Use a nice medium lube with lots of detergent, I like Marvel Lube 'cause it smells good
5) Apply the lube to the lower roller of your rear derailleur, so that it pushes the bits of crap from the inside of the chain towards the outside
6) Use WAY more lube than you need, the lube is being used as a solvent right now
7) Wipe down the chain really well with a new rag, make sure to get the excess off the chain rings, idlers and cog too

All this takes about 90 seconds if done regularly.









wipe it off with a rag, apply a bunch of lube, run it over the sprockets a few times and wipe it all down with a fresh rag.

On 08/12/13 12:47, Frank Krygowski wrote:

On a quick glance, the calculations look OK, except that bearing
pressure (pins, shafts, etc.) is always calculated based on projected
area. For any cylindrical surface like that, the projected area is
what's seen in the side view, so to speak: a rectangle, whose area is
simply diameter times length. No Pi term. For a pin 1/8" diameter
and 1/8" long (in the region it's supporting bearing stress) that
area is 1/8" x 1/8" = 0.016 in^2.

Bearing pressure is thus 1.6 times what's been calculated, or about
2200 psi.



Does the pin experience the same pressure over the entire bearing
surface of diameter x width?

I would have thought that the pressure would be much higher where the
force is at right angles to the pin surface, and zero where the force is
tangential to the pin surface.

This seems true from my observation of a disassembled worn chain, that
the wear is most over about half of one side of the pin.

http://sheldonbrown.com/images/chain_wornpin.gif


--
JS

James writes:

On 08/12/13 12:47, Frank Krygowski wrote:

On a quick glance, the calculations look OK, except that bearing
pressure (pins, shafts, etc.) is always calculated based on projected
area. For any cylindrical surface like that, the projected area is
what's seen in the side view, so to speak: a rectangle, whose area is
simply diameter times length. No Pi term. For a pin 1/8" diameter
and 1/8" long (in the region it's supporting bearing stress) that
area is 1/8" x 1/8" = 0.016 in^2.

Bearing pressure is thus 1.6 times what's been calculated, or about
2200 psi.



Does the pin experience the same pressure over the entire bearing
surface of diameter x width?

Theoretically, yes. Practically, I don't know.

I would have thought that the pressure would be much higher where the
force is at right angles to the pin surface, and zero where the force is
tangential to the pin surface.

This seems true from my observation of a disassembled worn chain, that
the wear is most over about half of one side of the pin.

http://sheldonbrown.com/images/chain_wornpin.gif

Assume the major wear occurs as the link exits the rear cog (that's when
it is under max tension and pivots). If the pivot is through an angle
theta, then 2*theta of the contacting half will experience less wear
than the remainer because it is loaded for only a portion of the pivot.
Alas, I'm thinking theta isn't big enough for that to be a complete
explanation.

--
Joe Riel

On Sunday, December 8, 2013 5:47:06 PM UTC-5, James wrote:

Does the pin experience the same pressure over the entire bearing
surface of diameter x width?

I would have thought that the pressure would be much higher where the
force is at right angles to the pin surface, and zero where the force is
tangential to the pin surface.

This seems true from my observation of a disassembled worn chain, that
the wear is most over about half of one side of the pin.

http://sheldonbrown.com/images/chain_wornpin.gif

Well, first, I know that for boundary lubricated plain bearings, bearing area is always calculated as diameter x length, and pressure based on that area is used to select bearing materials.

I think I understand what you're visualizing about the distribution of pressure. It may be that, initially, pressure is not evenly distributed, with the more perpendicular surfaces supporting more load. But...

.... but as soon as wear occurs, if wear increases with pressure (as is usually assumed), the area with the highest pressure would be expected to wear away faster than other areas. When that happens, the lack of material would probably transfer load to what was previously an area with less pressure. Thus it all tends to even out, literally.

To visualize the wear in the Sheldon Brown photo, try this: Using a circle template, draw one precise circle on paper, using (say) black ink. That represents the original pin surface.

Now shift your circle template about 1/10 radius to the right, and draw a partial circle, just the left half, within the original circle, using (say) red ink. Kind of like a partial Venn diagram. That represents the shift in the wearing surface due to abrasion from the roller, pushing from left to right.

The shape in the "intersection" portion of the Venn diagram - with a black circular arc on its right and a red circular arc on its left - is, I think, pretty much the shape shown by Sheldon's pin.

To get that shape, every red point shifted 1/10 radius rightward, w.r.t. it's mate on the original black arc. The rightward wear was the same all up and down the original left surface.

I hope that makes sense. This would be _so_ much easier to explain visually!

Again, I think the key is that as soon as you have microscopic wearing away of material in one spot, the surrounding spots see a corresponding increase in pressure. So I think it's reasonable to say the pressure is constant, at _least_ to a first (or second?) approximation.

- Frank Krygowski

On 09/12/13 13:10, Frank Krygowski wrote:
On Sunday, December 8, 2013 5:47:06 PM UTC-5, James wrote:

Does the pin experience the same pressure over the entire bearing
surface of diameter x width?

I would have thought that the pressure would be much higher where
the force is at right angles to the pin surface, and zero where the
force is tangential to the pin surface.

This seems true from my observation of a disassembled worn chain,
that the wear is most over about half of one side of the pin.

http://sheldonbrown.com/images/chain_wornpin.gif

Well, first, I know that for boundary lubricated plain bearings,
bearing area is always calculated as diameter x length, and pressure
based on that area is used to select bearing materials.

I think I understand what you're visualizing about the distribution
of pressure. It may be that, initially, pressure is not evenly
distributed, with the more perpendicular surfaces supporting more
load. But...

... but as soon as wear occurs, if wear increases with pressure (as
is usually assumed), the area with the highest pressure would be
expected to wear away faster than other areas. When that happens,
the lack of material would probably transfer load to what was
previously an area with less pressure. Thus it all tends to even
out, literally.

To visualize the wear in the Sheldon Brown photo, try this: Using a
circle template, draw one precise circle on paper, using (say) black
ink. That represents the original pin surface.

Now shift your circle template about 1/10 radius to the right, and
draw a partial circle, just the left half, within the original
circle, using (say) red ink. Kind of like a partial Venn diagram.
That represents the shift in the wearing surface due to abrasion from
the roller, pushing from left to right.

The shape in the "intersection" portion of the Venn diagram - with a
black circular arc on its right and a red circular arc on its left -
is, I think, pretty much the shape shown by Sheldon's pin.

To get that shape, every red point shifted 1/10 radius rightward,
w.r.t. it's mate on the original black arc. The rightward wear was
the same all up and down the original left surface.

I hope that makes sense. This would be _so_ much easier to explain
visually!

Again, I think the key is that as soon as you have microscopic
wearing away of material in one spot, the surrounding spots see a
corresponding increase in pressure. So I think it's reasonable to
say the pressure is constant, at _least_ to a first (or second?)
approximation.


Ah, yes I suppose after some wear in period that would be so.

--
JS.