What's the recommended torque of the front caliper bolt for a carbon fork?  The Fork is from a Felt F1. I don't want to damage the fork from over thightening.
123620
I've recently got a new bike with a . This comes with deep-drop calliper brakes (Shimano R451 Dual Pivot). The first time out I bumped one of the brakes and it twisted around the single attachment bolt and rubbed on the rim.  I managed to easily twist it back by hand and carry on, however I thought the bolt probably needs tightening. The question is: How tight can it be?  As the fork is part carbon I've so far only dared to take it up to 4Nm, but it still seems possible to twist the whole calliper around with a bump of the hand. Maybe that's normal? Perhaps it's easier to bump around because of the longer deep-drop lever. Should I be aiming to get it really secure?    I'm not sure whether the brake mount in the fork is all carbon, or some of it is alloy underneath, but either way I'd like to know what a sensible torque is. As far as I can see from the  there is no such specification. How can I find this out?
76298
I have been working my way through issues with noisy front cantilever brakes. After running through many different ways to solve the issue including new pads (Kool Stop Dura2 Duel Compound), setting the toe in and cleaning the rims.  I now believe the noise is being generated by the play in the pivot point of the cantilever arm. It is not possible to tighten the bolt any further so there will need to be a washer or shim placed in here.   Has anyone had any success in removing the play from the pivot point?  The bike is a cyclocross Kona Jake 2012.
106287
The spec sheet for the Shimano Alivio FD-M430 front derailleur contains a number of points that I don't understand.   What does "Maximum capacity: 22 teeth" mean? Why are "Top gear teeth" constrained to "44/48T"? Is there anything wrong with 46? What is "Cable Routing: dual-pull type"? And most puzzling, what is "Chain Stay Angle  63-66 / 66-69" and how do I measure mine?
68196
I just bought a pair of SRAM Powerlocks for 10-speed chains. The packaging says that they can only be used with SRAM chains. Will anything nasty happen if I use it with a Shimano chain and cassette?
61593
Would someone explain wheel sizes.  If I am not mistaken, the wheels on my Trek 520 touring bike are 700C. There are no markings, and of course, no documentation of any kind. I assume 700 means 700mm, so 700 / 25.4 = 27.56 inches.  If I measure the rim however, its diameter is approximately 25.1 inches.  If I buy Continental tires, the size is 28 inch, but I have to buy 27 inch Bontrager tubes.  To make matters worse, per this link, tires come in sizes of 26, 27, and 28 inch, and size 700.     How can I determine my wheel size, and are these various sizes interchangable. Also, is the 'C' suffix important, and what does it mean?  Of course the tires are marked, but I am under the impression (per the dealers comments) that Continental marks their tires differently than everyone else. Also, the tire markings don't help for ordering a new wheel.   With so many tire sizes available, how can I know if I am using the correct size.  After reading the various answers here, I should explain that my confusion on this matter stemmed from the fact that I used to work as an auto mechanic. In the auto industry, car tires are based on wheel size, 13", 14", 15" etc, and there is no possible way to mount a 15" tire on a 13" rim for example.  So when I saw all these different tire sizes, I assumed they all required a different wheel size, but in fact, there is essentially just one wheel diameter for my type of bike, which they refer to as 700C, which is actually 622mm.   I also initially assumed that the C referred to the wheel's width, but that isn't true either. The C is apparently just antiquated terminology.  So at least in theory, I can mount a 26", 27" 28" or 29" tire on my 700C (622mm) rim. I just need to verify the width of the tire is suitable my wheel width, which is what Sheldon Brown's article explains.  Also, I measured my Continental tires, which are marked as 28 inch. They actually measure 27 inches (fully inflated), which must be why my dealer calls their sizing non standard.
91194
Just bought my first bike and it was put together by the store. I was wondering if this little cable/wire is okay, it's hitting my pedal every time I’m riding. It doesn’t really bother me, but I don’t want it to end up messing something up really bad. Help please new bike owner
115134
I live in the UK and I expect my front brake lever to be operated by my right hand. However, I recently went on a friend's bike that had the brakes 'back to front', which would be the 'right way round' for people in Europe, North America and everywhere else that drives on the right.  I am left-handed and therefore used to finding things designed to be best operated with the right-hand, e.g. the controls on a lawnmower.  What are the reasons for the brake lever convention on a bicycle? It seems that the front brake is more important for controlled braking, in the UK we have that worked by the right hand, which makes sense. For countries that drive on the right that means the back brake has the 'handed-ness' advantage, rather than the front.  I also like to be controlling my speed with the front-brake when making a right-hand turn, however it is not possible to do this when signalling. A left turn is not so important to stick your arm out for - drivers coming the other way and behind don't really care if the sign is not made for a left-hand turn, whereas a right-hand turn really should be indicated. So, from that perspective, the brakes are the 'wrong-way-round' for me.  Can someone kindly guide me through why the brakes are wired up so and any historical origins of the convention?
50021
I'm building my first road bike, which means I'll have to find appropriate bar height through personal experience and cut the fork steerer in process at least once. I plan to start with the following combination, from down to top:    8mm conical spacer that came with headset. 20mm and 10mm spacers. 40mm high stem. 5mm spacer.   In summary, 38mm before stem, 5 after, resulting in total of 83mm of exposed steerer after cut. The fork has 1 1/8 inch (~28mm) tapered carbon fibre steerer, without metal liner inside. Headset is integrated, upper bearing is almost flush with head tube.  Will it be safe or I'll have to cut shorter? Fork manufacturer did not answer the question. I do not plan to ride out of saddle too much, at least during initial fitting, so bars won't experience significant load, and I'm pretty lightweight.
76718
I was wondering what advantages 29" wheels give. From what I understand they are well suited to mountain biking. Are they any use on road bikes perhaps?
20650
I have a Lappierre Zesty 314  (2012). I'm running XT rear mech, an SLX front mech, my crank is XT, rear cassette is XT and XT chain. I went to the Forest of Dean and after the day was over I snapped my chain! I picked up a SRAM chain for my bike and now when I pedal hard it slips with a big bang. I have new Eastern Havens so my hub is fine. Can you help with any ideas to help me solve this? Shall I get a new rear cassette?
26772
I have an intuition problem  calculating torque using the cross product formula.  As for example let the magnitude of the force be 50 lbs and length of the  wrench be one foot and you are exerting force in a clockwise motion and the angle you apply the force 60 degrees.  This is an example so I can ask my question.  Using the right hand rule the torque points  perpendicular to the force you are applying to the bolt. In this case since the sine of 60 degrees is about .86 it would be (.86)(50) foot lbs. How can the bolt turn clockwise if the force is concentrated perpendicular to where it needs to turn?  The cross product formula demands the torque be perpendicular. Obviously my mistake but I don't see where.
122198
I've decided I want a disc brake, at least on the front, which seems a good enough reason to go bike shopping, but am I overlooking the possibility of fitting a cable pull disc brake to my current bike?  I've been using a ten year old road bike as a commuter.  It has carbon front forks with road caliper/rim brakes and integrated Shimano road brake/gear levers.
62684
These are my current rims: Wheeltech TD26 by Alex Rims (can't find much info about these online).  I need to replace the rear wheel as it is cracked (cracks on the rims around a few spokes) It's a 700C disc brake rim with quick releases.  What do I need to consider other than the size of the wheel when choosing a replacement?  I'll likely migrate over the cassette and rotor (if need be).  My knowledge of rims is limited so any help is appreciated!  Here is my bike:  (Must select Whistler 70 when on the site)
96918
Changed back tyre 700x35c (WTB ALLTERRAINasaurus) to 700x23c (Schwalbe Lugano) and front rim from 18 to 13 inner width (photo below), is it better to change rim or tyre size to get lower rolling resistance ?  One of changes made significant improvement (>20% shorter travel time), but not sure which one.    Pressure around max 500kPa front, 700kPa back (max. >8bars)
74911
I just installed a new chain on my 2002 Trek 2200 Road Bike.  The old chain was a Shimano 9 speed that I installed 18 months and about 5000 miles ago.  It had stretched modestly and was full of winter cruft/grime.  The new chain is a SRAM 9 speed, my first time using the gold snap ring system.  I followed instructions and made sure I have the same link count as the chain that came off the bike.  The cassette was new with the old chain, so it has about 5000 miles on it and doesn't appear to have significant wear.  After installation and a clean/wipe/lube/wipe, I took it for a 15 mile easy ride at about 15-16 mph.  It shifts perfectly and I didn't have to make any adjustments to either derailleur.  Three times, however, when putting it under stain (once on a modest hill climb, once starting from a stop sign in a fairly big gear, and once on an acceleration coming out of a sharp turn) it skipped several teeth and caused my foot to slam down hard.  On the hill, it actually hurt as I had my weight off the saddle (not really standing) and I came down hard on the inside of my thigh.  I think all three were on the right side (power train side) pedal stroke, and I don't think they were all in the same gear.  I'm pretty sure it was the rear cassette that skipped, not the front chainring, but it did happen really fast each time.  I will also say that it's not like these were the only three times I put substantial pressure on the pedals in the ride - I wasn't going hard, but I wasn't being tentative or anything.  Also, in case it's relevant, it was cold, about 40 degrees F or about 4 degrees Centigrade.    My question:  Is this going to get me hurt, or will the new chain 'settle in' over a few rides?  I searched here and pretty much got confirmation of what I had previously thought, that you always get a new chain when you get a new cassette, but you should be able to get a few chains out of each cassette.  Thanks!
22402
I've learned that the torque on an object is just $ \text{perpendicular force}\times\text{radius}$ and that has worked well for things like seesaws but it doesn't take into account the torque of the object itself!    In the picture the rod has a total mass of 5kg, and by definition the torque on the right is greater than that on the left.  $$T=Fr$$   I am supposed to solve the question     What distance to the left of the fulcrum (triangle supporting the rod) would I have to put a 3kg weight to balance the rod?     I've tried calculating the net torque $T_{net} = I\alpha$ but I don't really know how to approach this. I'm looking for an explanation more than an answer.
77836
I am trying to learn physics by myself as I do not have a good teacher at school. I've been reading Feynman Lectures on Physics and I can't figure out how he found  out this number. Here is an excerpt of :      Let us now illustrate the energy principle with a more complicated problem, the screw jack shown in Fig. 4–5. A handle $20$ inches long is used to turn the screw, which has $10$ threads to the inch. We would like to know how much force would be needed at the handle to lift one ton ($2000$ pounds). If we want to lift the ton one inch, say, then we must turn the handle around ten times. When it goes around once it goes approximately $126$ inches. The handle must thus travel $1260$ inches, and if we used various pulleys, etc., we would be lifting our one ton with an unknown smaller weight $W$ applied to the end of the handle. So we find out that W is about $1.6$ pounds. This is a result of the conservation of energy.       If you divide $2000$ pounds by  $1260$ in you get $\frac{100}{63}$, which is $1.58$, a number close enough to $1.6$ pounds, which according to Feynman is the weight that we need to apply to the end of the handle. I believe, however, that this calculation is wrong given a dimensional analysis of such operation.  Here is my dimensional analysis:  $\frac{2000 pounds * 1 in}{ 1260 inches }$ = $\frac{[M] * [L] }{[L]}$ = ${[M]}$ And this is wrong, as we are suppose to get a force, not a mass.  I found this , where you can find what I think is an email from an student reporting the teacher that he believes he got the number diving  $\frac{5}{\pi}$ . Can you also explain me this?   Finally, in a thread in a Physics Forum question,they concluded that $1.6$ is just the force to keep the system at equilibrium, and you need $\text{a force} &gt; 1.6$ to actually lift the ton. After, someone said that Feynman would have said this answer is wrong. Why is that?