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How does a single pivot side-pull caliper bicycle brake work? The brake cable only pulls on one of the arms, so what is causing the force on the other arm? I understand that it is a normal force due to the cable housing, but why is this force developed?

EDIT

I'll elaborate on my questions a bit. Let $A$ be the point where the cable is attached, $B$ be the point where the housing pushes on the other arm, $P$ be the point of pivot, $C$ be the point where the first arm touches the rim and $D$ be the point where the second arm touches the rim. $APC$ is then the arm pulled on by the cable and $BPD$ is the arm potentially pushed by the housing.

Let's say the cable pulls at $A$ with some tension force $T$. A normal force $N$ from the wheel must then push on $C$, in order for the torque with respect to $P$ to be zero. We have $N = (|AP|_x/|CP|_yS) $, if $x$ and $y$ are in the horizontal and vertical direction respectively, and $T$ is vertical and $N$ horizontal.

The force $N$ gives a lateral force on the rim and a torque on the wheel. They could be balanced by a normal force of the same magnitude at the other brakepad. By symmetry, $B$ would then have to be pushed by a force of magnitude $T$.

But couldn't they also be balanced at the hub? Does it have to do with the construction of the wheel, i.e. because it's spoked? If the wheel were in the form of a solid cylinder (which would of course be very impractical!), would the situation be different?

enter image description here image source wikipedia

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  • $\begingroup$ " By symmetry, B would then have to be pushed by a force of magnitude T. " Exactly. The cable housing (white part) pushes down equal and opposite the the cable tension. $\endgroup$ – agentp Aug 4 '17 at 19:27
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Inside The pivot there is a spring holding the 2 arms apart. This adds a torque to the arm and keeps the cable under tension.

In normal state the brake pads are not in contact with the rim.

When the cyclist starts pulling $A$ and $B$ will start to move together until one of the pads contacts the rim. This will then prevent that arm from moving further and apply a little bit of drag on the rim.

Further pulling on the cable will still try and move the pads together but not only 1 arm can move because the other is blocked by the rim.

Once both pads are in contact with the rim any tension that isn't compensated for by the spring will put force on the pads squeezing the rim and applying the breaking force.

The force applied by both brakepads don't need to be equal. The arms are very rarely perfectly centered and balanced and the brake line will also apply a bit of force putting things further off balance. Any sideways force can indeed be compensated for by the hub.

If you look at a bicycle wheel head-on you will see that where the spokes are attached to the hub is much wider than the single line where they are attached on the rim. This creates a triangle shape that can compensate for the sideways force on the wheel and prevents buckling in turns. Any sideways force on the rim will result in a upwards force on one side of the hub and a downwards force on the other.

When dealing with force diagrams you have to take into account all forces which can be quite the rabbit hole in complex situations.

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  • $\begingroup$ "When the cyclist starts pulling A and B will start to move together until one of the pads contacts the rim". I am sorry i still don't understand where does the force which will push B towards A come from when u pull the cable $\endgroup$ – incas May 30 '18 at 13:00
  • $\begingroup$ @incas the brake cable in encapsulated in a sheath that will not compress. If you pull the steel cable out the other end (using the brake lever) then you will force the end of the cable to be pulled towards the end of the sheath. $\endgroup$ – ratchet freak May 30 '18 at 13:11
  • $\begingroup$ @ratchet freak in my understanding of this B should remain in a fixed position while A moves closer to B. Instead what is seen is that both A and B move closer to each other. $\endgroup$ – incas May 30 '18 at 13:58
  • $\begingroup$ @incas Because the sheath is flexible allowing B to move and the fixed point if the pivot P. $\endgroup$ – ratchet freak May 30 '18 at 14:00
  • $\begingroup$ @ratchet freak I am sorry I still don't get it. That B is allowed to move doesn't mean that it has to move. In my understanding of things, B moves closer to A because a force is applied to it and commands it to do so. However i fail to see how the pulling force exerted on A gets transmitted to B as a pushing force. $\endgroup$ – incas May 30 '18 at 14:28
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The brake cable doesn't apply a force to the arms. It reduces the size of the gap between the arms, and therefore between the two brake pads.

The cyclist can only to apply a force to the brake lever because the wheel rim prevents the arms moving together. If the bike wheel is removed, it requires very little force to pull the brake lever as far as it can move.

Since the brake assembly can pivot around the point where it is attached to the bike frame, by symmetry the force applied to brake pad on each side will be equal and opposite. The two arms and the pivot form a symmetrical lever arrangement.

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  • $\begingroup$ Thanks! I'm afraid I still don't quite understand. I've elaborated on my question a bit to explain what I mean. $\endgroup$ – Étienne Bézout Aug 4 '17 at 7:31
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I taught about it a lot and I think I have come to an answer . the housing of the cable is actually a tube, that tube pushes down the lever . suppose the pad break from the other side of the wheel is already pressing the rim so now when you pull- the magic starts to happen - the lever from the side of the cable is pushed down by the cablehouse . now when you squeeze the break - where does the cable come from ? (the cable that is attached to the lever doesn't move, so how come you succeed pulling the cable ? ) answer--> the cablehouse is curved - is not a strait line between the break handle and the lever.if the cable was bear with no cablehouse you would think that there is a lot of redundant cable .now you are pulling that "redundant" cable - now the cable house dosnt have enough space so it pushes the lever- its like trying to insert a 25 cm spring(not streched spring )to a 20 cm gap.the string has a "belly" trying to push the spring to make it without a belly will cause a pressure on the gap to open . I hope this is helpfull ....yoav Israel

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The problem in understanding why the B part is being pulled towards the A part lies mostly in the image from Wikipedia. It shows only the brakes and the arms, but not the whole cable and housing from the arms to the lever.

If you go and see a real bicycle, you will notice that there is always a curve in the cable and housing, of 90 degrees, or more sometimes. This curve is maintained by the brake noodle. It is because of this curve only that the B part is pulled towards the A part. (The housing of the cable is able to take on compression forces, and also tension forces. The compression forces are the ones that matter for this effect)

The cable is in tension and it exerts a force along its path. If you analyze the forces acting on the curved part (Free body diagram), than you will see that the resultant force has a component in the direction of the A part. Because the housing is able to take compression loads, and because of the noodle that keeps the shape curved, the cable-housing structure is being pulled to the A part.

I thought about it myself, and only thanks to a video mentioning this break noodle, it hit me that there is a curve there...

Edit: Now I see that Yoav wrote the same thing before me... I agree with his answer.

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