Spa Ti Audax - anyone using a disc fork?

Trull

Senior Member
Location
Aberdeenshire
I'm interested in replacing my old lightweight tourer with a Spa Ti Audax framed bike, but as a heavier rider I'd prefer disc braking at least at the front… has anyone else done this? I see Lynskey have an appropriate fork, and wonder if there might be some solutions I'm missing?

Any suggestions or thoughts would be welcome.
 

ianrauk

Tattooed Beat Messiah
You may want to drop Spa a line to see if the frame, especially the headtube will take the extra forces that a disc fork will make on the bike.
 

swansonj

Guru
You may want to drop Spa a line to see if the frame, especially the headtube will take the extra forces that a disc fork will make on the bike.
Ummm... Does the method of braking actually affect the forces on the frame? On the forks themselves, yes, absolutely, but once you're above the fork crown, does the origin of the braking force make any difference to its size or distribution?
 

andrew_s

Guru
Location
Gloucester
You can lift the back wheel with a decent rim brake, so there shouldn't be any stronger forces on the frame because you are using a disc brake.
It's the fork itself that has to be stronger - braze a disc mount on a standard rim brake fork, and you'll risk the LH blade bending back at the top of the mount.
 
Location
Loch side.
Really? Surely the length of the fork (lever) increases the torque therefore increasing the forces and centre point of the forces?

Also disc brakes are well known for extra power, surely this fact alone increases the the stress on the frame?
The length of the lever is constant for rim or disc brakes provided some obvious givens.

For a given deceleration, the retarding force is the same.

The lower headset is the fulcrum, not the other end of the lever.

For a given force, the fulcrum receives the same force no matter where it is applied.

Further, the old myth that a disc brake has more "power" is just that, a myth. Semantics first, it is not power, but force, we're talking here. Nevertheless, no matter how good the clamping force of the brake, the bike can only decelerate up to a point where the overturning momentum chucks you over the bars. This point is well below the traction level of the front wheel so a bike/motorbike will always overturn before the wheel skids. Effectively we have unlimited traction but limited overturning resistance.

Therefore, no matter how strong the brake is, all its force cannot be used. The only way to increase braking distance on al bicycle or motorcycle is to lower the centre of gravity to closer to the level of the hub, as it is in a car. A car's front wheels can skid, a bike's not. Cars brake quicker than bicycles and motorbikes.
 

Rohloff_Brompton_Rider

Formerly just_fixed
But the braking force is further away from the head tube...

Ok for arguments sake lets say all brakes exert the same force (which simply isn't true but that's another debate).

Place this force further away from the head tube the force exerts more pressure because of the leverage doesn't it? It also places the force further back due to flex, surely this moves the forces more towards the bottom of the head tube.

For example on my old croix der fer the the front disc brake yanked the wheel out of the drop outs in emergency stops. A change of skewer to security skewer and big spanners stopped this happening.

I tried it with old rim brake forks and in emergency stops not once did it yank out of the dropouts. And I used the original skewer for consistency.

Edit: and discs brakes are usually put onto bikes that take fatter tyres so the centripetal friction increases due to the contact being larger ergo more force.

Edit again: if the forces don't change, why have manufacturers gone down the tapered headset route on disc bikes, granted one or two have taken this over onto a few rim braked bikes but majority are on discs brake bikes.

Edit again again: semantics it is power due to the displacement (forks flex and the time for the effect of fork flex has on the time it takes to come into effect, I.e., the greater the flex the longer the delay of the force to the headset).
 
Last edited:
Location
Loch side.
The path that the retarding force at the front wheel uses to push against the bike is irrelevant. The road pushes against the bike and the bike pushes against the road and it all goes through the same place. The only way to alter this is to stop the bike by pushing on the frame/rider or putting a parachute at the back and pull.
The fact that for a disc brake and a rim brake the forces reach the bike via a different path before it lands at the headset bottom bearing, is irrelevant. Draw a line diagram and look at it. The headset pushes the fork with exactly as much force as the road pushes the tyre backwards.

I didn't say all brakes exert the same force. I said no matter how strong the brake, it can only use the retarding force up to a maximum point and that maximum point is when the rear wheel lifts and the bike starts to overturn.

It is easy to understand this for rear brakes because all I need to say is "no matter how strong your brakes, the wheel will skid at a certain given point and any extra brake force after that will have no effect." However, a similar scenario exists for the front brake. But instead of the wheel skidding at that maximum point, the bike overturns. Once the bike overturns, no amount of extra braking can make the bike slow down quicker - unless you face plant.
 
Location
Loch side.
Cut cut cut cut

For example on my old croix der fer the the front disc brake yanked the wheel out of the drop outs in emergency stops. A change of skewer to security skewer and big spanners stopped this happening.

I tried it with old rim brake forks and in emergency stops not once did it yank out of the dropouts. And I used the original skewer for consistency.
The way a disc brake is arranged on a fork causes an ejecting force usually directly inline with the drop-out's opening. That has nothing to do with stopping and the other issue we're discussing. Your observation is correct but not applicable to the problem at hand.

Edit: and discs brakes are usually put onto bikes that take fatter tyres so the centripetal friction increases due to the contact being larger ergo more force.
There is no such thing as centripetal friction, so I can't answer that. Disc brakes are also found on narrow tyre road bikes. The size of the contact patch is irrelevant, friction not being dependent on contact area.

Edit again: if the forces don't change, why have manufacturers gone down the tapered headset route on disc bikes, granted one or two have taken this over onto a few rim braked bikes but majority are on discs brake bikes.
Tapered headsets are not exclusive to disc brake bikes by any means. The evolution to tapered headset was driven by tube shape development. I'll address that in a separate post if you want, but the headset size has nothing to do with the issue at hand.

Edit again again: semantics it is power due to the displacement (forks flex and the time for the effect of fork flex has on the time it takes to come into effect, I.e., the greater the flex the longer the delay of the force to the headset).
No, it is force. F=MA and there is no time component in that formula. Power is work done which has a per second element to it.
 

Rohloff_Brompton_Rider

Formerly just_fixed
The path that the retarding force at the front wheel uses to push against the bike is irrelevant. The road pushes against the bike and the bike pushes against the road and it all goes through the same place. The only way to alter this is to stop the bike by pushing on the frame/rider or putting a parachute at the back and pull.
The fact that for a disc brake and a rim brake the forces reach the bike via a different path before it lands at the headset bottom bearing, is irrelevant. Draw a line diagram and look at it. The headset pushes the fork with exactly as much force as the road pushes the tyre backwards.

I didn't say all brakes exert the same force. I said no matter how strong the brake, it can only use the retarding force up to a maximum point and that maximum point is when the rear wheel lifts and the bike starts to overturn.

It is easy to understand this for rear brakes because all I need to say is "no matter how strong your brakes, the wheel will skid at a certain given point and any extra brake force after that will have no effect." However, a similar scenario exists for the front brake. But instead of the wheel skidding at that maximum point, the bike overturns. Once the bike overturns, no amount of extra braking can make the bike slow down quicker - unless you face plant.
Sorry but I think we need to disagree on this, I genuinely cannot understand why the flex of the fork (discs) which alters rake which in turn increases power due to the power equation (velocity). The altered rake puts the weight of the rider on to a different pivot point (virtual) than rim brakes. Taking these changes (granted delta changes but they all add up) into consideration, surely increases the stress on the head tube?
 
Location
Loch side.
Sorry but I think we need to disagree on this, I genuinely cannot understand why the flex of the fork (discs) which alters rake which in turn increases power due to the power equation (velocity). The altered rake puts the weight of the rider on to a different pivot point (virtual) than rim brakes. Taking these changes (granted delta changes but they all add up) into consideration, surely increases the stress on the head tube?
You are tying yourself into knots here. Now you have fork flex in here as well. And delta changes. Delta is change. You can't have change change.
 
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