Yellow Saddle
Guru
- Location
- Loch side.
The first article doesn't even have the word friction in it and the second discusses how friction keeps a car/bicycle from following a radius when turning. I don't see the relevance to the problem at hand.
Spa Ti Audax - anyone using a disc fork?
- Thread starter Trull
- Start date
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Yellow Saddle
Guru
- Location
- Loch side.
There is so much nonsense in those few paragraphs that I don't even know where to begin. If you want to start a head tube discussion, open one, I'll join in. However, head tubes size has nothing to do with disc brakes. Nothing. It would make it easier to debate this if you would take the trouble to extract the relevant sentences/formulas/quotes from web pages or text books rather than throw the whole caboodle at me. You can always reference it, but multiple links confuse the issue, slows down the debate and provides lots of red herrings. .
Yellow Saddle
Guru
- Location
- Loch side.
You can't change change....seriously? That's what delta is used for.
The upper-case letter Δ can be used to denote:
- Change of any changeable quantity, in mathematics and the sciences (more specifically, the difference operator[citation needed]); for example, in:
How shall I put this?
Delta is a fancy word for change. It is a Greek symbol that is nice and easy to draw and when scientists put it in front of a number or variable, we all know it refers to how much that number has changed.
Now, if you say, "Delta change" you are saying change change. Delta is not change squared, but just changed.
It is a little bit like crank arm. A crank is an arm and we don't want to say arm arm,
Rohloff_Brompton_Rider
Formerly just_fixed
Haha! YS is as predictable as ever, my way or the highway and trying to use clever quoting.
OP
OP
Trull
Über Member
- Location
- Aberdeenshire
Does not sound like anyone has fitted a disc fork to a Ti Audax bike then...
jayonabike
Powered by caffeine & whisky
- Location
- Hertfordshire
Andrew Br
Still part of the team !
- Location
- Chorlton Cum Hardy
Have a look here (scroll down):- https://www.cyclechat.net/threads/show-us-your-disc-braked-rides.157783/
Two Ti Enigmas with disc forks.
andrew_s
Legendary Member
- Location
- Gloucester
Probably not.
Most people would buy a bike that came with disc brakes rather than spend an extra £200 to change the forks afterwards, and the Spa Ti Audax hasn't been available for long enough for there to be many which have needed a fork replacement.
swansonj
Guru
A few diagrams always help.
In this diagram, the bike is braking at the limit, which as has already been explained upthread, means (unless the road is slippery or you're on a tandem) that the rear wheel is just about to lift of the ground. The rear wheel is now contributing no braking force and the entire braking force comes from the front wheel. F1 is the force exerted by the road on the front tyre by friction and if it was relevant we could work out the deceleration of the bike+rider from F1=ma.
F2 is the force exerted by the forks+steerer assembly to the frame of the bike via the headset bearings. F1 = F2 to a first approximation. Actually, F2 is less than F1 by the ratio of the mass of the front wheel+forks+steerer+stem+bars to the total mass of bike+rider, with a further adjustment for forces exerted on the rider's arms by the bars, but that doesn't change the principle and is likely to be only a few percent, so we can ignore that effect and say F1=F2.
The relation F1=F2 stems from simple mechanics and is true regardless of what happens between F1 and F2. Disk or rim? Makes no difference, F1=F2. Forks flex backwards under the forces? Makes no difference, F1=F2.
If you want a bit more detail, try this diagram.
F1 is still the force exerted by the road on the front tyre. But now consider what happens at the headset bearings. At the lower bearing, the steerer is pushing back through the bearing onto the frame. The force exerted on the frame by the steerer is F3. But the whole steerer is being twisted forwards, and at the top bearing, the steerer exerts a forwards force F4 on the frame. The net force on the frame is exactly the same as before: F3-F4=F1.
If we need to, we can work out F3 and F4. Consider now the forces exerted on the forks+steerer:
F4 has exactly the same magnitude (action and reaction are equal and opposite) but is in the opposite direction because it’s now the force on the steerer by the frame not the other way round. But now we know that the moments on the steerer+forks are balanced about the lower bearing (because the forks are not undergoing any angular association) so F1.x = F4.y, sufficient information now to work out F3 and F4.
The thing to note is that once again, F3 and F4, just like F2, don’t depend on the type of braking, or the flex of the forks, or anything else like that – just on the geometry and on the limiting value of the braking force F1. (Pedantry qualification: as the fork flexes, then because of the head angle, x would increase slightly and y decrease by an even smaller amount. But this is a small effect. And the flex of the fork doesn’t directly depend on the type of braking either when you’re at the limiting value. It might depend on it indirectly because disk brake forks are likely to be stiffer.)
(all these diagrams consider horizontal components of forces only. Comments about engineers, autistic spectrum, and general nerdishness taken as read, thank you.
)
In this diagram, the bike is braking at the limit, which as has already been explained upthread, means (unless the road is slippery or you're on a tandem) that the rear wheel is just about to lift of the ground. The rear wheel is now contributing no braking force and the entire braking force comes from the front wheel. F1 is the force exerted by the road on the front tyre by friction and if it was relevant we could work out the deceleration of the bike+rider from F1=ma.
F2 is the force exerted by the forks+steerer assembly to the frame of the bike via the headset bearings. F1 = F2 to a first approximation. Actually, F2 is less than F1 by the ratio of the mass of the front wheel+forks+steerer+stem+bars to the total mass of bike+rider, with a further adjustment for forces exerted on the rider's arms by the bars, but that doesn't change the principle and is likely to be only a few percent, so we can ignore that effect and say F1=F2.
The relation F1=F2 stems from simple mechanics and is true regardless of what happens between F1 and F2. Disk or rim? Makes no difference, F1=F2. Forks flex backwards under the forces? Makes no difference, F1=F2.
If you want a bit more detail, try this diagram.
F1 is still the force exerted by the road on the front tyre. But now consider what happens at the headset bearings. At the lower bearing, the steerer is pushing back through the bearing onto the frame. The force exerted on the frame by the steerer is F3. But the whole steerer is being twisted forwards, and at the top bearing, the steerer exerts a forwards force F4 on the frame. The net force on the frame is exactly the same as before: F3-F4=F1.
If we need to, we can work out F3 and F4. Consider now the forces exerted on the forks+steerer:
F4 has exactly the same magnitude (action and reaction are equal and opposite) but is in the opposite direction because it’s now the force on the steerer by the frame not the other way round. But now we know that the moments on the steerer+forks are balanced about the lower bearing (because the forks are not undergoing any angular association) so F1.x = F4.y, sufficient information now to work out F3 and F4.
The thing to note is that once again, F3 and F4, just like F2, don’t depend on the type of braking, or the flex of the forks, or anything else like that – just on the geometry and on the limiting value of the braking force F1. (Pedantry qualification: as the fork flexes, then because of the head angle, x would increase slightly and y decrease by an even smaller amount. But this is a small effect. And the flex of the fork doesn’t directly depend on the type of braking either when you’re at the limiting value. It might depend on it indirectly because disk brake forks are likely to be stiffer.)
(all these diagrams consider horizontal components of forces only. Comments about engineers, autistic spectrum, and general nerdishness taken as read, thank you.
)another thread destroyed and only a few answer the question directly
i have Ti disc road frame arriving tomorrow and there are many carbon disc forks available - enve, easton, lynskey, whisky, kinesis and the like.......nothing says within the information described online that a different frame/headset etc etc is need to fit the forks.............just make sure you buy the correct steerer to match your from whether it be tapered or straight
i have Ti disc road frame arriving tomorrow and there are many carbon disc forks available - enve, easton, lynskey, whisky, kinesis and the like.......nothing says within the information described online that a different frame/headset etc etc is need to fit the forks.............just make sure you buy the correct steerer to match your from whether it be tapered or straight
MacB
Lover of things that come in 3's
- Location
- Farnborough, Hampshire
Yellow Saddle
Guru
- Location
- Loch side.
A few diagrams always help.
View attachment 80855
In this diagram, the bike is braking at the limit, which as has already been explained upthread, means (unless the road is slippery or you're on a tandem) that the rear wheel is just about to lift of the ground. The rear wheel is now contributing no braking force and the entire braking force comes from the front wheel. F1 is the force exerted by the road on the front tyre by friction and if it was relevant we could work out the deceleration of the bike+rider from F1=ma.
F2 is the force exerted by the forks+steerer assembly to the frame of the bike via the headset bearings. F1 = F2 to a first approximation. Actually, F2 is less than F1 by the ratio of the mass of the front wheel+forks+steerer+stem+bars to the total mass of bike+rider, with a further adjustment for forces exerted on the rider's arms by the bars, but that doesn't change the principle and is likely to be only a few percent, so we can ignore that effect and say F1=F2.
The relation F1=F2 stems from simple mechanics and is true regardless of what happens between F1 and F2. Disk or rim? Makes no difference, F1=F2. Forks flex backwards under the forces? Makes no difference, F1=F2.
If you want a bit more detail, try this diagram.
View attachment 80856
F1 is still the force exerted by the road on the front tyre. But now consider what happens at the headset bearings. At the lower bearing, the steerer is pushing back through the bearing onto the frame. The force exerted on the frame by the steerer is F3. But the whole steerer is being twisted forwards, and at the top bearing, the steerer exerts a forwards force F4 on the frame. The net force on the frame is exactly the same as before: F3-F4=F1.
If we need to, we can work out F3 and F4. Consider now the forces exerted on the forks+steerer:
View attachment 80857
F4 has exactly the same magnitude (action and reaction are equal and opposite) but is in the opposite direction because it’s now the force on the steerer by the frame not the other way round. But now we know that the moments on the steerer+forks are balanced about the lower bearing (because the forks are not undergoing any angular association) so F1.x = F4.y, sufficient information now to work out F3 and F4.
The thing to note is that once again, F3 and F4, just like F2, don’t depend on the type of braking, or the flex of the forks, or anything else like that – just on the geometry and on the limiting value of the braking force F1. (Pedantry qualification: as the fork flexes, then because of the head angle, x would increase slightly and y decrease by an even smaller amount. But this is a small effect. And the flex of the fork doesn’t directly depend on the type of braking either when you’re at the limiting value. It might depend on it indirectly because disk brake forks are likely to be stiffer.)
(all these diagrams consider horizontal components of forces only. Comments about engineers, autistic spectrum, and general nerdishness taken as read, thank you.
Thank you. Nice work. But brace yourself for the following accusations.
1) Clever quoting (no, I have no idea what it is, but if your post has a quote in it, I suppose you are in the firing line).
2) Being a Google bunny.
3) Using cut and paste.
4) Plagiarism.
5) Thinking too much.
6) Using math to get your way.
And god forbid you make a spelling error. That's the worse professional embarrassment you can suffer.
Yellow Saddle
Guru
- Location
- Loch side.
I'll summarise the answer for you as follows:
1) Yes, you can fit disc brake fork and a disk brake up front. There are no structural issues with the bike (any bike) that prevent this.
2) The fork that you find will have big fat blades compared to your current skinny blades. This is to prevent fork flex at the lower part of the fork just above the brake itself.
3) A disc brake develops an ejection force on the drop-out. This isn't a problem because the fork you buy will definitely have "lawyers lips" on the end of the drop-out to prevent the QR from sliding down and out. But, your QR should still be a good quality one and it should be tight.
4) A mechanical disc brake is the worst of both worlds. It will not be an upgrade. Get hydraulic or leave things as is.
5) On a road bike you can get away with a disc as small as 140mm provided you use the new Shimano Freeza disc. This disc is an aluminium/stainless steel sandwich with cooling fins. If you don't want this, use a 160mm disc up front.
6) If your current brakes are good quality and in good condition, you won't experience better braking with the disc. You may experience a bit better braking control (called modulation by bike magazines) and a better feel.
7) Your handlebar setup may look a bit strange but if you have late model Shimano road shifters up front, you could buy just the one lever in a hydraulic version. That will give you similar looks. However, I suspect that the hydraulic version is also electronic. Start your research here, rather than at the fork. The fork may well be moot if you can't find the right handlebar options.
8) You will have to get a new front wheel.
