- Cotic Rocket?
Having cancelled my Enigma build I’m now looking for a frame to build into an all-day trail bike for lakes use and quite like the look of the Rocket.
I’ve searched this place and cotics site and contacted cotic directly ( only last night, not complaining) but can’t seem to find a geo table or details on colour schemes. Has anyone seen anything with more details?
Cheers,Posted 8 years ago
Interesting reading on his geek page, not sure I agree with this, tho he doesn’t really explain it:
By moving the caliper as far forward and above the rotor as possible, the brake force on the Hemlock pointed at the instant centre so the braking stabilised.
I can’t see a geo page, but it appears to be all about the tech – geo doesn’t matter!Posted 8 years agoahwilesMember
Al: you’re right – that doesn’t make any sense.
(pivot location determines braking response, not the location of the caliper on the seatstay (hemlock) or swingarm (rocket))
Ivan: the frames are months away at least – i guess Cy doesn’t want to get people too excited too early. he’s given us a few sneak peaks cause he knows we like him.Posted 8 years ago
Surely the braking response depends upon the forces acting on the pivot, and the location and direction of these will depend upon the caliper position. Geometry is alluded to in the ‘how to Rocket your Hemlock’ newletter ~1.5 deg slacker, slightly lower BB, steeper seat angle.Posted 8 years agosuperfastjellyfishSubscriber
The Enigma was going to be a custom design, similar to a brodie holeshot ti crossed with a Soul but slacker and the intention was to build it up into a fairly AM style hardtail. It’s not that big a jump to a trail full susser from there. Besides which there’s no other ht frame I want now.Posted 8 years ago
Impatience is the last thing you should take heed of when spaffing £1K+ on a frame.
Isn’t this droplink the same as used by Ventana at least?
chiefgrooveguru – Member
Surely the braking response depends upon the forces acting on the pivot, and the location and direction of these will depend upon the caliper position
Braking response depends primarily on chainstay growth doesn’t it?
In any event doesn’t a brake just create a torque on the bike via the frame member it’s attached to? (albeit the forces on by which it does so vary according to where it is attached)Posted 8 years agoahwilesMember
Al, it’s a bit too complicated to explain here, but i’ll try…
imagine a caliper bolted to a swingarm (standard single pivot), when the brake is applied, wheel rotation becomes swingarm rotation.
imagine a caliper bolted to the seatstay on something like an ellsworth dare, the seatstay doesn’t rotate as is moves, it just moves up and down, this means that braking forces can’t move the suspension.
back to the single-pivot, it doesn’t matter where the caliper is mounted – on top of the swingarm, underneath the swingarm, wheel rotation becomes swingarm rotation.
In any event doesn’t a brake just create a torque on the bike via the frame member it’s attached to?
albeit the forces on by which it does so vary according to where it is attached
err… this one probably needs a diagram, but in the case of a single pivot it’s mostly down to the height of the pivot above the ground.
anyway, if Cy ever makes a steel, 29er, ‘Rocket’, i’ll have his babies.Posted 8 years ago
Oh, and location of the brake calliper makes ALL the difference. I rode an early Hemlock prototype, before the calliper mount was moved to where it was on later prototypes and the production frames, and you had to stay off the back brake completely on steeper trails. My prod Hemlock doesn’t suffer from the same braking issues at all – the behaviour completely fixed by the changed calliper mount/position.Posted 8 years ago
Having just scribbled this out with pencil and paper, I believe that the reaction force from the calliper (looking from the drive side) placing the calliper:
at 3 o’clock (chain stay) would cause suspension jack
at 9 o’clock would cause suspension squat
at 6 o’clock would cause chain growth
at 12 o’clock would cause chain shortening
The reaction from the tyre would cause chain growth and suspension squat. Obviously there are numerous other force to consider, the most problematic being from the organic mass way above the bike’s CoG… 😉Posted 8 years ago
CGG can you explain your reasoning?
I don’t see how the brake produces anything other than a (net) torque and backwards force at the tyre contact patch, no matter where it is.
if this actually made a difference, we’d know about it, and all teh mfrs would have their own special patented location for it, which was way better than everyone elses.Posted 8 years agoMbnutMember
Frame Size Small Medium Large
Seat Tube (centre-top) 16″ 400mm 17.5″ 445mm 19.3″ 490mm
Top Tube Length 22.6″ 576mm 23.4″ 593mm 24.1″ 610mm
Head Angle 66.5° 66.5° 66.5°
Seat Angle 73° 73° 73°
Chainstay Length 428mm 428mm 428mm
BB Drop +8mm +8mm +8mm
Head Tube Length 100mm 105mm 120mm
Usual Height Range 5’5″ – 5’8″ 5’9″ – 6’0″ 6’0″ – 6’3″Posted 8 years ago
Stem Length 40-70mm 50-80mm 50-80mm
All measurements are static based on 150mm travel fork. 160mm suspension forks will slacken angles by 0.8 degrees. 140mm forks will steepen angles by 0.5 degrees.
The same reason that when you use a spanner to do up a nut, you create a torque on the nut through the application of a tangential force, and you receive a reaction force back through your hand. This continues through your body with reactions back from the ground etc. I highly doubt that even the idiotic US IP system would allow you to patent disc brake calliper locations with over a century of prior art…Posted 8 years ago
Forgot the forces through the axle! So you’re looking for the net vector force when calliper-disc, disc-axle, wheel-axle and wheel-ground forces are summed. Move the calliper and there is no doubt that the vector forces have to change. Change the vector force and you change the suspension action under braking. QEcan’tbebotheredtoD.Posted 8 years agojamesoSubscriber
Exactly. It’s about force vectors of the caliper and tyre contact patch forces and their influence at the axle – and how this relates to the pivot point. It’s more influential on a long travel / softer sprung, heavy-braking ability bike but moving the caliper can affect where the components of the brake reaction are directed. This means a variation in the directional forces at the axle that will have an effect on the suspension under braking. It’s been experimented with on DH bikes in the past and it’s a known effect on cars and motorbikes. You could get more variation by changing the pivot point some mm / degrees than the caliper position, but that’d affect more than just braking.Posted 8 years agocySubscriber
Most of the original questions seem to have been answered (including him buggering off to buy a Transition ;-)), but to summerise:
– Geometry is on the website. I’ll pop some reach/stack up soon if people want it. Never understand why wheelbase is such a big deal.Posted 7 years ago
– Delivery is mid-April for production frames. We’ll have a couple airfreighted earlier for photos and product launch at Bespoked Bristol show, plus our first demo at Calver on 14th April.
– The braking thing is evident from experiment, as Kelvin said. The Hemlock suffered horrendous brake jack in prototype form until I moved the caliper around. The Rocket is even better under braking because I’ve got the caliper even further forward, and the suspension works slightly differently. I’ll be honest that I don’t fully understand every last bit of physics involved, but just as with the whole front wheel ejection thing from a few years ago, the ‘ejection force’ vector is altered by the position of the caliper. It seems the further away from perpendicular to the suspension movement you can place that vector, the more neutral the braking response.
But “caliper forward” = “ejection force” pointing close to vertical?
If you look at the Cotic page (http://cotic.co.uk/product/rocket), it doesn’t look like the forces would be vertical. More like in line with the pivot point.Posted 7 years ago
So that the caliper would be acting on the rotor from directly above the axel. I could be wrong though! 🙂
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