BST Wheels, Suspension adjustment??

sushifly · Jul 30, 2020

So I finally bit the bullet and got a set of BST Rapid Tek wheels
Love them!!
but on first ride I noticed that my ride became more rough. Can feel a bit more chatter from the suspension
more from the back

MY stock V4 wheels were pretty heavy.
stock front: 8.75lbs , BST RapidTek front: 6.15lbs
stock rear : 10.7lbs , BST RapidTek rear: 5.3lbs

? do I decrease compression and rebound or increase it ?
or some combination??

vcyclenut · Jul 31, 2020

my non suspension expert brain says there is less weight to control so you need less dampening both ways

KurtP · Jul 31, 2020

lighten compression rebound slightly and if that doesnt solve it, increase rebound damping slightly. The lighter mass needs less control from the damper coming up, and needs more control from the damper going back out.

As the wheel approaches a bump, it has to compress the suspension to point X to compensate so the handle bars stay level. The lighter unsprung mass means the damper doesnt have to work as hard to handle the force coming in against the spring, hence lighten compression. On the return stroke, the lighter wheels mean the spring can now force the wheel back out more quickly, hence the damper has to slow the wheel down more.

Taz · Jul 31, 2020

Hmm.. I'm not a suspension expert, but given that less mass equals less inertia, the wheel will react quicker when external force is applied. Hence I would think both compression and rebound damping should be increased if one wants to maintain the same damping charecteristics.

KurtP · Jul 31, 2020

Taz said:
Hmm.. I'm not a suspension expert, but given that less mass equals less inertia, the wheel will react quicker when external force is applied. Hence I would think both compression and rebound damping should be increased if one wants to maintain the same damping charecteristics.

dont forget the spring in the equation. The damper is to control the motion/energy acting on the spring....In terms of compression, have to think equal and opposite reaction...the lighter wheel means less inertia, as you mentioned...but if the damper resistance stays the same, it means the wheel wont reach "point x" of the suspension stroke in time to respond to the bump and keep the handle bars level. so less damping resistance allows the lighter mass to stroke the spring to the compression required. For rebound, imagine 2 ball bearings, one double the weight of the other. launch them up from the same spring, the lighter ball goes higher and faster. thats the wheel hitting the pavement on the return stroke. more damping is required to slow the spring down and slow the return energy. Assuming the OP was happy with his suspension adjustments before the wheel change, then the end goal is to have the wheel hit "point x" on the up stroke and return to static sag on the down stroke at the same elapsed times as it did with the heavier wheels. so we need to resist compression less to allow it to hit point x in time y, and resist rebound more to prevent the wheel to return to static sag too quickly.

sushifly · Jul 31, 2020

KurtP said:
dont forget the spring in the equation. The damper is to control the motion/energy acting on the spring....In terms of compression, have to think equal and opposite reaction...the lighter wheel means less inertia, as you mentioned...but if the damper resistance stays the same, it means the wheel wont reach "point x" of the suspension stroke in time to respond to the bump and keep the handle bars level. so less damping resistance allows the lighter mass to stroke the spring to the compression required. For rebound, imagine 2 ball bearings, one double the weight of the other. launch them up from the same spring, the lighter ball goes higher and faster. thats the wheel hitting the pavement on the return stroke. more damping is required to slow the spring down and slow the return energy. Assuming the OP was happy with his suspension adjustments before the wheel change, then the end goal is to have the wheel hit "point x" on the up stroke and return to static sag on the down stroke at the same elapsed times as it did with the heavier wheels. so we need to resist compression less to allow it to hit point x in time y, and resist rebound more to prevent the wheel to return to static sag too quickly.

thank you! that makes a lot of sense!

KurtP · Jul 31, 2020

sushifly said:
thank you! that makes a lot of sense!

my pleasure!

start with a 1/4 or 1/2 turn on each and get it to where it feels as close to the same as you can remember. Thatll be good enough for the street until you can actually get everything dialed with a pro in terms of damping, preload, sag, and possibly spring rates.

sushifly · Jul 31, 2020

KurtP · Jul 31, 2020

sushifly said:
View attachment 33078

damn Thats smooth.....

Taz · Jul 31, 2020

I would still argue that compression damping should also be increased. When a force is applied that compresses the spring, for instance when the wheel hits a raised bump in the road, higher mass will slow down the rate of compression due to higher resistance to an external force. Ergo contributing to damping the rate at which the spring compresses. When mass is removed there will be less resistance to said external force and the spring will compress at a higher rate.

KurtP · Jul 31, 2020

Taz said:
I would still argue that compression damping should also be increased. When a force is applied that compresses the spring, for instance when the wheel hits a raised bump in the road, higher mass will slow down the rate of compression due to higher resistance to an external force. Ergo contributing to damping the rate at which the spring compresses. When mass is removed there will be less resistance to said external force and the spring will compress at a higher rate.

nope. The lighter mass of the wheel is carrying less energy into the spring, so...you need less damping to resist that movement. In this case, the velocity part of the vector of the compression event is staying the same.

the assertion you are making is that the speed of travel component of the compression vector moment is changing, but it hasnt, the mass changed, and thats where the energy change is coming from that we are accounting for. So with same velocity and less mass, we have less force, so we need less damping to control the compression vector.

does that help?

Jedclampet · Jul 31, 2020

I added BST's to my 959 that I was happy with at the time. It did feel a little off/busy and I added a click of rebound and all was great again. In my case it saved near 12 lbs since stock wheels were boat anchors, the switch backs were the most noticeable. E njoy

burton · Aug 1, 2020

KurtP said:
nope. The lighter mass of the wheel is carrying less energy into the spring, so...you need less damping to resist that movement. In this case, the velocity part of the vector of the compression event is staying the same.

the assertion you are making is that the speed of travel component of the compression vector moment is changing, but it hasnt, the mass changed, and thats where the energy change is coming from that we are accounting for. So with same velocity and less mass, we have less force, so we need less damping to control the compression vector.

does that help?

This is correct, Newton’s second law.... Force = mass x acceleration.

Since acceleration stays the same, but mass is reduced you are left with a smaller force.

Since you have less force on the spring you need less damping to control the smaller force.

craig bush · Aug 1, 2020

interesting discussion in here.

burton said:
This is correct, Newton’s second law.... Force = mass x acceleration.

Since acceleration stays the same, but mass is reduced you are left with a smaller force.

Since you have less force on the spring you need less damping to control the smaller force.

alternatively: a = F/m

mass and acceleration are inversely proportional - as mass increases, accel decreases if F is constant.

if the force (F) from a bump acting to displace the wheel is held constant (same bump, same vehicle speed), then wouldn't the resultant acceleration of the lighter carbon wheel be greater due to constant force acting on less mass?

KurtP · Aug 1, 2020

Taz said:
I would still argue that compression damping should also be increased. When a force is applied that compresses the spring, for instance when the wheel hits a raised bump in the road, higher mass will slow down the rate of compression due to higher resistance to an external force. Ergo contributing to damping the rate at which the spring compresses. When mass is removed there will be less resistance to said external force and the spring will compress at a higher rate.

also- here’s another way to think about it:

we need X amount of energy to compress the spring to position Y to compensate for the bump. That energy comes from the m

craig bush said:
interesting discussion in here.

alternatively: a = F/m

mass and acceleration are inversely proportional - as mass increases, accel decreases if F is constant.

if the force (F) from a bump acting to displace the wheel is held constant (same bump, same vehicle speed), then wouldn't the resultant acceleration of the lighter carbon wheel be greater due to constant force acting on less mass?

no. Because the force acceleration is mechanical and determined By the speed of the bike hitting the bump. The force on the wheel from the ground is the same, its the force against the spring from
the wheel that has changed.

craig bush · Aug 1, 2020

KurtP said:
also- here’s another way to think about it:

we need X amount of energy to compress the spring to position Y to compensate for the bump. That energy comes from the m

no. Because the force acceleration is mechanical and determined By the speed of the bike hitting the bump. The force on the wheel from the ground is the same, its the force against the spring from
the wheel that has changed.

i took a year of physics in college, but i'm no physicist so i'll accept that you're probably correct.

burton · Aug 1, 2020

KurtP said:
Because the force acceleration is mechanical and determined By the speed of the bike hitting the bump. The force on the wheel from the ground is the same, its the force against the spring from
the wheel that has changed.

Exactly, the only variable that has changed is mass, which results in less force to the spring

craig bush · Aug 1, 2020

burton said:
Exactly, the only variable that has changed is mass, which results in less force to the spring

ok, where am i going wrong, F = ma, right? and we're considering three separate but related force equations: F1 (force of bump deflecting wheel), F2 (force of heavy wheel compressing spring), and F3 (force of lighter wheel compressing spring), right?

if F1 is constant, then reducing wheel mass, results in greater acceleration of the lighter wheel by definition, right? so F2 = (M)(a), and F3 = (m)(A)

so without knowing the exact mass and accel, we cannot predict the relationship between F2 and F3. depending on the changing mass and accel, it's possible that F3 (lighter wheel) could be equal to, greater than, or less than F2. no?

craig bush · Aug 1, 2020

i did some additional reading. in the discussion thread below, the guy who seemed to be considered a suspension expert by other posters brought up an interesting point regarding low speed vs high speed damping, and the fact that compression/rebound clicks open/close needle orifices that control low speed damping, while high speed damping is controlled by the shim stacks. light weight wheels impact high-speed damping, not low, so if anything, they'd be more likely to require a shim stack change than comp/rebound adjuster changes.

interesting discussion either way.

https://www.speedzilla.com/threads/...id-you-change-after-lightweight-wheels.36444/

KurtP · Aug 1, 2020

we’re approaching page three of a physics discussion for what ultimately amounts to 1/4-1/2 turn of a damper adjuster.......

craig bush said:
ok, where am i going wrong, F = ma, right? and we're considering three separate but related force equations: F1 (force of bump deflecting wheel), F2 (force of heavy wheel compressing spring), and F3 (force of lighter wheel compressing spring), right?

if F1 is constant, then reducing wheel mass, results in greater acceleration of the lighter wheel by definition, right? so F2 = (M)(a), and F3 = (m)(A)

so without knowing the exact mass and accel, we cannot predict the relationship between F2 and F3. depending on the changing mass and accel, it's possible that F3 (lighter wheel) could be equal to, greater than, or less than F2. no?

youre looking at the right variables and on the right track.

we do know exact mass if we were to measure it. We do know exact velocity if we were to measure it.

this is the part that youre not going to like: The force actuating the spring is actually coming from the handle bars. the bump presses up on the wheel and the wheel presses back down on the bump. Equal and opposite reactions, right? The energy change from the wheel is coming from the handle bars pressing down on the wheel, and the wheel’s equal and opposite reaction back to the handle bars. The spring in between absorbs some of this. The damper controls it. So, now the wheels reduced mass pressing up means less energy, so less damping resistance is required.

again, its an interesting academic discussion; but in the end we’re only talking about a couple clicks of difference.

helpful?

BST Wheels, Suspension adjustment??

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