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How does damper energy conversion relate to damper openness?

1463 Views 12 Replies 6 Participants Last post by Dougal, Jan 30, 2022

mmmmtnbkr1

Discussion Starter · Jan 27, 2022

If a suspension's damper is all the way open (past the limit of a normal adjuster - thought experiment here), such that it produces no damping force during compression, then the damper will convert no energy to heat. A damper set up normally will convert some energy to heat. A damper fully closed, so strongly that no oil flow can occur, will again convert no energy to heat.

Is the above true, and how does the curve look between the two extremes? Below are some example curves, but what does it really look like?

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Dougal

· #2 · Jan 27, 2022

Energy dissipated by a linear damper is proportional to speed and distance.

Basically:
work energy = force * distance.
Force = damping coefficient * damper speed.
So work energy = damping coefficient * speed * distance.

You can break that down further by including power = work/time and distance/time = speed.

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wolfmw

· #3 · Jan 28, 2022

this could go down a very dark rabbit hole of thermo/fluid dynamics...but I'm pretty sure the basic answer is "it depends"

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brentos

· #4 · Jan 28, 2022

I've gone down this rabbit hole thought process with regards to damper set points and efficiency on XC bikes.

The theoretical conclusion I've drawn is that the middle/platform positions are likely the least efficient as they have the heaviest damping. Efficiency benefits of the middle position over the open position would be driven by biomechanics improvements (rider system) benefits rather than bike system benefits.

I even did an experiment with an SRM on a Scott Genius. Running it in three positions over nine repeated climbs. Any efficiency benefit between the three set points was overwhelmed by other factors/noise in the test. Additionally, I had broken / crushed my pinkie finger earlier in the day, I didn't yet realize I had shattered the bugger to bits. It was quite painful, so that didn't help.

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wolfmw

· #5 · Jan 28, 2022

energy dissipation (=power=heat generation rate) is proportional to the square of fluid velocity (which is related to the velocity of the piston in a generic linear viscous damper), but I don’t think that necessarily relates directly to damper "openness".

force generated by the damper is a function of the fluid velocity and the size of the restriction oil has to flow through. But that force acts to slow the velocity (whole point of having a damper), so velocity is constantly changing rather than constant as in the simple example above. If you change the size of the restriction (opening/closing damping circuit), it changes the flow characteristics which affects the force generated and therefore how the damper slows the velocity and therefore how much heat is generated during the stroke.

If we simplify by assuming the damper experiences the same velocity and just vary the orifice area, a “more open” damper will actually dissipate less energy because the force generated is lower than the “more closed” case. (But this is a little unrealistic, since if the damper is doing it’s job it should be slowing the velocity)

My gut tells me that the graph would actually look more like a mirrored image of what OP shows as Option 3, ie high energy dissipation when highly damped, less when less damped. Which makes intuitive sense based on what a damper is supposed to do. Although I would argue that the end cases are never 0: if fully shut (no flow) the shock is going to be hydraulically locked unless there is a bypass (well, ok, I guess 0 velocity makes it 0, but not a useful 0). If fully open, you still have some viscous fluid flow which will dissipate energy.

But I don’t think it’s necessarily linear - probably could be mostly linear in some portion of the curve with correct design choices. But keep in mind a real damper may also have some combo of separately tuneable high speed/low speed/compression/rebound circuits. Have at it.

Regardless, it isn’t something that is robbing any useful energy from your ride.

Disclaimer: I know enough to be dangerous but without breaking out some old fluids textbooks may not know enough to be right! And I’m not that bored 🤓

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Dougal

· #6 · Jan 28, 2022

No need to get into the fluid mechanics unless you're shim tuning. For energy dissipation you can treat the damper as a black box that turns kinetic energy into heat according to it's damper curves, velocities and stroke distance.

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wolfmw

· #7 · Jan 28, 2022

Dougal said:
No need to get into the fluid mechanics unless you're shim tuning. For energy dissipation you can treat the damper as a black box that turns kinetic energy into heat according to it's damper curves, velocities and stroke distance.

right, we're trying to understand the output of the box... any examples of damper curves? that would probably answer the OP's question

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Dougal

· #8 · Jan 28, 2022

wolfmw said:
right, we're trying to understand the output of the box... any examples of damper curves? that would probably answer the OP's question

For this exercise just pretend a shock is linear to 800N compression and 3000N rebound at 1m/s.

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GH28

· #9 · Jan 29, 2022

The rebound stroke is what produces the most heat out of a damper (usually).

If the damper isn't converting the compression stroke into heat from being too stiff, it's passing the force through the system as transient force.

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KoolGrandWizardLuke

· #10 · Jan 30, 2022

GH28 said:
The rebound stroke is what produces the most heat out of a damper (usually).

If the damper isn't converting the compression stroke into heat from being too stiff, it's passing the force through the system as transient force.

This sounds like hearsay and I think I know where you heard it.

From my calculations this is only really relevant to tarmac based vehicles. OEM vehicles, or Motorsport vehicles for track/tarmac use. This is because for sealed surfaces dampers are usually tuned with more rebound damping force than compression damping force. There are a two main reasons for this.
1. Spring rates are usually higher for sealed surface use. Higher spring rate = higher rebound damping forces to control it.
2. Bottom out resistance is much less of a concern. The higher spring rate also contributes to this. So we need less compression damping force.

For unsealed surfaces the damper is tuned differently for a number of reasons.
1. The spring rates are generally lower. Lower spring rate = lower rebound damping forces to control it. Couple this with the fact that for rough surfaces we deliberately sacrifice handling for traction. Basically this means we lower the rebound damping forces even more (relative to the spring rate).
2. Bottom out resistance is a major concern as there are often bumps which exceed the maximum travel of the suspension.

So as you can see the ratio for unsealed vehicles often goes to higher compression damping force than rebound damping force.

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Dougal

· #11 · Jan 30, 2022

KoolGrandWizardLuke said:
This sounds like hearsay and I think I know where you heard it.

From my calculations this is only really relevant to tarmac based vehicles. OEM vehicles, or Motorsport vehicles for track/tarmac use. This is because for sealed surfaces dampers are usually tuned with more rebound damping force than compression damping force. There are a two main reasons for this.
1. Spring rates are usually higher for sealed surface use. Higher spring rate = higher rebound damping forces to control it.
2. Bottom out resistance is much less of a concern. The higher spring rate also contributes to this. So we need less compression damping force.

For unsealed surfaces the damper is tuned differently for a number of reasons.
1. The spring rates are generally lower. Lower spring rate = lower rebound damping forces to control it. Couple this with the fact that for rough surfaces we deliberately sacrifice handling for traction. Basically this means we lower the rebound damping forces even more (relative to the spring rate).
2. Bottom out resistance is a major concern as there are often bumps which exceed the maximum travel of the suspension.

So as you can see the ratio for unsealed vehicles often goes to higher compression damping force than rebound damping force.

Bike shock damping ratios are 2-3x higher on rebound than compression. On badly tuned stock shocks it can be 5x higher on rebound.

Fork damping ratios can be 8x higher on rebound than compression.

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KoolGrandWizardLuke

· #12 · Jan 30, 2022

Dougal said:
Bike shock damping ratios are 2-3x higher on rebound than compression. On badly tuned stock shocks it can be 5x higher on rebound.

Fork damping ratios can be 8x higher on rebound than compression.

Hi Dougal.
How do we define a comp/rebound ratio?

GH28 said "The rebound stroke is what produces the most heat out of a damper (usually)"
Do you know how much heat is typically generated from Compression vs Rebound forces on bicycles?

While looking at comp/reb ratios it can be misleading as one would presume that having a 1:2 ratio means rebound creates more heat. But actually the compression forces often create more heat than the rebound forces.
This is due to the fact that compression forces are often significantly higher than rebound forces.

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Dougal

· #13 · Jan 30, 2022

KoolGrandWizardLuke said:
Hi Dougal.
How do we define a comp/rebound ratio?

GH28 said "The rebound stroke is what produces the most heat out of a damper (usually)"
Do you know how much heat is typically generated from Compression vs Rebound forces on bicycles?

While looking at comp/reb ratios it can be misleading as one would presume that having a 1:2 ratio means rebound creates more heat. But actually the compression forces often create more heat than the rebound forces.
This is due to the fact that compression forces are often significantly higher than rebound forces.

On a fork it's always going to be rebound dominating the forces and the heat. On a rear shock it's going to depend on the tune and the terrain. If you can keep the rear wheel on the ground and sustain a lot of fast impacts you can make compression dominate.

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