I would just like to say that it is obvious that I had my head up my ass in my previous posts and don't know what the hell I was thinking about. I'm enjoying the discussion though.
I'm with you brother. Basically it comes down to volume. The 25mm has a larger volume. So it takes longer to build pressure. There are many other factors that go into the equation such as piston design, needle designs, and of course shim stacks. In GENERAL terms all things being equal a 20m kit will create pressure quicker than a 25m kit because of the volume in the body.
(see bold). i don't think this statement is correct. especially when you're talking about rebound valves.... they dont care about the user's weight, they care about the weight of the spring and will ideally need to be modified to compensate for spring rate..... i think your statement is more applicable to compression valves, but even still it is quite a stretch to say that compression revalving is never necessary with a change in springs....
So why does it build less pressure? Frontal area of the piston? If you had a solid piston wouldn't it build pressure immediatly? The rod size wouldn't matter. This is part of answering why. This thread has all been about why since it was hijacked. Not what it does but why it does it.
yes, if you had a solid piston it would build pressure immediately. we're not talking about pistons and valving though. we're considering those factors to be equivalent. we're talking about the effect of a relatively smaller rod independent of other factors which could potentially affect pressure-building. it seems you're making this question more complicated than it actually is. it is just as simple as why a smaller orifice builds pressure faster than a larger orifice given equal force on each.... so to answer your WHY question: because flow is inhibited.
So if a solid piston will develop the same amount of pressure just as quickly if it was either a 20mm piston or a 25mm piston then why is the shim stack and the design of the piston ignored? I'm not making it more complicated than it actually is. I'm just not willing to accept someone telling me its this way just because. I want to know why it is one way or the other.
because although the piston will affect the pressure- it does not affect the pressure and the TRAVEL like the rod does. the pressure that the piston creates is sort of one-dimensional, while the pressure the rod creates is more two-dimensional since it's a function of TRAVEL as well as orifice-size.
But you just said that a solid piston would create equal amount of pressure no matter its size (20mm or 25mm).
ok.... ?? let's take pistons and shim OUT of this discussion for now. completely OUT. they only clutter the conceptualization. please explain what you DON'T understand about this. in a 25mm, it takes longer(more TRAVEL) to build pressure because the oil height is not raised as much for a given amount of time/TRAVEL as it would be for the same size rod in a 20mm cartridge.... the 20mm cartridge will build pressure faster because it will displace the oil higher than the 25mm would for the same amount of TRAVEL!!! oil higher faster = more pressure (less air) !!! what is so difficult about this?!
Ok lets make the pistons solid and of different sizes. The smaller one is pushing less oil. The larger one is pushing more oil. 25mm piston has 1963.495mm2 of surface area 20mm piston has 1256.63mm2 of surface area The 25mm piston displaces 4084.07 cubic mm of oil per 1mm of stroke The 20mm piston displaces 2638.93 cubic mm of oil per 1mm of stroke Now lets assume the other side of the cylinder the piston rides in is blocked off and there is a perfect seal between the cylinder and the piston which can stand all of the PSI generated from 1mm of travel. Which one is producing more pressure during 1mm of stroke?
dude, forget about the pistons. we're talking about the rods. in any event, for the sake of discussion, i would personally believe that they produce equal pressure with the parameters you specified. what do you reckon? we need pictures and/or videos. discussing this stuff with no images is giving me a headache. BTW, this is all i've been saying the whole time:
Needless to say I'm bored at work and want to redeem myself from the drivel I wrote earlier. Here is the math to back up Dr GoFast: Assumptions: Cartridge Length: 450mm Rod Diameter: 12.5mm Under compression of fork the rebound bypass valving has no effect on compression dampening. Compression valving does not release any pressure Rebound valve volume = negligible Compression of 25mm (~1 inch) Initial pressure = P1 = 100kPa (ambient pressure) Boyle’s Law = k = PV 20mm Cartridge: Volume of cartridge (Vc) = 3.14 * r^2 * L = 3.14 * 10mm^2 * 450mm = 141,372mm^3 Volume of displacement = 3.14 * r^2 * L = 3.14 * 6.25mm^2 * 25mm = 3068mm^3 Volume of cartridge after compression (Vc2)= 141,372mm^3 - 3068mm^3 = 138,304mm^3 A reduction of 2.17% P1Vc = k k = 100kPa * 141,372mm^3 = 0.1N/m^2 * 0.141372m^3 = 0.0141372N*m P2 = k / Vc2 = 0.0141372N*m / 0.138304m^3 = 0.102218N/m^2 = 102.218kPa 25mm Cartridge: Volume of cartridge (Vc) = 3.14 * r^2 * L = 3.14 * 12.5mm^2 * 450mm = 220,893mm^3 Volume of displacement = 3.14 * r^2 * L = 3.14 * 6.25mm^2 * 25mm = 3068mm^3 Volume of cartridge after compression (Vc2) = 220,893mm^3 - 3068mm^3 = 217,825mm^3 A reduction of 1.39% P1Vc = k k = 100kPa * 220,893mm^3 = 0.1N/m^2 * 0.220893m^3 = 0.0220893N*m P2 = k / Vc2 = 0.0220893N*m / 0.217825m^3 = 0.101408N/m^2 = 101.408kPa So, given all things equal except for cartidge size, the smaller cartridge develops less pressure therefore less pressure differential across the compression valve. This is only looking at the compression stroke. The principles for the rebound stroke are different.
I don't fully understand this discussion of how cartridge forks operate. Can anyone point me to a detailed description of the oil flows in cartridge forks during compression and rebound? The closest thing I can find is a Sportrider article, which is not very detailed. The diagram below is of the overal fork: The diagram of the rebound piston and compression valve in the article (shown below) does not make it clear how the check valves work. From the Traxxion article and Hordboy's input, it seems that the check valve in the rebound piston opens during the compression stroke, allowing oil to pass easily through the piston. I don't see much point for the rebound piston check valve unless this is true. This would imply that only the damping rod displaces the oil that flows through the compression valving. Can anyone explain why the rebound piston check valve exists? Why not use a compression valving stack on the piston instead (similar to most rear shocks)? This would allow the compression valving at the base of the cartridge to be just a fine tuning adjustment Alternately, why not just push the volume of oil swept by the entire piston area through the compression valve? This would reduce overall pressures in the cartridge, since to achieve the same damping force, you need lower pressure over the large piston area vs. just the shaft area. Finally, as far as how quickly the pressure rises in the cartridge (under compression), I would think that a larger diameter cartridge would have a slightly slower pressure rise. The pressure rise in the cartridge would be dependent upon the elasticity of the cartridge body, the compressiblity of the oil, and compressiblity of any air in the oil. My guess is that air in the oil is the most important factor. The larger volume of oil (+air) in the 25mm cartridge would tend to be more elastic and delay the pressure rise. I would think this delay would be pretty clear from the "hysteresis" shown on a shock dyno chart. Does anyone know where to see some representative dyno charts of forks? In any event, I'd appreciate any links to more technical information about cartridge forks! Thanks. :up:
Let me start off by saying that I am not an expert as I have already demonstrated in this thread and unfortunately I did not stay at a Holiday Inn Express last night. But I'm still going to give this a shot. The reason the rebound piston as the bypass valve for the compression is stroke is to prevent a vacuum. In theory, the forks would all but lock up if this was the case. Oil needs to get to the back side of the valve. Also, if there is no oil on the back side of the valve, then there is no oil to flow through on the rebound stroke to provide dampening. This also explains why the compression stack is not on the backside of the valve like a shock. A shock is different in that is gas charged and the gas is seperated from the oil by a bladder. Again, my thoughts, not proven facts. I'm not going to go into the reasons why unless you want me to, but the physical properties you described should not play a measurable role in the pressure build unless there is something wrong with the fork. The difference in pressure rise is directly attributable to the relastionship of pressure and affected volume. If anyone thinks I'm off base here, please let me know. Suspension facinates me and I'm trying to learn and understand as much as I can but I obviously have been haveing some problems.
That's correct. Some forks are built this way to an extent, with a valving stack built on the backside of the rebound valve, called a mid-valve. You can't get real aggressive with the stack, or cavitation becomes a problem, unless you pressurize the cartridge. I'm not sure I understand what you're saying, but I think that's been been covered above? I'm not much for book learnin', but it all comes down to the rod size/cartridge size ratio to me. I've been wrong plenty of times in my life though.
