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[Videos] MTB Rear suspension series

Udi

RM Chief Ornithologist
Mar 14, 2005
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1,213
@andrextr
a) I'm pretty sure that's not correct (if I understand your English correctly) - an air shock has a relatively high breakaway force (independent of the damper), because unlike a coil spring which at rest is at full extension, an air spring needs to generate an unbalanced force to extend the shock to the zero travel position. So you can't say "zero breakaway force" pointing to the 0 travel position, since that is completely untrue - if that were true, the shock would sink into its travel by itself. A force differential occurs because air is acting on a smaller surface area on the negative side of the piston compared to the positive side, and a pressure differential (+resultant force differential) occurs because the equalisation point isn't at topout.

On a coil shock, the damper (and spring preload, which is negligible) will be the only source of breakaway force. However on an air shock, you have the damper force AND a substantial force generated by the air spring itself, which together create the total breakaway force. There are also greater friction forces from the air seals on an air shock, which will add breakaway force at any point in the travel.

The high initial rate of the air shocks in this graph stems from this problem (yes I realise it's not showing breakaway force directly, but the high spring rate at 0 travel is a good indication):

Credit: Vorsprung Suspension

b) Where are you actually obtaining this data? I think you need to be more careful with the accuracy of data sources before claiming it as fact, there is a lot of misinformation out there on this topic currently (including from manufacturers). I think if you're going to post an air spring curve it should be referenced - and ideally verified as independently dyno tested.

In this example (actual measurements of specific cans), you can see the force increase (over linear) continues much deeper into the travel than in your graph:

Credit: Vorsprung Suspension

For example - you say "eg. Evol X2 / Debonair" - have you actually got measurement data for both of these cans? How do you know they aren't significantly different?

I understand you're posting basic data for the sake of educating, but if that is your goal, I think you need to use very accurate data (from valid sources) - because otherwise you can end up misinforming people on topics like these. Not having a go at you here, but while I think approximations are OK for personal use, they can be misleading for educational use.

Please correct me if I have misunderstood something, but when I read what you wrote, I understood "an air shock has the same breakaway force as a coil shock, since all breakaway force comes from the damper".
 
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Udi

RM Chief Ornithologist
Mar 14, 2005
4,918
1,213
@everyone else
PSA on Linkage. Before some genius decides to have a go at me/others for doing the same thing with Linkage data, please realise that the accuracy of that particular data is sufficiently high (for comparing *most* bikes) when an experienced user is creating the drawings - as they will know how to minimise inaccuracies.

To further aid this, if the analysis (separate from drawing) is done by an experienced user, they will know the error vectors for different designs (i.e. the potential magnitudes AND direction of error) - which makes it easy to notice and account for anomalies. For example, on a linkaged singlepivot, acceleration and braking performance is going to be represented with fairly high accuracy - as small variations in pivot point don't cause large variations in these graphs. In cases where there is a tendency for inaccuracy, it's quite easy to check large anomalies by physical link measurement (or other methods). At least personally, I do try to look out for potential anomalies before making public commentary on Linkage data.

There's a HUGE difference between someone like dw noticing the Sunday linkage graph has an anomaly in the leverage curve over a small portion of travel (since he has the real curve in front of him, from that one time he designed the bike himself), and someone like djjohnr/kidwoo writing off an accurate and valid theory about leverage appropriateness for air spring curves based on "I once read on the internet that Linkage is garbage in/garbage out".

No. Just because you opened the program three times and read something about it on the internet, doesn't mean you suddenly understand the program in its entirety + have an intimate understanding of the physics behind handling/chassis design on top. I'm just writing this so I can link to it next time some idiot throws a blanket statement like that. :)
 

andrextr

Chimp
Aug 6, 2007
63
67
Portugal
@andrextr
a) I'm pretty sure that's not correct (if I understand your English correctly) - an air shock has a relatively high breakaway force (independent of the damper), because unlike a coil spring which at rest is at full extension, an air spring needs to generate an unbalanced force to extend the shock to the zero travel position. So you can't say "zero breakaway force" pointing to the 0 travel position, since that is completely untrue - if that were true, the shock would sink into its travel by itself. A force differential occurs because air is acting on a smaller surface area on the negative side of the piston compared to the positive side, and a pressure differential (+resultant force differential) occurs because the equalisation point isn't at topout.

On a coil shock, the damper (and spring preload, which is negligible) will be the only source of breakaway force. However on an air shock, you have the damper force AND a substantial force generated by the air spring itself, which together create the total breakaway force. There are also greater friction forces from the air seals on an air shock, which will add breakaway force at any point in the travel.
Yes Udi, the main chamber of an air shocks produce a high breakaway force. And the role of the negative chamber is to reduce that breakaway. When the equalizing port is located on the correct spot in the main chamber, you can achieve a balance between the forces between negative and positive chamber, in that case they cancel each other and you have a ZERO breakaway for the Spring. And in fact this seems to happen! I already collected 8 breakaway values from people, and there is no statistically significant difference between Air and Coil shocks breakaways. I hope that more people can contribute with data to have stronger statistical significance. So, for a modern shock, the data suggests that the breakaway comes mainly from the IFP and not from the Spring (either coil or air). (PS: on the breakaway video, my air shock is an old model, and it needs about 35 kg of force to start compress, I believe that my shock don't have the equalizing port, at least I don't remember to saw it when I opened the shock few years ago).



The graph that I published was just for educational proposes (not real values were used, just an arbitrary example). The reference to EVOL and DEBONAIR was just to people understand and have an example of what is a big negative air chamber. In that graph I only changed the negative chamber size/volume, and kept the SAG constant. Just to show people how negative chamber volume affects the air spring curve. As you can see in the above graph (Font: Rockshox, I believe), it looks similar to mine in the initial part of the travel. However, at the final part they look different. This is because debonair air can also has a higher volume on the main chamber, so it has a lower compression ratio. On my graph I kept the compression ratio constant as mentioned on the legend.


I will talk about this on a future video, and then, as usual, I will try to be most rigorous as possible. Nevertheless, I did an excel file datasheet to allow to people to play with values (pressures, chambers sizes, diameters, port location, compression ratios, etc) and see how that affects the air spring curve. I also have the real values from a technical drawing of a Fox RP23 2009 shock. Anyway, here it is some graphs from the excel datasheet (using arbitrary, but plausible, values). I will share this excel file when everything is polished :) So, I think that more educative than this is quite difficult :) At least is the best I can do :)

Excel graph.png

(Graph obtained in my excel datasheet. The blue line is the force produced by the main chamber, and the orange line is the force produced by the negative chamber. Since they have an opposite direction, the net result is the green line, which is the final Air spring curve)
 
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Flo33

Turbo Monkey
Mar 3, 2015
2,146
1,376
Styria
Isn't the key factor in the Debon/Corset/Evol cans the relatively slower increase in volume of the neg chamber and therefore slower decrease in pressure(=resulting force at backside of piston)?

But the piston area on the neg side will always be smaller, as will the volume be too, compared to the pos side.

Can't think of a resulting force of cero at the piston.
 

Udi

RM Chief Ornithologist
Mar 14, 2005
4,918
1,213
When the equalizing port is located on the correct spot in the main chamber, you can achieve a balance between the forces between negative and positive chamber, in that case they cancel each other and you have a ZERO breakaway for the Spring.
Yes that may be possible - however - it's difficult to combine what you describe with a linear spring curve as well (since the spring curve is dependent on the same factors - chamber sizing and eq port location), and most manufacturers make some amount of sacrifice in breakaway performance to get a better spring curve (especially within the size / space restrictions for a can). From graphs like the one Vorsprung provided (a dyno output, not a theoretical one), it's also very obvious that manufacturers don't necessarily have the best algorithm for generating spring curves, so it's possible that you're actually ahead of some of them here.

I honestly don't think measuring this the way you're doing (with various scales, with varying leverage curves on different frames, with varying pressures in the shocks, by various people) is a sufficiently accurate way of comparing this.

Also - air shocks have numerous extra seals which each provide a breakaway (friction) force of their own, which is increased by a function of the pressure that is applied to them by the air spring. So they can't have ZERO breakway force, even in the spring component alone - and this value will almost always be noticeably higher than a coil shock. Obviously this is at every point in the travel - but it happens at 0 travel too.

As a sidenote, I've noticed most air shocks will have a very small amount of free movement (~1mm, which could be mistaken for the shock having successfully "broken away") however it is followed by a noticeable spike in spring force - and I think this is because of air seal flexure.

I also have the real values from a technical drawing of a Fox RP23 2009 shock.
I believe this data may be inaccurate (after discussions with Vorsprung, who obtained the same drawing if I remember correctly) about testing the same shock on the dyno and obtaining substantially different spring curves and forces. This is why I said often manufacturer data can be unreliable, and you really need to be using verified dyno measurements when making claims about very fine elements like breakaway force.

I don't think statistical evidence from multiple third parties is sufficient proof of anything to be honest, there's just too much sources of error at that point. This is my opinion and not to disrespect your work. I just think finer elements like this need to be measured directly at the shock (ideally in a way that normalization is not needed).

Anyway I'll stop hijacking your thread now, appreciate you sharing.
 

andrextr

Chimp
Aug 6, 2007
63
67
Portugal
@Udi I understand your concerns, I also have them. But as you can imagine, I don't have a dyno in my living room, and the only shock that I have access is my old DHX4... This is why I tried to recruit the help of the people, using the assumption that the higher the amount of values collected the more diluted are the external variables and the better the statistical significance... But I will only talk about this if, and when, I get a substantial amount of data.

Nevertheless, regarding the air spring, I will use theoretical graphs to explain the principles behind the negative chamber volume and pressure. I don't want to say that X brand has that X curve. I just want to say what are the components of the air spring curve (positive chamber minus negative chamber), and how manipulating the pressure & volumes in each compartment affects the air spring shape. This is a safe and correct way to explain it, and I have the tools to do it, either the Linkage software, or the excel datasheet which will be available to everyone for scrutiny :)

Bye :)
 

Udi

RM Chief Ornithologist
Mar 14, 2005
4,918
1,213
Isn't the key factor in the Debon/Corset/Evol cans the relatively slower increase in volume of the neg chamber and therefore slower decrease in pressure(=resulting force at backside of piston)?

But the piston area on the neg side will always be smaller, as will the volume be too, compared to the pos side.
Can't think of a resulting force of cero at the piston.
Yeah that's correct, however as @andrextr says, you can bring this force very low depending on where you place the equalisation port - because as the piston extends towards topout after passing the EQ port, the negative pressure increases (substantially near topout) as the positive pressure drops. Modern air cans place the EQ port deeper into the stroke to increase this effect, reducing initial spring force substantially.

However restrictions on can geometry amongst other things (like the need for spring curve linearity) mean in practice you can't have everything, which is why air spring curves (even on modern cans) have substantial nonlinearity compared to a coil spring, and thus require different leverage curves on frames etc.

I don't think air shocks in real life have zero breakaway force, and this could be easily verified on a shock dyno by removing the damper oil and IFP.
 

Udi

RM Chief Ornithologist
Mar 14, 2005
4,918
1,213
Nevertheless, regarding the air spring, I will use theoretical graphs to explain the principles behind the negative chamber volume and pressure. I don't want to say that X brand has that X curve. I just want to say what are the components of the air spring curve (positive chamber minus negative chamber), and how manipulating the pressure & volumes in each compartment affects the air spring shape. This is a safe and correct way to explain it, and I have the tools to do it, either the Linkage software, or the excel datasheet which will be available to everyone for scrutiny :)
Fair enough, I understand and respect that. :)
Also appreciate that you don't have a dyno in your living room, neither do I.

Look forward to your spreadsheets etc.
 

djjohnr

Turbo Monkey
Apr 21, 2002
3,122
1,811
Northern California
Udi

I don’t say this anywhere:

“someone like djjohnr/kidwoo writing off an accurate and valid theory about leverage appropriateness for air spring curves”

I said exactly this:

“1) Everyone in this discussion is using a graph someone created looking at pictures (super accurate yeah?)

2) I pointed out that based on the super accurate graphs data set, the delta between the point A and point B amounts to 2% over 25mm, which I propose no one could actually feel vs dead linear

AND

You can't get highly accurate measurements from a photo at the scales we're talking about. I think those graphs are decent approximations but not gospel.

AND

“ some of us feel a linear initial rate with an air shock using a new larger can isn't much of an impediment, YMMV obviously”

These comments were around a specific graph and bike, and reflect that 1) It’s very difficult to get highly accurate measurements from a photograph (due to a number of factors including perspective and lens distortion), and those inaccuracies can throw off the exact values of the graph 2) if you assume the values are correct in the graph in question, they only show a 2% change in rate over 25mm and 3) I doubt most people could feel the difference between 0% and 2% over 25mm.

One thing I haven’t done is use language to belittle you, or be otherwise disrespectful. Maybe you aren’t aware of this so I’ll point it out, phrases like “I'm just writing this so I can link to it next time some idiot throws a blanket statement like that” is a poor and disrespectful way of discussing things with others. Then again, maybe you just don’t care, because it’s just the internet right?
 

andrextr

Chimp
Aug 6, 2007
63
67
Portugal
Playing around with the datasheet (I still have to do a proofreading to the formulas, but so far it seems to be working good)...

PROOF OF CONCEPT:

Influence of negative chamber length (volume) on the spring curve. All parameters and pressure were kept fixed. As you can see, negative chamber length/volume affects mainly the initial suppleness and linearity. High negative chamber volume produces a very linear and coil-like effect. (Note: The SAG was NOT kept fixed on this simulation. If I kept the SAG fixed, you will notice that the higher negative volume produces a more mid-stroke support similar to a coil, and in similar way to Corset, Debonairs and EVOLs air spring curves).
Negative volume.png


While the main chamber volume affects the final progressivity (as everyone already know).
Main volume.png


@Udi I also thought that breakaway of air spring were far from zero, but I'm actually changing my mind on this point... By placing the transfer EQ port higher on the main chamber you can have a close to zero breakaway, even considering that the negative chamber has a smaller contact surface with the "piston".

Enough of nerdy things for today :D
 
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Udi

RM Chief Ornithologist
Mar 14, 2005
4,918
1,213
One thing I haven’t done is use language to belittle you, or be otherwise disrespectful. Maybe you aren’t aware of this so I’ll point it out, phrases like “I'm just writing this so I can link to it next time some idiot throws a blanket statement like that” is a poor and disrespectful way of discussing things with others. Then again, maybe you just don’t care, because it’s just the internet right?
In case you didn't notice, I usually respond to things in a fairly logical fashion - so if I belittle someone, it is because I intended to. Often it's in jest, but this is one of the very rare cases where it isn't.

Why? Because I find it "disrespectful" to use your own words, when someone waters down (i.e. "belittles") a scientifically proven point, verified by many engineers in this specific field (at least three of whom post on this very forum) to something of supposed "insignificance" based on pseudoscience combined with exaggerated claims about the software used.

While I respect rider opinions on what they can and cannot feel, these are subjective and unreliable measures when it comes to actual performance optimisation. I gave you specific examples in the thread - but to reiterate them - you kept writing off my claims based on "how can you feel a digression of x.xx, it's basically linear" and "linkage is inaccurate, I've used it myself" while I continued to try and explain that the flatline was the issue in itself (i.e. even if there was zero digression, it's still a problem). In fact it would appear that you STILL don't understand the problem - if you think that the 2% digressive change in rate is the issue - it's not. It's the fact that there isn't an aggressive progressive change in rate in the same portion of travel. This is not my opinion. This is engineering fact, and a simple application of a moment arm to reduce an undesirable, dyno-verified force peak.

Believe it or not, before making the comments about the Transition, I verified that the digressive spike was present on 4 other Transition frames of the same design, and actually knew (from a bike I mapped myself years ago) that the TR450 actually had the same trait as well - so it was a well established trend within the brand for whatever reason. Finally, the creator of the graphs then posted that they were taken from blueprints and thus perfectly accurate - which came as no surprise to me. I think you'll find blueprints are free of lens distortion. Finally, your generalisations to me suggest that you aren't aware of the specifics of potential inaccuracies in Linkage and how much effect they can have on each curve - it varies substantially between various points, links, and graphs.

Finally, as a practical measure - just so you're aware, every couple weeks, I get a random PM in my inbox from someone who owns one of the bikes I've potentially ripped on, either asking for a suggestion on how to alleviate one of the problems that they too have noticed, or thanking me after trying a solution that I suggested on the open forum and enjoying their ride more as a result. I post this stuff because I find it interesting, and it's potentially useful to people who are trying to solve the same problems I am.

However if I have to constantly defend myself because someone wants to argue scientific data of reasonable accuracy because someone wants to wank on about an issue over 2% of digression, while being ignorant of the fact that the real issue might be the fact that a 10% progression is missing, it becomes a huge strain on my time. This is what I referred to as psuedoscience in case you were wondering - it's misleading because people read the 2% and think "oh yeah, that is a small number, Udi is wanking on about nothing". Obviously not the case.

The bottom line here - just because you either don't feel an issue, or are happy with something less than perfect (which is reasonable) it doesn't mean that everyone else is - and watering down something that you don't understand 100% based on a few random facts and your subjective, non-quantitative testing - is something I find offensive. It's the sort of thing that makes me want to post less.
 
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djjohnr

Turbo Monkey
Apr 21, 2002
3,122
1,811
Northern California
Show me where I refute or water down a scientific point. Use quotes. While you like to post shit for fun on a forum, I've been running software product management for 15 years. When developing products you always need measures of what is acceptable, and often times that's based on user reactions to designs tested by a sample of your population. That's how we establish standards. We could aim to get a particular system response to zero milliseconds, but the reality is at sub one second it doesn't make a difference to the user. If you have no threshold point then your argument is useless, and is the kind of thing I'd hear from a pedantic junior level employee.
 

Udi

RM Chief Ornithologist
Mar 14, 2005
4,918
1,213
Show me where I refute or water down a scientific point. Use quotes. While you like to post shit for fun on a forum, I've been running software product management for 15 years.
Software product management /= engineering dynamics, physics, handling and chassis design. I've given you plenty of examples that you are making an active choice to be blind to. You're a waste of my time.

At least @kidwoo is funny. :)
 

Flo33

Turbo Monkey
Mar 3, 2015
2,146
1,376
Styria
Yeah that's correct, however as @andrextr says, you can bring this force very low depending on where you place the equalisation port - because as the piston extends towards topout after passing the EQ port, the negative pressure increases (substantially near topout) as the positive pressure drops. Modern air cans place the EQ port deeper into the stroke to increase this effect, reducing initial spring force substantially.

However restrictions on can geometry amongst other things (like the need for spring curve linearity) mean in practice you can't have everything, which is why air spring curves (even on modern cans) have substantial nonlinearity compared to a coil spring, and thus require different leverage curves on frames etc.

I don't think air shocks in real life have zero breakaway force, and this could be easily verified on a shock dyno by removing the damper oil and IFP.
Got it, thx.

In theory it would be possible, but at the current linkage designs floating around not very sensible. You would either bottom out all the time because of the missing amount of progressive behavior of the air spring's positive chamber or blow the seals out.

I had a fact sheet somewhere, where I started to do some calculations myself when I received my Corset, because I was thinking about this stuff back then too. If I remember correctly the piston area size difference was quite huge like 1/10 or something. @andrextr must have the numbers. So while the smaller volume of the neg chamber helps because of the higher rate of progression this stays the same.
 

Udi

RM Chief Ornithologist
Mar 14, 2005
4,918
1,213
I had a fact sheet somewhere, where I started to do some calculations myself when I received my Corset, because I was thinking about this stuff back then too. If I remember correctly the piston area size difference was quite huge like 1/10 or something.
Yeah there's a large piston area differential.
With regards to modelling, Vorsprung spoke to me about this in the early stages of the design of that can, and there was a problem where a lot of current designs were based around adiabatic assumptions and calculations, however in actual fact the dynamic heat transfer of the system play a part in the resultant air spring curves too - if I remember correctly a substantial part - which is why there's deviations from reality in even some of the Fox internal documentation.

Accurate calculation of this stuff requires a strong background in thermodynamics, and even then, I think you really need concurrent dyno testing to help ensure that your model is correct. It's very easy to build an incomplete or inaccurate model of these scenarios and assume it's correct, but building a complete model involved many months of work, possibly over a year if I remember correctly.

I think what we're discussing here is grossly simplified, which is okay for basic understanding and discussion, but I think it's important to note there's a lot more to it.

No data about your threshold? Not smart enough to know that it's applicable across product types? Get out of my office junior twat.
I'm sure the work you do is very important. :)

 

Flo33

Turbo Monkey
Mar 3, 2015
2,146
1,376
Styria
I think what we're discussing here is grossly simplified, which is okay for basic understanding and discussion, but I think it's important to note there's a lot more to it.
That's my savior line there - I'm an engineer myself, unfortunately from the wrong field of, which is civil works. But I'm always curious to know how stuff works, at a basic level at least. And I try to understand why.

Accurate calculation of this stuff requires a strong background in thermodynamics
That would have been my next question... :cheers:
 

andrextr

Chimp
Aug 6, 2007
63
67
Portugal
@Udi, yes, compression of air springs are an adiabatic process because the temperature rises during the compression. However, the amount of temperature rising is dependent on the compression velocity. This is why the air springs are also speed-sensitive, contrary to coil springs (under fast velocities they became more progressive due to the temperature increase). Nevertheless, if you assume a slow rate compression, the rise of temperature can be neglected, and you can assume the classical equations of a perfect gas under an isothermal expansion/compression (PV=nRT). So, there is no much need to over-complicate this topic at least for educational proposes.

@Flo33 Actually, its a kind of the other way around...I will try to convince you... Increasing the volume of negative chamber does not have a huge affect on the bottom-out force. Check-out my previous graphs. I changed the negative chamber length between 1mm and 30mm (30x variation), and the main effect is only on the initial stiffness of the spring. Indeed bigger negative air chambers smooth the initial part of the travel (which is characteristically stiffer on most of the air shocks out there). By making it more smooth, you need to apply more pressure on the main chamber to achieve the same SAG. Thus, as a side effect of more pressure in the main chamber, the bottom-out force also increases proportionally. This is why this new air cans (Debonairs, etc) besides using a larger negative volume, also uses a larger positive volume (to reduce the increase in bottom-out).

And yes, the breakaways of an air spring are very close to zero. Indeed in most cases you can start compressing the air shock just by applying pressure on the saddle with one finger (such also happens with coil shocks). So, again, the breakaway of modern shocks seems to came mostly from the IFP (as I showed on my video without the coil spring). I already did the test on a couple of bikes (air vs coil), and didn't find much difference in the force that I have to apply in the saddle in order to start compressing the shock. Some years ago this was not true. But now, it seems that most manufacturers nailed it.

I hope I could convinced you.
Bye.
 
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andrextr

Chimp
Aug 6, 2007
63
67
Portugal
Giant Reign vs Specialized Enduro linkage analysis



Demystifying some myths from Maestro Suspension....


If you want to laugh a bit, just watch this "great" video from Giant.... lol :D

 

Jm_

sled dog's bollocks
Jan 14, 2002
20,416
10,890
AK
Giant Reign vs Specialized Enduro linkage analysis
Something is missing there. The E29 I had pedaled just fine on level ground, pretty well, acceleration, etc, but uphill, it squatted like crazy, the steeper the hill, the worse the squat, to the point where it felt like a sponge.
 

andrextr

Chimp
Aug 6, 2007
63
67
Portugal
Something is missing there. The E29 I had pedaled just fine on level ground, pretty well, acceleration, etc, but uphill, it squatted like crazy, the steeper the hill, the worse the squat, to the point where it felt like a sponge.
Hi Jm! I will talk more about anti-squats in a future episode, but anti-squat only eliminates the squat (bob) caused by horizontal acceleration. If you have a bad pedaling technique, eg: your body is not stable under pedaling and moves up and down, this creates additional vertical forces, and in that case a 100% anti-squat is not enough to eliminate the bobbing. If you pedal on foot (not seated) this creates a lot of vertical body movement.

Or, in alternative, you had a very fast rebound setting on your shock (which amplifies the amplitude of low speed oscillations created by the pedalling). Thus, a fast rebound setting increases the amount of bob, which is annoying. (PS: I'm not saying that a correct rebound will eliminate pedal bob, it only dampen and decrease the oscillations). Check my Ep.5 about rebound here:

Bye
 

Kanye West

220# bag of hacktastic
Aug 31, 2006
3,768
501
Software product management /= engineering dynamics, physics, handling and chassis design. I've given you plenty of examples that you are making an active choice to be blind to. You're a waste of my time.

At least @kidwoo is funny. :)
I will back up Udi here on this one...this comes from many years of experience dealing with software product managers or program managers. I've found Udi's input to be relevant, factual and in proper context in the discussions that are being referenced.


Andre - your system of measurement with the scale is allowing a LOT of other parameters to come into play and move around and play with your breakaway force numbers. The tire, the bushings, the linkage for starters. While I appreciate the ingenuity and it's great for a gut-check, if you want accurate static friction data, and want to be able to compare SHOCK to SHOCK, you need to isolate the shock on a load frame and start breaking the system down component by component (full system, full system minus air spring, full system with damper drained/IFP removed, etc). It would be interesting data. I'd love to see those numbers published for every major shock on the market.

It is good you're seeing consistency with that method from other sources though.

Also, remember that the IFP "breakaway force" is still basically just working as a gas spring, and the air pressure has a minimum level at full extension to prevent the low pressure side of the piston from dropping below 0psi. This is separate from friction. If you were going to quantify the exact static friction from the IFP itself, you'd need to take a F-d curve of the shock itself, with no charge behind it (probably remove valve core or endcap and only measure on the compression stroke), and assume that the static friction from the main piston band, and the shaft seal/bushing were much much less than that of the IFP seal. Pretty sure that isn't always the case, depending on the shock.

It would be a lot of fun to validate static loading versus dynamic loading of these shock air springs too. I suspect their curves would be a tad different under higher frequency loading due to the adiabatic element of the system.

If I get really bored one day maybe I'll machine some eyelet adapters to run some of these things on our Instron/MTS load frames at work. I've been meaning to make some hardware for testing moto fork/shock springs and full moto forks to validate inner/outer chamber static forces.


I still remain VERY skeptical of the Arma shock that claims to only require 40psi or so of charge. There must be a LOT of bleed in both ends of the main chamber if it's supposed to prevent cavitation, in which case it would be pretty useless working with 250#-600# springs in rebound. The springrates size the system in this case. The springrates call for X amount of maximum rebound damping at the top end of the range, which calls for X bleed and valve stiffness, which sizes the valve and valve diameter, which sizes the pressure requirements for a given oil. It almost looks like the bridge of the shock is built with a super restrictive port at the end of the main chamber, which would be a ghetto way of helping with the pressure drop across the main piston. Or their compression damping forces are just WAY low in the system as a whole, and the pressure drop isn't that large. I have no idea how low they would have to be, but I expect it would have to be almost non-existent.
 

Happymtb.fr

Turbo Monkey
Feb 9, 2016
2,084
1,469
SWE
I did some experiment with my monarch rt3 + (tune M/M) since it was time for a service. I used the same method as Andre with a scale but with only the rear shock.
- after realising the pressure and equalizing both chambers, open comp: 12-13 kg
- same as above but after removing the debonair air can: 15-16 kg
- same as above but with firm comp: 18 kg
- same as above but with open comp and released IFP pressure: 4 kg

I would say that it is difficult to get accurate readings with this method but the order of magnitude should be ok.
 
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Jm_

sled dog's bollocks
Jan 14, 2002
20,416
10,890
AK
Something is missing there. The E29 I had pedaled just fine on level ground, pretty well, acceleration, etc, but uphill, it squatted like crazy, the steeper the hill, the worse the squat, to the point where it felt like a sponge.
I think I figured it out (with some days of riding).

In a static world, those bikes that have 130-100% of AS would be just fine pedaling uphill, this means that you have a completely smooth surface with no grade changes and the suspension never activates.

The issue is many of those bikes have a falling AS curve/rate that drops off rapidly, as you go through say half the travel, the amount of AS goes way way down, 30% or less (just throwing that out, but common with many linkage bikes).

As you ride over bumps, roots, etc, your suspension compresses, if you try to keep pedaling, all of a sudden it's a wet-mattress effect as your pedaling input keeps the shock compressed. Probably not much of an affect downhill, but uphill pretty huge. In reality, the suspension is working all the time, maybe a soft 2" bump, or a 4" root, the suspension activates to a large extent, maybe going through 60-75% of travel on rocky uphills, but even on more moderate ones it probably still uses at least half the travel frequently. Possibly a little bit of weight-transfer "bob" will even start to affect this, depending on the bike of course. That's where it's not really telling the full story, because our rides are not static and this is where I notice that I can just "keep pedaling" and accelerating over the bumps on a bike with a 100-130% flat AS line.

From the video and website, the reign isn't a great example, because it has a significantly falling AS curve as well, so both the reign and the Enduro would suffer from the same basic issue, with the enduro being a little less through the travel according to the figures. Look at Yeti or modern DWs to see the difference.

The linkage website seems to only be qualifying the "pedaling" or "efficiency" at the sag point.
 
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andrextr

Chimp
Aug 6, 2007
63
67
Portugal
Hi @Jm_

Yes, you are right, the ideal AS should be stable on the first 50% of travel at least (pedaling zone), and then drop down (like Santa Cruz VPP bikes for instance). The reason that AS should fall at the end of the travel is to decrease the amount of pedal kickback (and because you don't need too much AS on the final part of the travel).

However, a 100% AS will only eliminate the bob created by the horizontal acceleration forces. A 100% AS is not enough to eliminate the bob caused by the vertical movement of the rider on the saddle (such as pedaling with bad technique or pedaling on foot / sprint). With good pedaling technique you can reduce this vertical component, but it's not possible to fully eliminate it, since it's part of our pedaling bio-mechanic motion. There are some companies that defend higher anti.squats, such as 150% because of that.

Another very important aspect is the rebound speed. A fast rebound can amplify the oscillations of the pedal bob. With a rebound near the critical damping (see my Ep.5 video), you can dampen the pedal bob oscillations. (this might not make the bike more pedal efficient, but reduces the annoying bob).

The last aspect is the gear combination. With traditional drivetrains, the AS changes between rear cogs, and specially between chainrings.

Bye
 

andrextr

Chimp
Aug 6, 2007
63
67
Portugal
Just to share my new video, about the instant center (virtual pivot). Although it's a bit long, I think it's a quite important video to understand better the kinematics of the suspension :) I tried to make it interesting. Hope you like it.


Bye :)
 

Jm_

sled dog's bollocks
Jan 14, 2002
20,416
10,890
AK
Hi @Jm_

Yes, you are right, the ideal AS should be stable on the first 50% of travel at least (pedaling zone), and then drop down (like Santa Cruz VPP bikes for instance). The reason that AS should fall at the end of the travel is to decrease the amount of pedal kickback (and because you don't need too much AS on the final part of the travel).

However, a 100% AS will only eliminate the bob created by the horizontal acceleration forces. A 100% AS is not enough to eliminate the bob caused by the vertical movement of the rider on the saddle (such as pedaling with bad technique or pedaling on foot / sprint). With good pedaling technique you can reduce this vertical component, but it's not possible to fully eliminate it, since it's part of our pedaling bio-mechanic motion. There are some companies that defend higher anti.squats, such as 150% because of that.

Another very important aspect is the rebound speed. A fast rebound can amplify the oscillations of the pedal bob. With a rebound near the critical damping (see my Ep.5 video), you can dampen the pedal bob oscillations. (this might not make the bike more pedal efficient, but reduces the annoying bob).

The last aspect is the gear combination. With traditional drivetrains, the AS changes between rear cogs, and specially between chainrings.

Bye
Well, it seems like a lot of the bikes with less than optimal numbers are being referred to as "good pedaling bikes", and according to all of this, that would only exist on level ground with no suspension activation. But yes, the graphs and numbers are very helpful in figuring it all out or making sense of what is felt.
 

Flo33

Turbo Monkey
Mar 3, 2015
2,146
1,376
Styria
High low speed compression damping (Propedal, climb switchs etc.) might help here and play a significant role. I have a Spesh Pitch with a Corset equipped RP2, pedals pretty well with propedal on. Put a DHX RC4 in it and all is shitty wobbly wiggle waggle, even with LSC nearly full closed. I'm riding an oval abomination ring too which may cause even more problems, varying AS numbers I think are the cause.
 

Happymtb.fr

Turbo Monkey
Feb 9, 2016
2,084
1,469
SWE
I wonder about something :
Anti squat and brake squat are based on the position of the CoG and are always calculated with the bike on flat ground. I have a hard time figuring out how these values change when the bike is tilted upward or downward...
Does somebody know?
 

andrextr

Chimp
Aug 6, 2007
63
67
Portugal
Hi, the CoG height is what determines the 100%. If the line of anti-rise (aka brake-squat) or anti-squat crosses at the CoG heigth rightly above the front wheel axle them you have 100% (see image below). If it crosses on the ground it's 0%.


(note: the AR value on the image is 50% for this bike, but the real value is ~60%).

By tilting the bike forward (fully compressing the front suspension), the IC moves a bit downward, so the values change a bit (they decrease by around 10-15%). Compressing the rear wheel is completely different, since the IC can move in a very different ways depending on the type of bike/linkage design. For instance, in Nomad 3 the anti-rise decreases from +60% to -70% in the last 20mm of travel (because both links became parallel at the end of the travel and the IC goes to a very very faraway distance below the ground...). This is why anti-squat and anti-rise values should be looked at SAG zone (the zone where you spent most of the time). Bye

Bye :)
 
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Happymtb.fr

Turbo Monkey
Feb 9, 2016
2,084
1,469
SWE
Hi André, thanks for your answer! :thumb:
By tilting the bike upward or downward I ment riding uphill or downhill but this also results in compressing the rear suspension or the front since we put more weight on the rear wheel while climbing and more on the front wheel while descending.
Then I wonder about the validity of the model when riding uphill or downhill since the CoG height line starting at the contact point of the front tire should follow gravity and no be perpendicular to the floor.
 

andrextr

Chimp
Aug 6, 2007
63
67
Portugal
I'm not an expert, and I'm not sure about this, but according to a quick search some people say it doesn't change others say it does. If it does, it should be in a similar way than compressing or extending the front suspension (a basic tilt). If true, thus, per each 10º of uphill incline the AS and AR would increase 15%. So, in a very steep incline with 20º (36% slope grade) the AS should be 30% higher. And this somehow matches with Pole bikes description:

"What we’ve learned from our testing is that, if the AS is more than 100% it needs more effort to pedal uphill since the pedaling effect wants to lift rider weight." - https://www.polebicycles.com/what-is-anti-squat-and-pedal-kickback/?v=35357b9c8fe4

By "lifting the rider weight" it means that the suspension wants to extend (so it matches with an increasing AS).

Using the same rational, on downhill, then the AS and AR will decrease in the same proportion.

Keep it mind that over a certain uphill incline, you start to loose rear tire traction and the wheel starts to spin. In this case you neither accelerate (so, you don't squat) neither you anti-squat (since when the wheel spins and you loose the chain tension needed to extend the suspension). Thus, these AR and AS vs inclines are somehow limited by the rear wheel traction :)

I hope this is correct. Keep it mind that I'm not an expert and this might be wrong.

Anyway, independently of the incline effect, it's a good principle to analyse and compare all bikes with each other on the flat ground.

Bye :)
 
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rbx

Monkey
Just to share my new video, about the instant center (virtual pivot). Although it's a bit long, I think it's a quite important video to understand better the kinematics of the suspension :) I tried to make it interesting. Hope you like it.


Bye :)
The IC is only used for braking analysis.

The CC is used for acceleration behaviour.
 

andrextr

Chimp
Aug 6, 2007
63
67
Portugal
Hi. I never used CC (center of curvature) for that...

What I know is that the IC is used both for braking and for acceleration behavior (anti-squat). The anti-squat % is a result of the IC position, the rear axle position, and the chain-line. In the case of the image below, the chainline angle is pushing the wheel down, so it's fighting against suspension squat.

Over-100precent-Anti-Squat-635x286.jpg

https://www.polebicycles.com/what-is-anti-squat-and-pedal-kickback/?v=35357b9c8fe4

When the wheel is placed on the chainstay the CC is the same as IC. So in that case you can use CC, otherwise I don't know how to use the CC to calculate Anti-Squat... If you know it please share :)

Bye
 
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Happymtb.fr

Turbo Monkey
Feb 9, 2016
2,084
1,469
SWE
Thanks for your answer again! There was some food for thought there but if it does change then it seems good that it goes toward more AS when riding uphill and less AS and AR when riding downhill!
 

Vrock

Linkage Design Blog
Aug 13, 2005
276
59
Spain
Thanks for your answer again! There was some food for thought there but if it does change then it seems good that it goes toward more AS when riding uphill and less AS and AR when riding downhill!
It doesn't change, when you are on a slope the bike gets a bit deeper in the travel, but you still need 100% AS to stop the bob. If you increase the AS the bike sags a bit less, but you start to have a bit of bob because now you are extending the shock too much...
 

andrextr

Chimp
Aug 6, 2007
63
67
Portugal
It's not easy to understand if it changes or not, already saw some contraditory info on the net and from suspension designers... But anyway, regarding @rbk comment about the center of curvature (CC) vs IC for anti-squat determination. I said that you need to use IC, but indeed you also can use CC, and it will give you the same result. That happens because the CC is located on the same line that crosses the wheel axle and the IC. So, although I use the IC, which is the standard method, it seems that you can also use the CC with the same result.
 

Vrock

Linkage Design Blog
Aug 13, 2005
276
59
Spain
CC is a very old definition and it's very misleading, so it's better if we stop using it.
And I don't care what other people say, even if it's DW. I did the math, I did a couple of experiments and I'm 100% sure about it.