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single piv?

WheelieMan

Monkey
Feb 6, 2003
937
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kol-uh-RAD-oh
Lumpy_Gravy said:
I am progressively regressing at a rapidly rising rate.
Isn't it confusing?

A progressively progressive design is progressive, yet a regressively progressive design is still progressive?

And a regressively regressive design is regressive, but a progressively regressive design is still regressive?
 

OGRipper

back alley ripper
Feb 3, 2004
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ncrider said:
YES! So a falling rate suspension curve is a curve that becomes more progressive through it's travel and a rising rate suspension curve becomes less progressive (regressive) through it's travel. Then the graph of the v10 would show that it's travel is increasing in progresivness for about the first 3/4 of travel and then it backs off on progresiveness for the last 1/4 ( or so) of travel. Hmm interesting. Can anybody map a Vpfree?

Well, pardon my digression during this session on compression, but like we were clarifying above, a "falling" rate suspension can (theoretically) have a constant falling rate, so it doesn't necessarily have to become "more progressive" as you move through the travel, even if (as I understand it) most do.
 

ncrider

Turbo Monkey
Aug 15, 2004
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Los Angeles
OGRipper said:
Well, pardon my digression during this session on compression, but like we were clarifying above, a "falling" rate suspension can (theoretically) have a constant falling rate, so it doesn't necessarily have to become "more progressive" as you move through the travel, even if (as I understand it) most do.
I'm barely keeping up with this entire discussion and now...
I'm not sure I understand how a falling rate suspension can have a constant falling rate and not become more progressive. It can be falling at a constant rate, but it will still be more progressive at one end than the other. In that case the red line would be going from the bottom left corner to the top right in a straight line. If it's constantly falling for example starts at 6:1 and ends at 2:1, but goes from point A to B in a constantly falling path, then it will end more progressive than it began. That's my understanding anyway.
 

dw

Wiffle Ball ninja
Sep 10, 2001
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ncrider said:
YES! So a falling rate suspension curve is a curve that becomes more progressive through it's travel and a rising rate suspension curve becomes less progressive (regressive) through it's travel.
reverse that, a falling rate gets regressive, a rising rate gets progressive, generally speaking.


ncrider said:
Then the graph of the v10 would show that it's travel is increasing in progresivness for about the first 3/4 of travel and then it backs off on progresiveness for the last 1/4 ( or so) of travel. Hmm interesting.
yup, thats what the graph shows. the most important part is in the #s though.
 

OGRipper

back alley ripper
Feb 3, 2004
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ncrider said:
I'm barely keeping up with this entire discussion and now...
I'm not sure I understand how a falling rate suspension can have a constant falling rate and not become more progressive. It can be falling at a constant rate, but it will still be more progressive at one end than the other. In that case the red line would be going from the bottom left corner to the top right in a straight line. If it's constantly falling for example starts at 6:1 and ends at 2:1, but goes from point A to B in a constantly falling path, then it will end more progressive than it began. That's my understanding anyway.
Well I'm barely keeping up too.

I am talking about the shock rate created by the frame, without a shock. The rate is the amount of wheel travel:shock compression. If the rate is constant, the rate doesn't change, the rate doesn't get "more" progressive. So in your example, it would stay 2:1 the whole time.

I am speaking theoretically to better understand the terminology. I have no idea if any bike is truly constant this way.
 

ncrider

Turbo Monkey
Aug 15, 2004
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Ok I see what you're saying OG. I thought we HAD been talking about the suspension rate without the shock. Aren't we using the shock shaft just as a reference point to calculate the rate curve? My understanding is that if it remains 2:1 throughout the entire travel then it is a constant rate design, horizontal line. I don't think it can be a falling or rising rate curve if it remains 2:1 throughout the travel. I could be totally wrong though.
 

OGRipper

back alley ripper
Feb 3, 2004
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ncrider said:
Ok I see what you're saying OG. I thought we HAD been talking about the suspension rate without the shock. Aren't we using the shock shaft just as a reference point to calculate the rate curve? My understanding is that if it remains 2:1 throughout the entire travel then it is a constant rate design, horizontal line. I don't think it can be a falling or rising rate curve if it remains 2:1 throughout the travel. I could be totally wrong though.
Whoops, I think you are right. But if it gets consistently progressive, wouldn't the rate be a straight line, just not horizontal (and not curved like in the graph)? I still don't know what that blue line represents but that's pretty much what I mean.

(I hope the enginerds think this is fun in that special way, like watching a train wreck or something.)
 

ncrider

Turbo Monkey
Aug 15, 2004
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Los Angeles
OGRipper said:
Whoops, I think you are right. But if it gets consistently progressive, wouldn't the rate be a straight line, just not horizontal (and not curved like in the graph)? I still don't know what that blue line represents but that's pretty much what I mean.

(I hope the enginerds think this is fun in that special way, like watching a train wreck or something.)
Yeah the rate would be a straight line, but it would look like the image below.

Man the enginerds must be just laughing their arses off at us trying to figure this out. Oh well.
 

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Lex

Monkey
Dec 6, 2001
594
0
Massachusetts
LOOnatic said:
Yeah i give up.
Everything i thought i knew about axle paths and curves blah,blah,blah wasn't really anything.... :nopity:
I actually have devised a very complicated and specific model for testing various suspension designs. Here it is:

suspension travel x height of mountain x number of runs
divided by the number of crashes =
how much fun a given bike is to ride
 

OGRipper

back alley ripper
Feb 3, 2004
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ncrider said:
Yeah the rate would be a straight line, but it would look like the image below.

Man the enginerds must be just laughing their arses off at us trying to figure this out. Oh well.
Yep, I think that's got it. It's just theory anyway, like I said I have no idea if any bikes actually have a constant rising rate like that. When people say their rate is "linear" I start asking questions because even though it's straight it could still be rising or falling. Wheee this is fun!!!

I think most people use "rising" and "falling" to describe curved rates (those that get more regressive or progressive). And like DW said, linear should mean horizontal, as in no progression or regression, constant or otherwise.
 

Lumpy_Gravy

Monkey
Sep 16, 2003
194
0
It makes me feel better by remembering that all this would only make any difference if I were as fast as Sam Hill.
 

Inclag

Turbo Monkey
Sep 9, 2001
2,773
457
MA
I'm not even bothering reading all the posts in this thread because any thread about single pivots that goes for 10 pages is most likely a trainwreck.

Since I've concluded with a high certainty that this is indeed a trainwreck, I'd just like to point out that technicaly all "single-pivots" are four-bars just to add some fuel to the fire :devil: :devil:
 

OGRipper

back alley ripper
Feb 3, 2004
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Inclag said:
I'm not even bothering reading all the posts in this thread because any thread about single pivots that goes for 10 pages is most likely a trainwreck.

Since I've concluded with a high certainty that this is indeed a trainwreck, I'd just like to point out that technicaly all "single-pivots" are four-bars just to add some fuel to the fire :devil: :devil:
I have nothing to say, but I wanted to see something I typed actually get posted. Yippeeee!!! :rolleyes:
 

WheelieMan

Monkey
Feb 6, 2003
937
0
kol-uh-RAD-oh
dhkid said:
could some one post pics of the axel path and the rate of a 223, sunday and dhr? pls? :help:
I haven't found a good picture of the Sunday yet because the bottom pivots are so well hidden, but here are the 223 and DHR. The bikes are complete opposites, wheelpaths are completely different as well as the shock rate.

DHR is the blue line, 223 is the green line for wheelpath comparison.

DHR is the red line, 223 is the pink line for shock rate comparison.

Remember, the data I am using is taken from a picture so these graphs are not 100 percent accurate, but pretty dang close.
 

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Rory

Chimp
Oct 9, 2003
10
0
Expanding on what ohio said with regard to ‘s’ shaped and modified axle paths, what they look like on a graph is defined by the reference point and coordinate system being used. Generally axle paths are shown with the centre of the wheel when the suspension is fully extended as the reference point (origin of the graph) with the axle path plotted in x and y coordinates relative to this point as the suspension compresses. For example if you were to draw a circle on a graph taking its centre as the reference point/origin of the graph with x and y axes of distance from the centre of the circle and distance along the circumference of the circle the graph would be a straight line not a circle. This rough picture shows how a single axle path can appear different when represented graphically using different reference points and coordinate systems to define the position of the wheel. In this case the centre of the BB and the point where the chain comes onto the chainring are used as reference points and the distance from the reference point through the travel is plotted.

As has already been said, the gradient/derivative of the red line (second derivative of the blue line) can be used to determine if the suspension behaves in a progressive or regressive manner. A positive gradient/derivative of the red line indicates a progressive rate and a negative gradient/derivative indicates a regressive rate.

As dw said the design of a single pivot or multi link bike can be done well. The design of all bikes are a compromise in some way and when set up properly all good ones perform well, look at the different bikes people have ridden to podiums at world cups, world championships and norbas, its not like f1 was a couple of years ago when you needed a Ferrari.
 

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ncrider

Turbo Monkey
Aug 15, 2004
1,564
0
Los Angeles
Wow, wheelieman and Rory just about scrambled my brain, but I think I was able to follow along and learn something. Could you guys map a Vpfree. please?
 

WheelieMan

Monkey
Feb 6, 2003
937
0
kol-uh-RAD-oh
Rory said:
Expanding on what ohio said with regard to ‘s’ shaped and modified axle paths, what they look like on a graph is defined by the reference point and coordinate system being used. Generally axle paths are shown with the centre of the wheel when the suspension is fully extended as the reference point (origin of the graph) with the axle path plotted in x and y coordinates relative to this point as the suspension compresses. For example if you were to draw a circle on a graph taking its centre as the reference point/origin of the graph with x and y axes of distance from the centre of the circle and distance along the circumference of the circle the graph would be a straight line not a circle. This rough picture shows how a single axle path can appear different when represented graphically using different reference points and coordinate systems to define the position of the wheel. In this case the centre of the BB and the point where the chain comes onto the chainring are used as reference points and the distance from the reference point through the travel is plotted.

As has already been said, the gradient/derivative of the red line (second derivative of the blue line) can be used to determine if the suspension behaves in a progressive or regressive manner. A positive gradient/derivative of the red line indicates a progressive rate and a negative gradient/derivative indicates a regressive rate.

As dw said the design of a single pivot or multi link bike can be done well. The design of all bikes are a compromise in some way and when set up properly all good ones perform well, look at the different bikes people have ridden to podiums at world cups, world championships and norbas, its not like f1 was a couple of years ago when you needed a Ferrari.
Why would you want to use the initial point of contact between chain/chainring as a measure of chaingrowth?
 

OGRipper

back alley ripper
Feb 3, 2004
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WheelieMan said:
Why would you want to use the initial point of contact between chain/chainring as a measure of chaingrowth?
I was about to ask too. Maybe because chaingrowth depends in part on what ring you're in, and by looking at different points you can better understand actual growth?
 

ohio

The Fresno Kid
Nov 26, 2001
6,649
26
SF, CA
Jm_ said:
What the heck is the BB-to-dropout distance if it's not the axle path? The two go hand in hand. The distance determines the axle path.
Uh, dude, I really shouldn't have to explain this. One is the radial distance, the other is cartesian. Axle path is a cartesian measurement. Chain growth is radial.

As an example, if an axle path is perfectly straight and forms the chord of a circle about the bottom bracket... the chain growth will be a U shape (or C shape depending on your orientation), that is decreasing initially and increasing at the end.
 

WheelieMan

Monkey
Feb 6, 2003
937
0
kol-uh-RAD-oh
ohio said:
Uh, dude, I really shouldn't have to explain this. One is the radial distance, the other is cartesian. Axle path is a cartesian measurement. Chain growth is radial.

As an example, if an axle path is perfectly straight and forms the chord of a circle about the bottom bracket... the chain growth will be a U shape (or C shape depending on your orientation), that is decreasing initially and increasing at the end.
And?

Your example is true, but it doesn't have anything to do with your claim that axle-path and BB to axle distance are somehow unrelated.
 

ohio

The Fresno Kid
Nov 26, 2001
6,649
26
SF, CA
WheelieMan said:
And?

Your example is true, but it doesn't have anything to do with your claim that axle-path and BB to axle distance are somehow unrelated.
Huh? Who said they were unrelated?
 

ohio

The Fresno Kid
Nov 26, 2001
6,649
26
SF, CA
WheelieMan said:
I don't even know where to begin.
Try. My post wasn't complicated.

Jm_ stated that the axle path was not an S-shape. To achieve what the VPP patent claims you don't need the axle path to be an S-curve, you need the chain growth to be an s-curve. I pointed out that the linkage can still match the claim and not have an s-shaped axle-path.

Now, the system completely ignored the affects of rider mass under the acceleration that results from pedalling, but that's a seperate issue. What was under dispute was whether the bike does what the marketing claims. In this case, it does.

Whenever you feel like "beginning," have at it.
 

WheelieMan

Monkey
Feb 6, 2003
937
0
kol-uh-RAD-oh
ohio said:
Try. My post wasn't complicated.

Jm_ stated that the axle path was not an S-shape. To achieve what the VPP patent claims you don't need the axle path to be an S-curve, you need the chain growth to be an s-curve. I pointed out that the linkage can still match the claim and not have an s-shaped axle-path.

Now, the system completely ignored the affects of rider mass under the acceleration that results from pedalling, but that's a seperate issue. What was under dispute was whether the bike does what the marketing claims. In this case, it does.

Whenever you feel like "beginning," have at it.
Axle-path and chaingrowth are mutual. If the axle-path is "s shaped" the chaingrowth will also be "s shaped". Of course this all depends on the scale of the graphs/axis etc...
 

thaflyinfatman

Turbo Monkey
Jul 20, 2002
1,577
0
Victoria
WheelieMan said:
Axle-path and chaingrowth are mutual. If the axle-path is "s shaped" the chaingrowth will also be "s shaped". Of course this all depends on the scale of the graphs/axis etc...

No, Ohio is right. Consider this: would a wheel with a perfectly straight, vertical axle path have no chaingrowth?
 

thaflyinfatman

Turbo Monkey
Jul 20, 2002
1,577
0
Victoria
WheelieMan said:
No, not sure where you're going with this. Explain more...
So you can see that axle path and chaingrowth AREN'T exactly the same thing. What Ohio is saying is that if you graphed BB-axle distance with respect to time, you can have an S-shaped CHAIN GROWTH curve (by curve, that means graph, not an actual physical curve) without having an S-shaped axle path, and vice versa. The "s-shape" claim in the VPP patents is, as Ohio said, related to chain growth rather than axle path. Chain growth is only measured as a linear (radial) distance from the BB, and as such you could get an S-shaped chain growth graph by having a system (if it was possible) that moved the axle horizontally back, then horizontally forward, then horizontally back again with NO movement along the vertical axis. Axle path however, is described with cartesian coordinates (X-Y plane where perpendicular ordinates are independent).
 

WheelieMan

Monkey
Feb 6, 2003
937
0
kol-uh-RAD-oh
thaflyinfatman said:
So you can see that axle path and chaingrowth AREN'T exactly the same thing. What Ohio is saying is that if you graphed BB-axle distance with respect to time, you can have an S-shaped CHAIN GROWTH curve (by curve, that means graph, not an actual physical curve) without having an S-shaped axle path, and vice versa. The "s-shape" claim in the VPP patents is, as Ohio said, related to chain growth rather than axle path. Chain growth is only measured as a linear (radial) distance from the BB, and as such you could get an S-shaped chain growth graph by having a system (if it was possible) that moved the axle horizontally back, then horizontally forward, then horizontally back again with NO movement along the vertical axis. Axle path however, is described with cartesian coordinates (X-Y plane where perpendicular ordinates are independent).
Yeah, and in that case, the axle-path would just be going backwards instead of upwards, no difference.

Axle-path is what causes chaingrowth. So if the VPP parents are related to chaingrowth, they must also be related to axle-path.

It's the term "s shape" that is throwing a monkey wrench into all of this. Regardless of what the shape of the chaingrowth curve is called, the shape is determined by the axle-path. Which is why JM and myself are confused about Ohio's original comment.
 

dw

Wiffle Ball ninja
Sep 10, 2001
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Ohio's comment is dead on. Pick up a geometry text book and read up on cartesian geometry. That should give you a good basis to understand what he is trying to explain to you.

Dave