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dw, gear box?

S.K.C.

Turbo Monkey
Feb 28, 2005
4,096
25
Pa. / North Jersey
Me:

hmmmmm...

dw speaks with much secrecy...

That means he's probably gonna drop something HUGE if it's ready in time for Interbike...

Can't wait to check it out. If it is gonna look and work as cool as I think it is (it's from dw after all) I've got a plasma cutter on standby to retro-fit it to my Demo 8... :D
escapeartist:
E-speculation at its E-finest.

...good Lord...can't anyone take a joke?

I doubt dw has anything built yet, and who in their right mind is gonna take a plasma cutter to their ride?...

...speaking of who knows more about suspension,(except for dw - he's the real deal :D ) I heard that the Honda RN01 actually has a crank-powered food processor instead of a gear box, and that the rear suspension dosen't even EXIST. Greg Minaar uses special telekinetic powers to create the illusion of a rear wheel while levitating the rear end.

...and it has ZERO chain-growth.

He's really just nose-manualing the entire course... and after he crosses the finish line?

He enjoys a refreshing strawberry smoothie.
 

thaflyinfatman

Turbo Monkey
Jul 20, 2002
1,577
0
Victoria
dw said:
The fact is that chain force is ALWAYS the MINOR of the two internal forces in a chain driven suspension system that make up a squat amount. The major force is always driving force.
Ok this may seem like a dumb question, but when you say that driving force is the major of the two internal forces that contribute to squat, do you mean that the driving force is of a greater raw magnitude than chain tension (which I'm pretty sure is impossible), or do you mean that the driving force, due to its point of application, has a much greater effect on sqaut?
 

dhkid

Turbo Monkey
Mar 10, 2005
3,358
0
Malaysia
dw said:
For me, all I stated is that what was previously written is incorrect. The statement that you cut and pasted assumes that chain force is the only factor that contributes to "pedal bob" (or the lack of a sufficient amount of anti-squat to counteract mass transfer under acceleration). The fact is that chain force is ALWAYS the MINOR of the two internal forces in a chain driven suspension system that make up a squat amount. The major force is always driving force. You would have to have an infinitely small rear sprocket to have them be equal and that is not feasable as you can imagine.

Hope this helps

dw
i think what he is saying is that forces like the rear wheel having to push the whole bike and riders mass forward (which is transfered throught the rear suspension linkages) has a far greater effect on the rear suspension than what the chain growth does to the rear suspension.
ie: in high single pivot bikes, the rear suspension would not react well to the terrain coz it has loads of chain growth. but then if you have no chain growth or very little chain growth, the suspension would have a inch worm effect (bob). hope someone gets what i am trying to say, and correct me if i am wrong... :monkey:
 

dw

Wiffle Ball ninja
Sep 10, 2001
2,943
0
MV
thaflyinfatman said:
Ok this may seem like a dumb question, but when you say that driving force is the major of the two internal forces that contribute to squat, do you mean that the driving force is of a greater raw magnitude than chain tension (which I'm pretty sure is impossible), or do you mean that the driving force, due to its point of application, has a much greater effect on sqaut?
Think of your wheel and sprockets as a lever, a special lever, but still a lever. Draw it out.
 

rbx

Monkey
Just taking into consideration chain forces on the suspension will not tell you the final and complete picture of how the suspension will react.
Frame Geometry(HA,BB height,CG rider) will all effect how suspension will react to pedal input.
 

skinny mike

Turbo Monkey
Jan 24, 2005
6,415
0
all i gotta say is arguing with dw about suspension design is like arguing with einstein about his theory of relativity, you just cant win.
 

WheelieMan

Monkey
Feb 6, 2003
937
0
kol-uh-RAD-oh
atrokz said:
^ add in clips vs flats.
Nahh, I don't believe there is a big difference between clips and flats when it comes to how force is applied. I don't know of anyone who pedals in "perfect circles"; not only is it impossible, but it's not practical in downhill because you are standing and pedaling 99% of the time. So basically you are pedal mashing in downhill no matter if you are using clips or flats. Cross country riding is a little bit different however because you are able to sit and pedal.
 

atrokz

Turbo Monkey
Mar 14, 2002
1,552
77
teedotohdot
Actualy, there is a difference. I dont need to explain it because you did yourself.

"So basically you are pedal mashing in downhill no matter if you are using clips or flats. Cross country riding is a little bit different however because you are able to sit and pedal."

When we discuss suspension, all types are being discused. DW link is on XC bikes too no? Besides, I was adding to a moot list if you cared to notice. Whether you're force pulls and pushes or just pushes is going to have a effect as great as the factors given in what I added too.

No one is debating Dave's findings and developments, but there is more than one way to produce similar results, and I guess I should have known debating the wording of something would lead to confusion, but I can assure you what was said is technicaly correct, but did not include other factors. It's important to notice however, that that was just a quote out of pages and that issues such as downward force are not necesarily new findings. The 'DW' link however, has steped forward and created a platform bike that accomplishes what we've been relying on shocks for, but does it with linkage and for that I realy respect him. Understand?
 

WheelieMan

Monkey
Feb 6, 2003
937
0
kol-uh-RAD-oh
atrokz said:
Actualy, there is a difference. I dont need to explain it because you did yourself.

"So basically you are pedal mashing in downhill no matter if you are using clips or flats. Cross country riding is a little bit different however because you are able to sit and pedal."

When we discuss suspension, all types are being discused. DW link is on XC bikes too no? Besides, I was adding to a moot list if you cared to notice. Whether you're force pulls and pushes or just pushes is going to have a effect as great as the factors given in what I added too.

No one is debating Dave's findings and developments, but there is more than one way to produce similar results, and I guess I should have known debating the wording of something would lead to confusion, but I can assure you what was said is technicaly correct, but did not include other factors. It's important to notice however, that that was just a quote out of pages and that issues such as downward force are not necesarily new findings. The 'DW' link however, has steped forward and created a platform bike that accomplishes what we've been relying on shocks for, but does it with linkage and for that I realy respect him. Understand?
Yes, there is a difference in the pedaling forces between xc and dh bikes because of how you ride uphill in xc and strictly downhill in dh. (But I thought we were talking about downhill bikes as a gearbox doesn't make as much sense for xc...)

I still don't feel that the power difference is very large between clips and flats for downhill.
 

thaflyinfatman

Turbo Monkey
Jul 20, 2002
1,577
0
Victoria
dw said:
Think of your wheel and sprockets as a lever, a special lever, but still a lever. Draw it out.
Yeah, so it's because of the point of application then... because chain tension force has to be a higher force than the actual driving force (balancing moments about the axle).
 

zedro

Turbo Monkey
Sep 14, 2001
4,144
1
at the end of the longest line
thaflyinfatman said:
Yeah, so it's because of the point of application then... because chain tension force has to be a higher force than the actual driving force (balancing moments about the axle).
well the point of application defines the magnitude, so it shouldnt of been an either/or question :D
 

dw

Wiffle Ball ninja
Sep 10, 2001
2,943
0
MV
thaflyinfatman said:
Yeah, so it's because of the point of application then... because chain tension force has to be a higher force than the actual driving force (balancing moments about the axle).
No, the chain force is always LESS than the actual driving force. Draw out the lever. We are talking about seperating out the forces in the system, basically the kiematical breakdown of the suspension.

Lets focus on the wheel. Draw a simple free body diagram of the wheel. Say you choose a cog with a 4" pitch diameter. Say your wheel OD is 26 inches. Assume that the Cf at the ground to tire is 1. The lever is anchored at the ground in our free body diagram. This gives a 13inch distance from the fixed ground end to the center of the wheel (where the driving force is reacted) and a 15 inch distance to the chain force line. In our free body diagram driving force will be 15/13 of the chain pull, or 115%.

dw
 

thaflyinfatman

Turbo Monkey
Jul 20, 2002
1,577
0
Victoria
dw said:
No, the chain force is always LESS than the actual driving force. Draw out the lever. We are talking about seperating out the forces in the system, basically the kiematical breakdown of the suspension.

Lets focus on the wheel. Draw a simple free body diagram of the wheel. Say you choose a cog with a 4" pitch diameter. Say your wheel OD is 26 inches. Assume that the Cf at the ground to tire is 1. The lever is anchored at the ground in our free body diagram. This gives a 13inch distance from the fixed ground end to the center of the wheel (where the driving force is reacted) and a 15 inch distance to the chain force line. In our free body diagram driving force will be 15/13 of the chain pull, or 115%.

dw
Ahhhh, gotcha - by driving force, you mean at the axle. I was thinking of it as being at the contact patch, sorry.
 
Jul 5, 2002
52
0
thaflyinfatman said:
Ahhhh, gotcha - by driving force, you mean at the axle. I was thinking of it as being at the contact patch, sorry.
It's easy to get confused when DW defiines terms like "driving force" differently from anyone else.

The "orthodox" way of looking at his example would be to say that the driving force, which is at the ground, is 1 or 2/2. Since the ratio of cog radius to wheel radius is 2/13, the chain force must be 13/2. Both the chain force and the driving force can be taken as acting at the axle which gives a total force of 15/2 at the axle. 15/2 = 15/13 * 13/2. So if you call the total forces acting on the axle the "driving force" then DW is right.

But the force that makes the bike go is the one at the ground, whatever you want to call it.
 

dw

Wiffle Ball ninja
Sep 10, 2001
2,943
0
MV
Steve from JH said:
It's easy to get confused when DW defiines terms like "driving force" differently from anyone else.
FYI, Cossalter and Foale use the term "driving force" and "swingarm force" interchangeably, and in the identical manner that I do. I agree it may be confusing to someone not versed in the subject, but its not an issue for me.
 
Jul 5, 2002
52
0
dw said:
FYI, Cossalter and Foale use the term "driving force" and "swingarm force" interchangeably, and in the identical manner that I do. I agree it may be confusing to someone not versed in the subject, but its not an issue for me.
Cossalter does not use the term the way you do. I believe you know that to be true. He uses S for driving force and T for chain force and states that S=T*c/r, where c is the radius of the cog and r is the radius of the wheel. I changed those last two from what he uses in order to avoid subscripts.
 

dw

Wiffle Ball ninja
Sep 10, 2001
2,943
0
MV
Steve from JH said:
Cossalter does not use the term the way you do. I believe you know that to be true. He uses S for driving force and T for chain force and states that S=T*c/r, where c is the radius of the cog and r is the radius of the wheel. I changed those last two from what he uses in order to avoid subscripts.
There is an italian to english translation of the glossary of terms (which I do not have a photocopy of anymore). Cossalter uses S as his designation for tractive driving force. I can't remember the exact translation, but its pretty close to that. Nobody that I know of uses swingarm force as a term for tractive driving force, but the term driving force is quite commonly used as a term for the second component force when discussing internal chassis forces. It would be nice to have some universally agreed upon nomenclature, but there are really only a few people in the world who use this every day, and even fewer who actually understand it. I started writing a book a few years back, but there is just no time. Someday.
 
Jul 5, 2002
52
0
dw said:
There is an italian to english translation of the glossary of terms (which I do not have a photocopy of anymore). Cossalter uses S as his designation for tractive driving force. I can't remember the exact translation, but its pretty close to that. Nobody that I know of uses swingarm force as a term for tractive driving force, but the term driving force is quite commonly used as a term for the second component force when discussing internal chassis forces. It would be nice to have some universally agreed upon nomenclature, but there are really only a few people in the world who use this every day, and even fewer who actually understand it. I started writing a book a few years back, but there is just no time. Someday.
Cossalter's S stands for spinta which means "push" or "impelling force". T stands for tiro di catena or "chain throw".

I don't like "drive force" when talking about internal forces because they obviously do not drive the bike or make it go. Otherwise an idea I had in the first grade would work. Attach a rope to the back of a rowboat, sit in the boat, pull on the rope, and across the pond you go.
 
Jul 5, 2002
52
0
dw said:
There is an italian to english translation of the glossary of terms (which I do not have a photocopy of anymore). Cossalter uses S as his designation for tractive driving force. I can't remember the exact translation, but its pretty close to that. Nobody that I know of uses swingarm force as a term for tractive driving force, but the term driving force is quite commonly used as a term for the second component force when discussing internal chassis forces. It would be nice to have some universally agreed upon nomenclature, but there are really only a few people in the world who use this every day, and even fewer who actually understand it. I started writing a book a few years back, but there is just no time. Someday.
S stands for spinta. The word basically means "push" or "impelling force". From the Italian Wikipedia, I translate the techical definition as "the force produced by the drivetrain which is opposed by the resistance".

T stands for tiro di catena which means "chain throw".
 

Wilhelm

Monkey
Aug 10, 2003
444
19
dw said:
I've been working on some new and very different things in my spare time, testing components of the mechanism and applying for patents. I am not rushing it, The ideas are protected now, so I am just working on doing it right.

I am no longer working on the G-BOXX system actively, from a structural standpoint and manufacturing standpoint I learned what I needed from the development and implementation.

There are a couple of new and interesting gearbox developments coming from all kinds of places. It will be a cool couple of years. I still don't see gearboxes replacing conventional drivetrains anytime soon.

Thanks guys!

dw

Did somebody see or hear something of DW´s new gearbox developments at Interbike 2005?