Future Pedal?

[yim]

It seems smart.

[martianbait]

Looks fragile. If made of the right materials, though, could be a winner.

[coylifut]

why does it remind me of Dynadrive. It doesn't have the proprietary axel, but looks similar.

[Chad Engle]

What are they fixing/accomplishing with this design over what is currently being used? See thread about jacked up frame geometry.

[J.Greene]

Do I see it right? It looks like the pedal effectively extends the length of the crank.

JG

[rpm]

These remind me of Hi-E pedals, from way back. They also had the stubby axle and the dropped platform, which did indeed seem to be more ergonomically efficient. I had a pair and felt they gave me more power. There was a problem, though. One of them literally exploded while I was climbing a hill. The problem was that the short axle wasn't strong enough to take the force of a standing climb. You need a really strong axle to make this design work.

[David Kirk]

What's old is new again. Shimano's "direct drive" patent must have run out.

I'd like to try them.

Dave

[CIOCC FAN]

Those look interesting.

I'm looking at pedals for my new bike, the old Dyna Drive worked well for me (until they dropped them and I became a Campy fan).

[David Kirk]

Do I see it right? It looks like the pedal effectively extends the length of the crank.

JG

It doesn't actually make the crank longer but it puts the base of your foot at the center of rotation of the pedal spindle.

With a normal pedal system your foot doesn't actually travel in a true circle. The path is actually an oval do to the fact that your foot is actually above the center of rotation of the pedal.

Most manufacturers have made efforts to lower the foot on the pedal as much as possible to lower the effort required to keep you foot on top of the pedal. Think of a silly system (like the original Looks) that holds your foot far from the center of rotation. When pushing down on the pedal you need to resist the foot rotating forward or back while it's on it's little stilt. if the foot is on center (or even below) then then foot is more stable and can just push down without having to stabilize itself.

I didn't say that well. Hope it makes some sense.

Dave

[CIOCC FAN]

According to their website it does move the ball of the foot forward as well, extending the effective crank length from 12 - 6 0'clock. Then from 6-12 o'clock it decreases effective crank length on the upstroke. Pretty cool idea (to me).

[flydhest]

It doesn't actually make the crank longer but it puts the base of your foot at the center of rotation of the pedal spindle.

With a normal pedal system your foot doesn't actually travel in a true circle. The path is actually an oval do to the fact that your foot is actually above the center of rotation of the pedal.

Most manufacturers have made efforts to lower the foot on the pedal as much as possible to lower the effort required to keep you foot on top of the pedal. Think of a silly system (like the original Looks) that holds your foot far from the center of rotation. When pushing down on the pedal you need to resist the foot rotating forward or back while it's on it's little stilt. if the foot is on center (or even below) then then foot is more stable and can just push down without having to stabilize itself.

I didn't say that well. Hope it makes some sense.

Dave

Dave,
The stabilization part makes sense, but the oval part has me scratching my head. If the pedal spindle is going in a circle, and the cleat is -- say -- 5 cm above the pedal spindle, then isn't the cleat going in a circle with a center 5 cm's above the bb?

y^2 + x^2 = 1 is a circle with center at zero

(y-1)^2 + x^2 = 1 is a circle with center at 1 but identical radius, just the first one translated up.

[davids]

Dave,
The stabilization part makes sense, but the oval part has me scratching my head. If the pedal spindle is going in a circle, and the cleat is -- say -- 5 cm above the pedal spindle, then isn't the cleat going in a circle with a center 5 cm's above the bb?

y^2 + x^2 = 1 is a circle with center at zero

(y-1)^2 + x^2 = 1 is a circle with center at 1 but identical radius, just the first one translated up.
I'm pretty sure you're right - you'd have to lower your saddle a bit, but everything else remains in alignment.

[David Kirk]

Dave,
The stabilization part makes sense, but the oval part has me scratching my head. If the pedal spindle is going in a circle, and the cleat is -- say -- 5 cm above the pedal spindle, then isn't the cleat going in a circle with a center 5 cm's above the bb?

y^2 + x^2 = 1 is a circle with center at zero

(y-1)^2 + x^2 = 1 is a circle with center at 1 but identical radius, just the first one translated up.

I stand corrected. I brain farted. it's still a circle but it's just shifted up the same distance that the sole of the foot is above the spindle.

I think?

Dave

[flydhest]

I'm just so happy I can still remember 8th grade. Mrs. Kramer and the room was at the end of the hallway. As it turns out, college is a bit fuzzier.

[michael white]

can you factor in ankling with that equation?

[SoCalSteve]

Dave,
The stabilization part makes sense, but the oval part has me scratching my head. If the pedal spindle is going in a circle, and the cleat is -- say -- 5 cm above the pedal spindle, then isn't the cleat going in a circle with a center 5 cm's above the bb?

y^2 + x^2 = 1 is a circle with center at zero

(y-1)^2 + x^2 = 1 is a circle with center at 1 but identical radius, just the first one translated up.

Here we go again!

I always wondered why I should have taken all those math classes (now I know).

I thought cycling was supposed to be fun...

Math + Cycling = NOT fun

Steve

Just joking! More power to the people who understand if 2 trains running in opposite directions..yada, yada, yada...

[stevep]

Dave,
The stabilization part makes sense, but the oval part has me scratching my head. If the pedal spindle is going in a circle, and the cleat is -- say -- 5 cm above the pedal spindle, then isn't the cleat going in a circle with a center 5 cm's above the bb?

y^2 + x^2 = 1 is a circle with center at zero

(y-1)^2 + x^2 = 1 is a circle with center at 1 but identical radius, just the first one translated up.

no, dave is right. it is an oval.
think of the position of yr foot on the pedal in the roatation.
at top d center your foot sole is 5mm above the pedal axle ctr= radius +5mm
at 90 degrees and 270 degrees you are exactly lined up with the pedal axle ctr ( but 5mm above it )...=radius

picture what would happen if you put a large pedal block on top of the pedal for exaggeration purposes...would end up with a huge oval.
..and the sort of rocking balance required would be an exaggerated version of what you get w/ a regular pedal ...this is why every pedal manufacturer tries to reduce the height of the pedal/ cleat.

[flydhest]

no, dave is right. it is an oval.
think of the position of yr foot on the pedal in the roatation.
at top d center your foot sole is 5mm above the pedal axle ctr= radius +5mm
at 90 degrees and 270 degrees you are exactly lined up with the pedal axle ctr ( but 5mm above it )...=radius

picture what would happen if you put a large pedal block on top of the pedal for exaggeration purposes...would end up with a huge oval.
..and the sort of rocking balance required would be an exaggerated version of what you get w/ a regular pedal ...this is why every pedal manufacturer tries to reduce the height of the pedal/ cleat.

as it turns out, dave is right the second time when he said I was right. Think of the position of your foot on the pedal in the roatation. at top d center your foot sole is 5mm abover the pedal axle center = radius + 5 mm. at 90 degrees and 270 degrees you are exactly lined up with the pedal axle ctr but 5 mm above it = radius . . . yes, but not the radius away from the bb spindle. at 90 degrees, the distance between the pedal spindle and the bb can form the base of a triangle--a right triangle. it is of length r. The stack height is 5, so the distance from the cleat to the bb spindle is the hypotenuse, so it is (r^2 + 25)^(1/2), which must be more than r. However, draw a line parallel to the crank arm that goes through the cleat. it will be 5 mm higher than the crank arm. from a point 5 mm above the bb spindle over to the cleat, this line segment will be length r. indeed, it will be the radius of the circle that the cleat is describing as you pedal.

at dead center bottom, the cleat will be r away from that same center point. same with at dead center top.

[stevep]

as it turns out, dave is right the second time when he said I was right. Think of the position of your foot on the pedal in the roatation. at top d center your foot sole is 5mm abover the pedal axle center = radius + 5 mm. at 90 degrees and 270 degrees you are exactly lined up with the pedal axle ctr but 5 mm above it = radius . . . yes, but not the radius away from the bb spindle. at 90 degrees, the distance between the pedal spindle and the bb can form the base of a triangle--a right triangle. it is of length r. The stack height is 5, so the distance from the cleat to the bb spindle is the hypotenuse, so it is (r^2 + 25)^(1/2), which must be more than r. However, draw a line parallel to the crank arm that goes through the cleat. it will be 5 mm higher than the crank arm. from a point 5 mm above the bb spindle over to the cleat, this line segment will be length r. indeed, it will be the radius of the circle that the cleat is describing as you pedal.

at dead center bottom, the cleat will be r away from that same center point. same with at dead center top.

dont buy it.
yes the hypotenuse you describe is the exact measure of the 90/270 points. no, it is not r+5mm.
it is very slightly more than r...very slightly. .
( depending on crank length 175=175.07)


or call this r if you wish/
if we call r the length at tdc or bdc as you wish
then 90 and 270 will be the hypotenuse you describe. it will be r-4.9mm or some such.

not a circle. an oval. could be a circle with that new pedal or with shimano dyna drive.

[Jeff Weir]

it's hip to be square

[flydhest]

dont buy it.
yes the hypotenuse you describe is the exact measure of the 90/270 points. no, it is not r+5mm.

I didn't say that was r + 5. You will note how much of your text I quoted.

it is very slightly more than r...very slightly. .
( depending on crank length 175=175.07)

ok, we're talking math, so greater than is all I care about, not "slightly."

or call this r if you wish/
if we call r the length at tdc or bdc as you wish
then 90 and 270 will be the hypotenuse you describe. it will be r-4.9mm or some such.

not a circle. an oval. could be a circle with that new pedal or with shimano dyna drive.
I was using your definition of r.

So, let me try again, and I'll type slowly. At 90 and 270, how far away is the cleat from a point that is 5 mm above the bb spindle? For the rest of this discussion, call that point--5 mm above the bb spindle -- x. x is r away from the cleat. Why? Go out r from the bb spindle r and you get to the pedal. Go up 5 mm and you get to the cleat. So far, your out r and up 5. Go back r and your 5 mm above the bb spindle, at x.

So, at 90 and 270, the cleat is r away from x. Now, at the top, the cleat is r + 5 mm above the bb spindle, or r above x. At the bottom, the cleat is r - 5 away from the bb spindle, which makes it r away from x.

The cleat is r away from the point x at each of these points of rotation.

OK, so far, we've only defined 4 points, which doesn't make a circle, I realize. Nevertheless, if you can follow that, I hope it's clear that the cleat can't follow an oval.

[David Kirk]

Ok............since I'm laid up with a busted toe and can't work for a day or two I thought I'd play with this.......and since I'm not a math guy but a pencil and paper guy I drew the whole thing out.

First I drew a big x/y graph type deal with the BB center being at the intersection of the x/y lines. I then drew a circle with the center at the center (duh) of the x/y and made the radius 170mm to simulate a typical crankarm length. So now I have a 340mm circle.

Next I drew lines at 45* angles so as to segment the circle into 8 equal pie shaped segments. Now I wanted to add the pedal stack height into the deal. So....where each line intersected the circle I drew a vertical line. So these vertical lines are perpendicular to the circle at 12 and 6 o'clock and tangental at 3 and 9 o'clock. I measured up 15mm on each of these lines and made a mark to show where the actual sole of the shoe would be relative to the center of the pedal spindle if your pedal has a stack of 15mm.

Next I measured up 15mm from the x/y and made a mark. Now I took my compass and using this new spot 15mm above x/y center and center I drew the same 170mm radius circle.

Now guess what happened..........go on guess. OK, if you aren't going to guess I'll tell you what I found. This new circle passed right through all my shoe sole spots exactly, meaning that the shoe circle was also truly round. In fact it's a perfectly round circle just 15mm above the pedal spindle circle.

So it looks to me like the both the shoe and the pedal spindle both travel in circles.........just not the same ones.



So now on to part B -

One of the things I noticed is that your effective crank arm length changes a good bit during the rotation. This is of course due to the pedal stack. Again using a 170mm crank arm and a 15mm stack here is what the effective crank arm lengths end up being (as measured off my drawing, if these aren't exactly what you get doing the math cut me some slack) -

At 0* - 185mm
At 45* - 180mm
At 90* - 170mm
At 135* - 160mm
At 180* - 155mm.

Stands to reason when one thinks of it but it was surprising nonetheless. So a "Direct Drive" type pedal should keep the effective crank arm length constant. It seems like this would be a big deal really. The section from 45* to 135* is where we make the power and our lever just gets shorter as it goes through this zone. Doesn't seem optimal to me.

Interesting stuff......at least to me. Makes me want to try some of those pedals to see how they feel.

Dave

[stevep]

So, let me try again, and I'll type slowly. At .

thanks for typing slowly. you are so much smarter than i am.
a veritable genius, atmo.
to measure directly at 90 degrees from the ctr of the spindle to the ctr of the cleat will measure 175.07. thats how far the point is from the crank ctr.
thats the hypotenuse of the triangle that you cited earlier...i assume that you were typing slowly because that made sense to me.
from the ctr of the spindle to the same point on the cleat at tdc will be 175+ the 5mm depth of the pedal.
do it , measure it.

[mosca]

If your foot always remains at the same angle relative to the ground, it will be moving in a perfect circle regardless of pedal stack height, per flydhest's equation. If the angle of your foot changes throughout the pedal stroke, its path will become more ovalized depending on the stack height above the pedal spindle. This ovalization would seem to be negligible with any normal pedal as the stack heights are relatively small. That's my theory, which is mine.

[flydhest]

stevep,
No need for me to measure, I understand math. I take it, however, between the last elaboration and Dave's post that it is all clear to you now?

mosca, that's right, but it isn't clear to me that an oval is the shape that would be defined by the cleat. It won't be a circle, but I'm not sure it's an oval.

[gt6267a]

i am imagining something like an egg.

[Chad Engle]

Math is hard. Any ideas what this pedal would do for me that my current ones don't?

[mosca]

mosca, that's right, but it isn't clear to me that an oval is the shape that would be defined by the cleat. It won't be a circle, but I'm not sure it's an oval.I agree, it would all depend on how the rider's ankle rotated during the stroke.

There is something that just seems "right" about having the ball of the foot perfectly aligned with the pedal spindle axis, but in reality I'm not sure if it would provide any noticeable advantage. I do recall the Shimano pedal (circa 1980, iirc) that was a clip and strap version of the same idea. Whatever happened with that?

[swoop]

the lowering the saddle that comes with this... lowers your center of gravity... and that tends to feel really good in a turn. even just a few mm's are noticeable. but then you gotta lower your stem too.

so part of the hidden benefit is in lowering your mass on the bike.... even just a little.

atmo.

i'd love to try it.. but i am a q factor queen... so it would be cool to see where that figures too. and my stem is already 80 degrees and slammed.

[mosca]

It seems like you'd lose cornering clearance and increase q-factor. To do it proper, the frame geo might need to be adjusted to compensate. And I question whether it could be as rigid as with a standard full length spindle. Still intrigued, though - who's gonna be the guinea pig?

[CIOCC FAN]

Still intrigued, though - who's gonna be the guinea pig?


:D

Bernie already was...

[pdxmech13]

them pedals are for sissy's.
toe straps are what makes
a real man

sean kelly has even gone
back toe them.

[Elefantino]

I must have fallen asleep in class.

Who makes/will make these?

[mosca]

Shimano used a dedicated crank so that they could move the bearings inward to reduce Q. Looks like these new ones do the same thing, so you need their crank as well.

[CIOCC FAN]

They are available with regular 9/16" threads as well as integrated with their own cranks (which lowers Q factor).

My set arrives tomorrow! :p