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Jack Mott/ Aeroweenie question
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Dialed in the fit for my "new" P3-SL last weekend. Got the UBC DU in a few weeks. What is the better tyre/wheel combo for the front, HED 3 with 20 Conti Supersonic, or Zipp 404 F/C w' 23 Conti 4000. Race is not that windy or hilly. Thnx for your advice (s).

http://www.fitspeek.com the Fraser Valley's fitness, wellness, and endurance sports podcast
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Re: Jack Mott/ Aeroweenie question [Hydrosloth] [ In reply to ]
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The HED3, no contest (both the tire and wheel are faster). Make sure to use a latex tube.

ECMGN Therapy Silicon Valley:
Depression, Neurocognitive problems, Dementias (Testing and Evaluation), Trauma and PTSD, Traumatic Brain Injury (TBI)
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Re: Jack Mott/ Aeroweenie question [Titanflexr] [ In reply to ]
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Edit: here's a graph from Zipp showing the H3 as 5 grams (.5 watts) of drag faster than the 404FC at zero yaw:


so the math:

H3 + .5 watts aero
H3 + 6 watts to spin*
H3 - 3 watts for 20mm SS vs. 23mm SS
___________

H3 is 3.5 watts to the good over the 404 FC, at zero yaw

*if my assumptions are correct.



I think the question is, "which is faster, H3 w/ 20mm SS or 404FC with 23mm SS?", because that's the comparable tire for the 404FC, and it's not quite as easy as "no contest"

For comparison's sake the Flo 90 w/ SS 23mm comes within half a watt of the benchmark Roval wheel on Specialized's own wind tunnel, same as the H3 with 20mm SS

A GP4000s that the OP might use is giving up 3.5 watts in Crr to a GP SS 23mm. A GP SS 20mm is giving up 3 watts to a Flo, but maybe more to a 404FC.

So, is the watts to spin of an H3 worth the 3 watts of Crr and any aero penalty over a Zipp 404FC? And is the occasional high yaw savings of a FC wheel worth the loss of watts to spin? In the H3 Deep thread awhile ago, I posited that it's 6 watts savings to spin a H3 over a spoked wheel.

So yes. If all the wheel companies that put their wheels at 1-3 watts faster at zero yaw than the H3 are right. If it's even slightly windy, or the athlete has poor crosswind handling skills, or the road surface is below average, then maybe not.

Titanflexr wrote:
The HED3, no contest (both the tire and wheel are faster). Make sure to use a latex tube.

Eric Reid AeroFit | Instagram Portfolio
Aerodynamic Retul Bike Fitting

“You are experiencing the criminal coverup of a foreign backed fascist hostile takeover of a mafia shakedown of an authoritarian religious slow motion coup. Persuade people to vote for Democracy.”
Last edited by: ericM40-44: Feb 12, 16 5:37
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Re: Jack Mott/ Aeroweenie question [ericM40-44] [ In reply to ]
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Latex tubes??

Have never used them before.

Maybe some free speed there?

http://www.fitspeek.com the Fraser Valley's fitness, wellness, and endurance sports podcast
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Re: Jack Mott/ Aeroweenie question [Hydrosloth] [ In reply to ]
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Latex tubes??

Where have you been?

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Re: Jack Mott/ Aeroweenie question [ericM40-44] [ In reply to ]
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In the H3 Deep thread awhile ago, I posited that it's 6 watts savings to spin a H3 over a spoked wheel.

Maybe. But I don't see that happening unless there is some yaw that allows the blades to do their airfoil thing and provide favorable lift. There won't be any lift at zero yaw. Then you are comparing the drag on those big fat blades to whatever 16 wee thin spokes would generate.

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Re: Jack Mott/ Aeroweenie question [rruff] [ In reply to ]
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Re: Jack Mott/ Aeroweenie question [jeffp] [ In reply to ]
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I've seen that study but after doing some rough calculations, I'm skeptical of such high numbers for rotational drag. The rotational drag is coming from the same place as translational, it's just the portion that results in negative torque on the wheel. If I remember correctly, all of the drag would need to concentrated at the maximum radius of the wheel and acting against forward rotation for the two to be equal.

I can't believe that this isn't routinely measured in wheel tests. You should be able to calibrate the motor that turns the wheel well enough to get a decent estimate.
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Re: Jack Mott/ Aeroweenie question [rruff] [ In reply to ]
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seems it might be somewhat in line with what btr might have on their pay for info
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Re: Jack Mott/ Aeroweenie question [rruff] [ In reply to ]
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rrr wrote

Where have you been?

...been here, on ST, since the GordoWorld days.

Just thought latex tubes were a silly way of throwing away $.

Sort of like buying a 35 dollar of merlot when a 15 dollar bottle will taste just as yummy

http://www.fitspeek.com the Fraser Valley's fitness, wellness, and endurance sports podcast
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Re: Jack Mott/ Aeroweenie question [Hydrosloth] [ In reply to ]
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Hydrosloth wrote:
rrr wrote

Where have you been?

...been here, on ST, since the GordoWorld days.

Just thought latex tubes were a silly way of throwing away $.

Sort of like buying a 35 dollar of merlot when a 15 dollar bottle will taste just as yummy

oh my, not even close to the same thing. Less rolling resistance, better ride comfort. FYI, they're cheaper us Canadians anyway if sourced through Wiggle or Ribble. Watch for sales and buy extras.
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Re: Jack Mott/ Aeroweenie question [ericM40-44] [ In reply to ]
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ericM40-44 wrote:
H3 + 6 watts to spin*

*if my assumptions are correct.

Which assumptions are those?

http://cds-0.blogspot.com
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Re: Jack Mott/ Aeroweenie question [Epic-o] [ In reply to ]
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Extrapolated from some data linked from Jack Mott's aeroweenie.com site.

https://docs.google.com/...htmlview?pli=1#gid=0



Epic-o wrote:
ericM40-44 wrote:
H3 + 6 watts to spin*

*if my assumptions are correct.

Which assumptions are those?

Eric Reid AeroFit | Instagram Portfolio
Aerodynamic Retul Bike Fitting

“You are experiencing the criminal coverup of a foreign backed fascist hostile takeover of a mafia shakedown of an authoritarian religious slow motion coup. Persuade people to vote for Democracy.”
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Re: Jack Mott/ Aeroweenie question [rruff] [ In reply to ]
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The local wind for a small area of a wheel is going to be proportional to r x omega+U where each term is a vector. The rotational component would be strictly the parts perpendicular to r: |r*omega|*r/|r|+(U - U(dot)r/|r|).

This value is zero at the floor and, U at hub height, and 2*U at max height, which means that the negative torque on the wheel produced below hub height is much smaller than the positive torques produced above hub height.

An actual calculation is too messy and requires CFD, but by order of scale, the lower half forces would be something proportional to the area of a triangle with b=U^2 and height r, and a trapezoid of height r and bases (2U)^2 and U^2 for something like a disc. (Ok this should be a proper integral or at least surface area weighted as well but I'm not gonna do that here for an order-of-magnitude thing :)

So in dynamic pressure scaling: 0.5r(U+2U)^2/0.5r(U)^2 = 9, the top side should have about 9 times more impact in terms of forces, while the torque arm for an equivalent center load would be about about r/3 for the bottom and about 0.6r for the top side. This doesn't include shear stress but that should be a much smaller contribution.

This also means that simply rotating a wheel in a wind tunnel "wind off" is not equivalent to a forward travelling drag induced torque.
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Re: Jack Mott/ Aeroweenie question [codygo] [ In reply to ]
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Now guesstimate what you'd expect for a relationship between translational drag and rotational, how big the later would be relative to the former.

Now that you mentioned CFD, I bet the Flo guys could shed some light on this.
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Re: Jack Mott/ Aeroweenie question [rruff] [ In reply to ]
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Pretty rough estimation but assuming that the Cd of all the wheel slices parallel to the ground is constant, rotational power is 50% of translational one.

http://cds-0.blogspot.com
Last edited by: Epic-o: Feb 13, 16 15:19
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Re: Jack Mott/ Aeroweenie question [Epic-o] [ In reply to ]
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Did you account for the horizontal slices getting smaller as you move away from the axle? I get a ratio of ~40%, assuming drag is uniform over the area of the wheel. I might have screwed up though.

I was thinking there wouldn't be as much drag once you get to the top rim section because there are no more spokes and it's a single nicely shaped profile. There is more surface area for part of it though. Moving towards the axle you have spokes, plus the rim presents two obstacles, rather than one.

Someone told me off the record (proprietary data) that rotational is a lot less than what is shown in the chart linked above. Since that agreed with what I already believed, that's where I sit on this. Would be nice to know though. I also don't get how the Trispoke would have so much lower rotational drag vs an 808 at 0 yaw. There isn't any blade lift then, so where could it come from? At yaw, it would make sense. Kraig Willett claims he's seen *negative* rotational drag on the Trispoke, but I don't know the details on that.
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Re: Jack Mott/ Aeroweenie question [jeffp] [ In reply to ]
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seems it might be somewhat in line with what btr might have on their pay for info

Magnitudes at zero yaw?

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Re: Jack Mott/ Aeroweenie question [rruff] [ In reply to ]
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Hi rruff,

I can't share details, but according to our recent investigations, you're on the right track - most of a wheel's drag appears to come from the middle part. Top & bottom contribute less.

Cheers,
Damon

Damon Rinard
Engineering Manager,
CSG Road Engineering Department
Cannondale & GT Bicycles
(ex-Cervelo, ex-Trek, ex-Velomax, ex-Kestrel)
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Re: Jack Mott/ Aeroweenie question [rruff] [ In reply to ]
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rruff wrote:
I also don't get how the Trispoke would have so much lower rotational drag vs an 808 at 0 yaw. There isn't any blade lift then, so where could it come from? At yaw, it would make sense. Kraig Willett claims he's seen *negative* rotational drag on the Trispoke, but I don't know the details on that.

perhaps because the entirety of the blades are behind the frontal profile of the leading edge of the wheel? Can there be a "virtual" airfoil across the wheel like at the top and bottom portions of the wheel?

There has to be some difference... at zero yaw all the spokes on a normal wheel are exposed... if not rotational than additional translational near the hub / fork dropouts

Eric Reid AeroFit | Instagram Portfolio
Aerodynamic Retul Bike Fitting

“You are experiencing the criminal coverup of a foreign backed fascist hostile takeover of a mafia shakedown of an authoritarian religious slow motion coup. Persuade people to vote for Democracy.”
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Re: Jack Mott/ Aeroweenie question [rruff] [ In reply to ]
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rruff wrote:
Did you account for the horizontal slices getting smaller as you move away from the axle? I get a ratio of ~40%, assuming drag is uniform over the area of the wheel. I might have screwed up though.

I was thinking there wouldn't be as much drag once you get to the top rim section because there are no more spokes and it's a single nicely shaped profile. There is more surface area for part of it though. Moving towards the axle you have spokes, plus the rim presents two obstacles, rather than one.

Someone told me off the record (proprietary data) that rotational is a lot less than what is shown in the chart linked above. Since that agreed with what I already believed, that's where I sit on this. Would be nice to know though. I also don't get how the Trispoke would have so much lower rotational drag vs an 808 at 0 yaw. There isn't any blade lift then, so where could it come from? At yaw, it would make sense. Kraig Willett claims he's seen *negative* rotational drag on the Trispoke, but I don't know the details on that.


True, take a look to M. N. Godo CFD work for some nice plots of the radial distribution of drag. He also carried out a CFD study (http://enu.kz/...1/AIAA-2011-1237.pdf) on the rotation moment that supports that the H3 has the lowest rotational drag of the bunch but the values are very low (0.5-0.6W at 20mph) so I would not trust that too much. Kraig's test was at 30mph and post-stall (20ş yaw).

You may not be aware but there is a Zipp document in the Aeroweenie page about spoke count that includes watts to spin data. I think that the tares have not been removed but the rotational power values seem to be in line with what others are measuring, 35-45% of total power is due to wheel rotation.

http://cds-0.blogspot.com
Last edited by: Epic-o: Feb 14, 16 5:49
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Re: Jack Mott/ Aeroweenie question [ In reply to ]
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some links to stuff i've written previously:

http://biketechreview.com/...-torquewatts-to-spin


negative watts to spin of a three spoke (ask questions if something isn't clear):


http://biketechreview.com/...og/511-watts-to-spin


lumping all betas and tires for both the 1080 and three spoke during these tests:


http://biketechreview.com/...trispoke-tire-choice


http://biketechreview.com/...458-1080-tire-choice


puts watts to spin at about 20% of total watts. there are some interactions at play with these data, though.





=================
Kraig Willett
http://www.biketechreview.com - check out our reduced report pricing
=================
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Re: Jack Mott/ Aeroweenie question [damon_rinard] [ In reply to ]
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I can't share details, but according to our recent investigations, you're on the right track - most of a wheel's drag appears to come from the middle part. Top & bottom contribute less.

Thanks a bunch, Damon! Really appreciate all your contributions.

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Re: Jack Mott/ Aeroweenie question [Epic-o] [ In reply to ]
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I think that the tares have not been removed but the rotational power values seem to be in line with what others are measuring, 35-45% of total power is due to wheel rotation.

Very old data that I think was taken straight from electrical power input to the motor turning the wheel. I also believe reducing spokes has a bigger effect than they show.

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Re: Jack Mott/ Aeroweenie question [ericM40-44] [ In reply to ]
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perhaps because the entirety of the blades are behind the frontal profile of the leading edge of the wheel? Can there be a "virtual" airfoil across the wheel like at the top and bottom portions of the wheel?

I must be getting senile. I forgot how narrow the H3 is. At zero yaw there would be a lot of drafting.

But I'm still skeptical of a large reduction in rotational drag, when the translational is about the same as an 808.

Last edited by: rruff: Feb 14, 16 9:33
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