That's exactly what he means...otherwise, there would be no need to place shipping boxes in the future "2nd control room" location to minimize the flow asymmetry he mentioned happens without it ;-)



According to Andy F's quote above, even those could be considered "closed circuit" :-)

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http://bikeblather.blogspot.com/

"That's exactly what he means"

again, i don't think it matters in the end. but, it had been my understanding that open and closed circuit were terms of art and that by any definition this is an open circuit tunnel, aka open return tunnel, with open circuit and open return being synonyms.

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Dan Empfield
aka Slowman

Yeah, having the tire Crr in there is a confounder...but, seeing as how the drums would need to be mounted on the balance, you really only need just enough pressure on the tire to prevent slipping, so that should help minimize any Crr effects. On the front that shouldn't be a huge issue...doing so on the rear, especially if the rear drum is applying a pedaling resistance, might be problematic. I guess that's why Mavic mentioned they'd be building an electric brake into a Comete disc...although I haven't completely thought through exactly how that would work...



How about all 3? Translational only, rotational (while subjected to translational air flow), and combined. All in newtons, of course ;-)

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http://bikeblather.blogspot.com/

See post #96 above...from the guys who basically have "written the book" on low speed wind tunnel design (emphasis added by me):

"...remember that an open-circuit tunnel in a room is really a closed-circuit design with a poorly designed return leg." ;-)

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http://bikeblather.blogspot.com/

Hi Dan,

Yup, it's a question of semantics...traditionally when discussing wind tunnel design, an open-return tunnel is a tube in a space (indoors or outdoors) and a closed-return tunnel is a continuous tube (e.g. the exhaust is continuously and directly connected to the inlet). However, as Tom and others have pointed out, strictly speaking every tunnel is in reality a "closed-return"; it's just a matter of how much control you have over the return path (ranging from a warehouse space to the entire planet's atmosphere). With a "open-return" tunnel sitting outside, if you run it long enough eventually the same air molecule of air will go through again =).

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Chris Yu
Applied Technology
Specialized Bicycle Components
@chrisyuinc

It's a matter of the area and design of the return. There are fully sealed closed-loop designs where the static pressure inside the tunnel is vastly different from outside the test section, there are closed-loop designs that are open to atmopheric pressure, and then there are non-recirculating designs. By virtue of the fact that Specialized tunnel and A2 are located inside warehouses, they do recirculate to a certain extent, but through an area an order of magnitude larger than the test section itself, so the return flow speed is reduced to an order of magnitude below the test section speed also. It's still an open design, but it's smart for bike testing. For one thing, closed loop designs take up substantially more space--double or even triple the footprint of a tunnel like A2 or Specialized. They're substantially more expensive because there's more of everything--more roof overhead, more wood, more paint, more design work, more flow correction materials, etc. Flow visualization media tend to junk up the tunnel walls too, forcing extra cleaning work. As sort of a final concern you have heating--our 3x4 closed-loop, vented tunnel would heat up from room ambient 72 to 80+ at Q=10psf (around 60mph maybe) within 10 minutes. A tunnel with more area, lower flow velocity, and larger vents would probably not suffer this effect as much. Enclosing a tunnel for bike testing also has the benefit of not freezing the rider as a blowdown tunnel with an outdoor inlet would. I'm a bit curious how A2 fares in the summer, with temps here in the 90's and 25-50hp of waste heat generation, but for the winter it's comfy.

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__________________________

I tweet!

Yup, good points on the Crr measurement. We'll definitely spend a bit of time playing with it to determine the most robust way of extracting the data.

All 3 huh? Ok, I think we can do that. Newtons though....that's pushing it! (just kidding...it's not too hard to convince me to use the "correct" units)

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Chris Yu
Applied Technology
Specialized Bicycle Components
@chrisyuinc

Right...and it's also why, even though they refer to their tunnel as "closed circuit", the CFME wind tunnel in Geneva is actually closer in design to what is usually called "open circuit", like the A2 and Specialized tunnels, than it is to a traditional closed-circuit design like the San Diego LSWT.

Semantics aside, it's really all a matter of understanding the pluses and minuses of each configuration and doing whatever is necessary to get good data out of the test section :-)

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http://bikeblather.blogspot.com/

Yeah, my thought is that you'd need to keep the "translational only" plots around (at least for a while, since that's what folks are used to seeing - pretty much the same reason you gave for reporting in gF for drag), but obviously the combined plot is really where the rubber meets the road (pun-ny analogy intended ;-)

Then, also having the rotational drag plot would also help to put that into perspective for folks...as in how much it can vary, both between setups and across yaw, and how much of the total drag it contributes. The "power to rotate" being as large as 15-30% of translational drag that Mavic quotes is nothing to sneeze at...

Of course, ultimately it would mean coming up with a way to combine the aerodynamic drag with the rolling resistance drag to create an overall "resistance to forward motion plot", sort of like what I've done with the combined aero drag + Crr plots I've put up on my blog.

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http://bikeblather.blogspot.com/

I would advice against using Newtons for the rotational drag for the same reason that rolling resistance shouldn't be given as a force. They both are moments so please use N·m

Translational drag, spin moment and total power can be a good combination. And you should give data about the effect of wheel angular velocity in both translation drag and spin moment for additional points ;-)

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http://cds-0.blogspot.com

There is a way to do it. You have to set up the resistance of the electric brake to a value high enough to prevent wheel acceleration. Knowing rider power and power dissipated in rolling resistance, you can calculate power to spin as the difference between rider power and the sum of rolling resistance and braking power

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http://cds-0.blogspot.com

Well, technically that moment is the result of a difference in force of compression in the leading and trailing sections of the contact patch ;-)

I don't have any problem with resolving the moment into a force at the contact patch...after all, that's what the rider "feels" as a resistance. In fact, it's why some folks confuse high Crr with high mass...and one can think of changes in Crr as being analogous to changes road grade.

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http://bikeblather.blogspot.com/

We are getting deep into dynamics here but technically a force at the contact patch doesn't dissipate power due to the RWS condition ;-). That force brought to the wheel axe is the force that is tipically defined as rolling resistance force. Due to the definition of Crr (the ratio between the shift of the center of pressure in the contact patch and the rolling radius), both powers are the same

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http://cds-0.blogspot.com

Okay, I admit I'm lost. This is almost surely because I've never been to a full-size wind tunnel testing full-size bikes (hint hint) but doesn't that mean you're going to have confounding with the wheel and drum bearings, and the tire/drum interface? If your drum and drum motor are attached to the balance surface why do you have to worry about the power the drum motors are drawing, or their efficiency?

You can know how much power you are using to drive the drum but you don't know how much of that power goes to the wheel, so you need to know the efficiency. Knowing that the wheel turns at constant velocity and power dissipated due to rolling resistance, you can calculate power to spin

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http://cds-0.blogspot.com

Hi Robert,

Epic-O basically hit it above. You're right that we'll also be measuring rolling resistance (tire/drum interface) and bearing drag on the drums, but that's pretty hard to avoid. But basically, we measure how much power it takes to turn the drum (and thus the tire and wheel) at a constant speed. Then if we want to extract out rotational aerodynamic drag, we need to know Crr to take out that contribution from the measurement.

By the way, you're also pretty local so you'll have to stop by for a tour sometime too!

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Chris Yu
Applied Technology
Specialized Bicycle Components
@chrisyuinc

For everyone following along, I know we're really into the nitty gritty here (which is awesome) but please (please!!) don't be scared away. No questions too basic or simple!

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Chris Yu
Applied Technology
Specialized Bicycle Components
@chrisyuinc

Right, but why would that be better than measuring the difference in forces between a stationary wheel and a wheel rotating at airspeed? Wouldn't that difference be the net effect of the rotation, both from the wheel but also any downstream effect on the frame? (I'm sure the reason is sensible -- it's just that because I'm only familiar with field testing I just get the net effect anyway).

Hi Chris,

Do you have any comparison data on your new Roval wheels with other brands?

Thanks,
Alex

Chris- 2 questions:

1. Have you done much research into how a reduced q factor would reduce drag on the cyclist? Obree and Ekimov both used narrow q bikes to great success and recently a British fellow did a study that efficiency of the rider was slightly higher with low q values. Seems like alot of cranksets have ridiculosly high q factors, and there is almost no way to get a narrow one without going back to square taper.

2. Have you shown boundary layer trip strips to be very helpful (on rider, bike or helmet) ?

Thanks,
BJ

You have to enforce wheel velocity using a drum to get something meaningful (if not drag won't be constant). To make the wheel rotate at airspeed you need to raise the bike to avoid contact (if there is contact, the angular velocity would be very low) with the WT ground but the wheel would accelerate until infinity and the direction of spin is wheel-dependent

The only possible way to avoid acceleration is to reduce the normal load on the wheel to equilibrate rolling resistance moment and power to spin for the desired wheel velocity but this is very complex and would need special wind tunnel fixtures. This could work for the H3 but not for the rest of wheels

With field testing you get a mixed CdA-moment coefficent so when manufacturers start to give power to spin values, it can become a source of error that hasn't been taken into account until now

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http://cds-0.blogspot.com

Oh, I completely understand the wheel has to be driven to match airspeed. What I don't quite understand is why you have to monitor the power drawn by the drum motors and their efficiency (and the bearing losses) in order to get the "power to rotate." Why can't you do it backwards by measuring the drag difference between rotating and non-rotating wheels. What does measuring the power at the motor (and its efficiency and the losses in the bearings) tell you that you don't get the other way?

With your proposed method you can only calculate the differences in translational drag. You don't get any info about the power to rotate

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http://cds-0.blogspot.com

Air molecules don't know the name you've given the path, as they try to return to the low-pressure zone you call the inlet. Even in what humans call "closed loop" return paths, the air molecules sometimes double-back a little bit. Really, they don't care -- they'll eventually get back around the merry-go-round.

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AndyF
bike geek

Forgive me, I'm used to field tests so this is not something I'm used to thinking about. I'm sure you're right but I'm still confused. If you hold the wheel fixed then don't you get the translational aero piece, and then if you drive the wheel at airspeed don't you get the total aero drag from the whole thing? Then isn't the difference due to the aero bit due to rotation, and it excludes the bearing losses?