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  #1  
Old 10-26.-2003
Andrew Price
 
Posts: n/a
Default Alternate and simpler measurement of power output?

Assume measurement of power output is a useful aid to training.

Assume an ergo (simple supported bike frame, fixed wind resistant panels attached to the spokes of a
driven wheel, with a conventional derailleur of x selectable gears and the ability to measure
cadence ).

Wouldn't the various gear and cadence combinations possible for any rider give distinct differential
power outputs ?

I appreciate the graphics may not be stunning (set the gear, watch the cadence!) and you may have to
calculate the output in watts with the assistance of a formula, but it seems to me on the simple and
relatively inexpensive equipment described you are setting repeatable power outputs and at a
considerably lesser cost than that offered by srm, power tap and other like devices.

or am I missing something ?

ps is there any easy way to express the formula to convert to watts? - presumably you need some
measure of the resistance created by the panels in the driven wheel?

best, Andrew


Alternate and simpler measurement of power output?







  #2  
Old 10-26.-2003
Phil Holman
 
Posts: n/a
Default Re: Alternate and simpler measurement of power output?

"Andrew Price" <arathorn@bigpond.net.au.x1> wrote in message
news:7JNmb.165991$bo1.72119@news-server.bigpond.net.au...
> Assume measurement of power output is a useful aid to training.
>
> Assume an ergo (simple supported bike frame, fixed wind resistant
panels
> attached to the spokes of a driven wheel, with a conventional
derailleur of
> x selectable gears and the ability to measure cadence ).
>
> Wouldn't the various gear and cadence combinations possible for any
rider
> give distinct differential power outputs ?
>
> I appreciate the graphics may not be stunning (set the gear, watch the cadence!) and you may have
> to calculate the output in watts with the assistance of a formula, but it seems to me on the
> simple and
relatively
> inexpensive equipment described you are setting repeatable power
outputs and
> at a considerably lesser cost than that offered by srm, power tap and
other
> like devices.

Tacx has a unit that does this with a mag trainer. The shortcomings of doing this with a wind
trainer are: Spoke mounted blades are real noisy (velodyne) and for drag alone have to be fairly
large but this still results in hill climbing workouts being somewhat limited. The Cateye
cyclosimulator encorporates a mag with the fan but this is a hopelessly inaccurate unit when it
comes to absolute power readout. I'm positive that the algorythm of this unit could be improved but
for some reason Cateye are not motivated to do so.

>
> or am I missing something ?
>
> ps is there any easy way to express the formula to convert to watts? - presumably you need some
> measure of the resistance created by the
panels in
> the driven wheel?

Power is roughly 1/2Cd* rho*V^3*A which for a rotating blade on a fan could be a neat integral
except the interaction of the blades has a significant lessening effect on the actual power
required. It's far easier to calibrate the power as a function of speed with another accurate
measuring device. A while back someone was selling mag trainer to PC hookup kits where a power
readout could be obtained.

Phil Holman
  #3  
Old 10-26.-2003
Carl Fogel
 
Posts: n/a
Default Re: Alternate and simpler measurement of power output?

"Andrew Price" <arathorn@bigpond.net.au.x1> wrote in message
news:<7JNmb.165991$bo1.72119@news-server.bigpond.net.au>...
> Assume measurement of power output is a useful aid to training.
>
> Assume an ergo (simple supported bike frame, fixed wind resistant panels attached to the spokes of
> a driven wheel, with a conventional derailleur of x selectable gears and the ability to measure
> cadence ).
>
> Wouldn't the various gear and cadence combinations possible for any rider give distinct
> differential power outputs ?
>
> I appreciate the graphics may not be stunning (set the gear, watch the cadence!) and you may have
> to calculate the output in watts with the assistance of a formula, but it seems to me on the
> simple and relatively inexpensive equipment described you are setting repeatable power outputs and
> at a considerably lesser cost than that offered by srm, power tap and other like devices.
>
> or am I missing something ?
>
> ps is there any easy way to express the formula to convert to watts? - presumably you need some
> measure of the resistance created by the panels in the driven wheel?
>
> best, Andrew

Dear Andrew,

Luddite!

If your kind of heresy were rampant, the thermometer by my kitchen window wouldn't read in tenths
of a degree.

Carl Fogel
  #4  
Old 10-27.-2003
Robert Chung
 
Posts: n/a
Default Re: Alternate and simpler measurement of power output?

Andrew Price wrote:
> Assume measurement of power output is a useful aid to training.
>
> Assume an ergo (simple supported bike frame, fixed wind resistant panels attached to the spokes of
> a driven wheel, with a conventional derailleur of x selectable gears and the ability to measure
> cadence ).
>
> Wouldn't the various gear and cadence combinations possible for any rider give distinct
> differential power outputs ?
>
> I appreciate the graphics may not be stunning (set the gear, watch the cadence!) and you may have
> to calculate the output in watts with the assistance of a formula, but it seems to me on the
> simple and relatively inexpensive equipment described you are setting repeatable power outputs and
> at a considerably lesser cost than that offered by srm, power tap and other like devices.
>
> or am I missing something ?
>
> ps is there any easy way to express the formula to convert to watts? - presumably you need some
> measure of the resistance created by the panels in the driven wheel?
>
> best, Andrew

http://www.geocities.com/almost_fast/trainerpower/
  #5  
Old 10-27.-2003
Andy Coggan
 
Posts: n/a
Default Re: Alternate and simpler measurement of power output?

"Andrew Price" <arathorn@bigpond.net.au.x1> wrote in message
news:7JNmb.165991$bo1.72119@news-server.bigpond.net.au...
> Assume measurement of power output is a useful aid to training.
>
> Assume an ergo (simple supported bike frame, fixed wind resistant panels attached to the spokes of
> a driven wheel, with a conventional derailleur
of
> x selectable gears and the ability to measure cadence ).
>
> Wouldn't the various gear and cadence combinations possible for any rider give distinct
> differential power outputs ?
>
> I appreciate the graphics may not be stunning (set the gear, watch the cadence!) and you may have
> to calculate the output in watts with the assistance of a formula, but it seems to me on the
> simple and relatively inexpensive equipment described you are setting repeatable power outputs
and
> at a considerably lesser cost than that offered by srm, power tap and
other
> like devices.
>
> or am I missing something ?
>
> ps is there any easy way to express the formula to convert to watts? - presumably you need some
> measure of the resistance created by the panels
in
> the driven wheel?

It's been done before, e.g., Concept II rowing ergometer, Kingcycle cycle ergometer. Not as
accurate as you might think, due to variations in flywheel/fan characteristics, environmental
conditions, etc.:

http://www.ncbi.nlm.nih.gov/entrez/q...=11039654&dop-
t=Abstract

http://www.ncbi.nlm.nih.gov/entrez/q...=10834352&dop-
t=Abstract

One limitation people may not immediately recognize is the fact that how close you place the fan
relative to other objects (e.g., walls) can affect the power requirement.

Andy Coggan
  #6  
Old 10-27.-2003
Andy Coggan
 
Posts: n/a
Default Re: Alternate and simpler measurement of power output?

"Phil Holman" <philjud@earthlink.not> wrote in message
news:rJSmb.3028$RQ1.715@newsread3.ne....earthlink.net...
>
> "Andrew Price" <arathorn@bigpond.net.au.x1> wrote in message
> news:7JNmb.165991$bo1.72119@news-server.bigpond.net.au...
> > Assume measurement of power output is a useful aid to training.
> >
> > Assume an ergo (simple supported bike frame, fixed wind resistant
> panels
> > attached to the spokes of a driven wheel, with a conventional
> derailleur of
> > x selectable gears and the ability to measure cadence ).
> >
> > Wouldn't the various gear and cadence combinations possible for any
> rider
> > give distinct differential power outputs ?

> Tacx has a unit that does this with a mag trainer. The shortcomings of doing this with a wind
> trainer are: Spoke mounted blades are real noisy (velodyne)

You must be thinking of the Schwinn Airdyne - the Schwinn Velodyne (originally developed by
Frontline Technologies) is electromagnetically braked.

Andy Coggan
  #7  
Old 10-27.-2003
Phil Holman
 
Posts: n/a
Default Re: Alternate and simpler measurement of power output?

"Andy Coggan" <acoggan@earthlink.net> wrote in message
news:tVenb.4293$RQ1.2355@newsread3.n....earthlink.net...
> "Phil Holman" <philjud@earthlink.not> wrote in message
> news:rJSmb.3028$RQ1.715@newsread3.ne....earthlink.net...
> >
> > "Andrew Price" <arathorn@bigpond.net.au.x1> wrote in message
> > news:7JNmb.165991$bo1.72119@news-server.bigpond.net.au...
> > > Assume measurement of power output is a useful aid to training.
> > >
> > > Assume an ergo (simple supported bike frame, fixed wind resistant
> > panels
> > > attached to the spokes of a driven wheel, with a conventional
> > derailleur of
> > > x selectable gears and the ability to measure cadence ).
> > >
> > > Wouldn't the various gear and cadence combinations possible for
any
> > rider
> > > give distinct differential power outputs ?
>
> > Tacx has a unit that does this with a mag trainer. The shortcomings
of
> > doing this with a wind trainer are: Spoke mounted blades are real noisy (velodyne)
>
> You must be thinking of the Schwinn Airdyne - the Schwinn Velodyne (originally developed by
> Frontline Technologies) is
electromagnetically
> braked.

Yes, the one with the arm levers and a noisy fan/wheel. The velodyne was/is a very good unit except
when racing the pack and cresting a hill, the pack took off much faster than your average rider. I
quickly learned to get near the front. This was 10 years ago and I don't see or hear about this
unit anymore.

Phil Holman
  #8  
Old 10-28.-2003
Phil Holman
 
Posts: n/a
Default Re: Alternate and simpler measurement of power output?

"Andy Coggan" <acoggan@earthlink.net> wrote in message
news:2Senb.4292$RQ1.4204@newsread3.n....earthlink.net...
> "Andrew Price" <arathorn@bigpond.net.au.x1> wrote in message
> news:7JNmb.165991$bo1.72119@news-server.bigpond.net.au...
> > Assume measurement of power output is a useful aid to training.
> >
> > Assume an ergo (simple supported bike frame, fixed wind resistant
panels
> > attached to the spokes of a driven wheel, with a conventional
derailleur
> of
> > x selectable gears and the ability to measure cadence ).
> >
> > Wouldn't the various gear and cadence combinations possible for any
rider
> > give distinct differential power outputs ?
> >
> > I appreciate the graphics may not be stunning (set the gear, watch
the
> > cadence!) and you may have to calculate the output in watts with the assistance of a formula,
> > but it seems to me on the simple and
relatively
> > inexpensive equipment described you are setting repeatable power
outputs
> and
> > at a considerably lesser cost than that offered by srm, power tap
and
> other
> > like devices.
> >
> > or am I missing something ?
> >
> > ps is there any easy way to express the formula to convert to
watts? -
> > presumably you need some measure of the resistance created by the
panels
> in
> > the driven wheel?
>
> It's been done before, e.g., Concept II rowing ergometer, Kingcycle
cycle
> ergometer. Not as accurate as you might think, due to variations in flywheel/fan characteristics,
> environmental conditions, etc.:

The Concept II recalibrates on every return stoke (coast down). Assuming environmental fluctuations
aren't exactly in phase with the return stroke, damper settings, gear selection or air density make
little difference (within detectable limits) to the power readout accuracy and the unit
automatically compensates for distance. Accurate times for a set distance is the main area of
interest as people compare results from all over the world. They even hold world championships and
results seem to be very consistent (no fast machines or ideal locations). The fan is even drilled
out to calibrate the mass moment of inertia accurately. IMO the unit is exceptionally well
engineered.

Phil Holman
  #9  
Old 10-28.-2003
Andy Coggan
 
Posts: n/a
Default Re: Alternate and simpler measurement of power output?

"Phil Holman" <philjud@earthlink.not> wrote in message
news:uFknb.4392$Px2.1254@newsread4.n....earthlink.net...
>
> The velodyne was/is a very good unit except when racing the pack and cresting a hill, the pack
> took off much faster than your average rider. I quickly learned to get near the front. This was 10
> years ago and I don't see or hear about this unit anymore.

www.velodynesports.com

I agree that racing with/against one of the "packs" makes for a less-than-perfect simulation of real
life. However, as I have habitually found myself between levels (can pretty easily solo away from
the level 7 pack, but can't get away from the level 8 group...yet just "sitting in" is too easy),
I've always just TTed different "courses", or used it in ergometer mode.

(BTW, CyclingPeaksSoftware reads Velodyne data files now.)

Andy Coggan
  #10  
Old 11-01.-2003
Ryan Cousineau
 
Posts: n/a
Default Re: 1 Horsepower [1 Clydesdale] = 1.25 Lances

In article <20031101001837.08266.00000006@mb-m18.aol.com>, bretcahill@aol.com (BretCahill) wrote:

> Find a steep mountain with altitude markings and remember:
>
> 1 hp = 550 foot-lbs/sec.
>
> If you + your bike = 183 lbs and you can climb 3600 feet in one hour, you are putting out 1/3 hp
> just in overcoming gravity.
>
> If your speed is less than 10 mph, the air friction is pretty negligible in comparison.

If you really care about this stuff, go read Dr. Michele Ferrari. He's come up with a formula he
calls "VAM" which offers comparative performance evaluation during climbs (it is designed with
correction factors for the steepness and duration of the hill, etc.)

He's quite hot on it, and while I haven't investigated it much, Lance seems to think it
has some use.

I think cyclingnews published his VAM-oriented diaries of the GC races this year, and he also has
his own site (google it yourself) featuring a mix of free and for-pay content oriented around VAM
and other training concepts.

Share & Enjoy,
--
Ryan Cousineau, rcousine@sfu.ca http://www.sfu.ca/~rcousine President, Fabrizio Mazzoleni Fan Club
  #11  
Old 11-01.-2003
Carl Fogel
 
Posts: n/a
Default Re: 1 Horsepower [1 Clydesdale] = 1.25 Lances

bretcahill@aol.com (BretCahill) wrote in message
news:<20031101001837.08266.00000006@mb-m18.aol.com>...
> Find a steep mountain with altitude markings and remember:
>
> 1 hp = 550 foot-lbs/sec.
>
> If you + your bike = 183 lbs and you can climb 3600 feet in one hour, you are putting out 1/3 hp
> just in overcoming gravity.
>
> If your speed is less than 10 mph, the air friction is pretty negligible in comparison.
>
>
> Bret Cahill

Dear Bret,

You may already have visited it, but there's a wonderfully detailed web page for putting in all
sorts of values for wind, slope, power, weight, surface roughness, streamlining, and probably credit
rating at:

http://www.analyticcycling.com/ForcesPower_Page.html

These people know how to pander to my love of lurid, quickly calculated numbers. A conversion of
watts to horsepower would probably be fairly easy.

Carl Fogel
  #12  
Old 11-02.-2003
David Kerber
 
Posts: n/a
Default Re: 1 Horsepower [1 Clydesdale] = 1.25 Lances

In article <8bbde8fc.0311011648.11092b56@posting.google.com>, carlfogel@comcast.net says...
> bretcahill@aol.com (BretCahill) wrote in message
> news:<20031101001837.08266.00000006@mb-m18.aol.com>...
> > Find a steep mountain with altitude markings and remember:
> >
> > 1 hp = 550 foot-lbs/sec.
> >
> > If you + your bike = 183 lbs and you can climb 3600 feet in one hour, you are putting out 1/3 hp
> > just in overcoming gravity.
> >
> > If your speed is less than 10 mph, the air friction is pretty negligible in comparison.
> >
> >
> > Bret Cahill
>
> Dear Bret,
>
> You may already have visited it, but there's a wonderfully detailed web page for putting in all
> sorts of values for wind, slope, power, weight, surface roughness, streamlining, and probably
> credit rating at:
>
> http://www.analyticcycling.com/ForcesPower_Page.html
>
> These people know how to pander to my love of lurid, quickly calculated numbers. A conversion of
> watts to horsepower would probably be fairly easy.

Yes, it is: divide by 746.

--
Dave Kerber Fight spam: remove the ns_ from the return address before replying!

REAL programmers write self-modifying code.
  #13  
Old 11-02.-2003
Phil Holman
 
Posts: n/a
Default Re: 1 Horsepower [1 Clydesdale] = 1.25 Lances

"David Kerber" <ns_dkerber@ns_ids.net> wrote in message
news:MPG.1a0e3f1a2c3557d9897ac@news.ids.net...
> In article <8bbde8fc.0311011648.11092b56@posting.google.com>, carlfogel@comcast.net says...
> > bretcahill@aol.com (BretCahill) wrote in message
news:<20031101001837.08266.00000006@mb-m18.aol.com>...
> > > Find a steep mountain with altitude markings and remember:
> > >
> > > 1 hp = 550 foot-lbs/sec.
> > >
> > > If you + your bike = 183 lbs and you can climb 3600 feet in one hour, you are putting out 1/3
> > > hp just in overcoming gravity.
> > >
> > > If your speed is less than 10 mph, the air friction is pretty negligible in comparison.
> > >
> > >
> > > Bret Cahill
> >
> > Dear Bret,
> >
> > You may already have visited it, but there's a wonderfully detailed web page for putting in all
> > sorts of values for wind, slope, power, weight, surface roughness, streamlining, and probably
> > credit rating at:
> >
> > http://www.analyticcycling.com/ForcesPower_Page.html
> >
> > These people know how to pander to my love of lurid, quickly calculated numbers. A conversion of
> > watts to horsepower would probably be fairly easy.
>
> Yes, it is: divide by 746.

And just to clarify what a watt is........ 1watt = 1Newton-meter/sec = 1Joule/sec 1 Newton will
accelerate a 1kg mass at 1meter/sec/sec.

Phil Holman
  #14  
Old 11-02.-2003
Carl Fogel
 
Posts: n/a
Default Re: 1 Horsepower [1 Clydesdale] = 1.25 Lances

David Kerber <ns_dkerber@ns_ids.net> wrote in message
news:<MPG.1a0e3f1a2c3557d9897ac@news.ids.net>...
> In article <8bbde8fc.0311011648.11092b56@posting.google.com>, carlfogel@comcast.net says...
> > bretcahill@aol.com (BretCahill) wrote in message
> > news:<20031101001837.08266.00000006@mb-m18.aol.com>...
> > > Find a steep mountain with altitude markings and remember:
> > >
> > > 1 hp = 550 foot-lbs/sec.
> > >
> > > If you + your bike = 183 lbs and you can climb 3600 feet in one hour, you are putting out 1/3
> > > hp just in overcoming gravity.
> > >
> > > If your speed is less than 10 mph, the air friction is pretty negligible in comparison.
> > >
> > >
> > > Bret Cahill
> >
> > Dear Bret,
> >
> > You may already have visited it, but there's a wonderfully detailed web page for putting in all
> > sorts of values for wind, slope, power, weight, surface roughness, streamlining, and probably
> > credit rating at:
> >
> > http://www.analyticcycling.com/ForcesPower_Page.html
> >
> > These people know how to pander to my love of lurid, quickly calculated numbers. A conversion of
> > watts to horsepower would probably be fairly easy.
>
> Yes, it is: divide by 746.

Dear Dave,

Or, as I was taught, by half of 1492, when Columbus sailed the ocean blue.

Carl Fogel
 

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