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  #151  
Old December 21st 18, 12:42 AM posted to rec.bicycles.tech
John B. Slocomb
external usenet poster
 
Posts: 805
Default Power on hills.

On Fri, 21 Dec 2018 07:57:03 +1100, James
wrote:

On 21/12/18 4:29 am, jbeattie wrote:
On Wednesday, December 19, 2018 at 10:04:58 PM UTC-8, James wrote:
On 20/12/18 4:19 pm, jbeattie wrote:
On Wednesday, December 19, 2018 at 6:42:50 PM UTC-8, James
wrote:

The iBike Newton does the maths & physics of what you theorise,
but it isn't perfect according to the reviews I've read.

https://cyclingtips.com/2013/09/ibik...-meter-review/



That's a power estimator. If you want a power meter, you have to
pay.

For a start, the iBike Newton measures a number of inputs and
*calculates* the input power according to the laws of physics. It
does measure power, just not in the same way. And there are others
like it, mentioned here;
https://www.dcrainmaker.com/2016/09/...6-edition.html



The Stages device, like most other power meters, uses strain gauges
(resistors) that detect elastic deformation in a part of the drive
train that is expected to have a predictable force/strain
relationship, and from that and other measured inputs like crank
position, for example, can calculate the input power.

Make no mistake, the Stages device does not measure power directly,
and is not infallible. The electronics are not perfect and even if
the device is calibrated in the beginning, it will not remain
calibrated for ever. For this reason people often compare power
meters to try to find the most accurate and stable.

https://www.cyclingweekly.com/news/l...-meters-330322



Yes, but a meter calculating power based on wind speed and gradient
is useless for indoor training and unreliable outdoors.


That depends on your indoor setup. On this inclined cycling mill, it
would work just fine. https://www.youtube.com/watch?v=PVxGFOb1KTY


The narrator commented that as they were indoors there was no wind and
thus not exactly the conditions as actually on the road but didn't I
read some years ago about Armstrong being tested in a wind tunnel?

(but it must have been dangerous as the test rider wasn't wearing a
helmet :-)

cheers,

John B.


Ads
  #152  
Old December 21st 18, 01:50 AM posted to rec.bicycles.tech
Andre Jute[_2_]
external usenet poster
 
Posts: 10,422
Default Power on hills.

On Thursday, December 20, 2018 at 6:17:03 PM UTC, wrote:
On Thursday, December 20, 2018 at 6:29:43 PM UTC+1, jbeattie wrote:
On Wednesday, December 19, 2018 at 10:04:58 PM UTC-8, James wrote:
On 20/12/18 4:19 pm, jbeattie wrote:
On Wednesday, December 19, 2018 at 6:42:50 PM UTC-8, James wrote:

The iBike Newton does the maths & physics of what you theorise, but
it isn't perfect according to the reviews I've read.

https://cyclingtips.com/2013/09/ibik...-meter-review/

That's a power estimator. If you want a power meter, you have to
pay.

For a start, the iBike Newton measures a number of inputs and
*calculates* the input power according to the laws of physics. It does
measure power, just not in the same way. And there are others like it,
mentioned here;
https://www.dcrainmaker.com/2016/09/...6-edition.html

The Stages device, like most other power meters, uses strain gauges
(resistors) that detect elastic deformation in a part of the drive train
that is expected to have a predictable force/strain relationship, and
from that and other measured inputs like crank position, for example,
can calculate the input power.

Make no mistake, the Stages device does not measure power directly, and
is not infallible. The electronics are not perfect and even if the
device is calibrated in the beginning, it will not remain calibrated for
ever. For this reason people often compare power meters to try to find
the most accurate and stable.

https://www.cyclingweekly.com/news/l...-meters-330322


Yes, but a meter calculating power based on wind speed and gradient is useless for indoor training and unreliable outdoors. https://www.bikeradar.com/us/road/ge...wton-15-49649/ You get what you pay for. Personally, I wouldn't pay for any of them since I'm not training for anything.

-- Jay Beattie.


You can argue about the need for a powermeter or if you want to spend the money but buying a device that measures windspeed and then have to estimate frontal area, rolling resistance etc etc to calculate power just to save money I don't get. Either you use a decent powermeter or just rely on the numbers Strava spit out. I can tell you that it makes quite a difference whether your average power is 200 Watt or 220 Watt during a 2 -3 hour ride.

Lou


I have little faith in working out power indirectly, especially on a bicycle and human body on the road in the wind. There are just too many crazy variables.

In the design stages of automobiles, where calculating the frontal area is a science, and very good guesses can be made about air resistance because the effects of adaptations like rounding edges are very well known, it is still easy to be 10 or even 20% out on the apparently simple question, "How many horses are required to move this car at 160kph/100mph?" It is usually far easier to calculate the torque required for acceleration because even in an unbuilt car the mass, the main determinant of acceleration, can be calculated to much closer margins.

Andre Jute
Ultimate precision is a chimera
  #153  
Old December 21st 18, 02:29 AM posted to rec.bicycles.tech
James[_8_]
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Posts: 6,153
Default Power on hills.

On 21/12/18 10:42 am, John B. Slocomb wrote:
On Fri, 21 Dec 2018 07:57:03 +1100, James
wrote:

On 21/12/18 4:29 am, jbeattie wrote:
Yes, but a meter calculating power based on wind speed and gradient
is useless for indoor training and unreliable outdoors.


That depends on your indoor setup. On this inclined cycling mill,
it would work just fine.
https://www.youtube.com/watch?v=PVxGFOb1KTY


The narrator commented that as they were indoors there was no wind
and thus not exactly the conditions as actually on the road but
didn't I read some years ago about Armstrong being tested in a wind
tunnel?


It matters not. Imagine you are out riding on the road up a hill with a
following wind such that your air speed is zero.

Or you could turn on a fan to induce a non stationary air stream.

Either way, because the iBike device *measures* air speed, it will (or
at least should) calculate the power from speed (measured by the wheel
sensor) and gradient (measured by the device built in accelerometer).

Whether it would work or not I cannot say, but there is enough
information available for it to work just fine. (I guess if it measures
elevation change instead of gradient, it's stuffed ;-)


Most power meters use strain gauges (special resistors) to measure
elastic deformation of a part of the drive train, and use that to
calculate a force or torque. I remember seeing a mates polar power
meter that consisted of what I think was a guitar pickup on the chain
stay. I believe it worked by monitoring the chain vibrations and
understanding that with higher chain tension the natural frequency would
be higher. It seemed a bit smoke and mirrors, but it did seem to work.
The point is that what we think of as more direct measures, really
aren't any more direct.

At least the Polar device and the iBike device don't measure one crank
and assume both legs contribute the same power. (I know the more
expensive Stages setup has a strain gauge on each crank, but these other
systems don't need that level of complexity.)

--
JS
  #154  
Old December 21st 18, 08:14 PM posted to rec.bicycles.tech
[email protected]
external usenet poster
 
Posts: 1,261
Default Power on hills.

On Wednesday, December 19, 2018 at 10:04:58 PM UTC-8, James wrote:
On 20/12/18 4:19 pm, jbeattie wrote:
On Wednesday, December 19, 2018 at 6:42:50 PM UTC-8, James wrote:


The iBike Newton does the maths & physics of what you theorise, but
it isn't perfect according to the reviews I've read.

https://cyclingtips.com/2013/09/ibik...-meter-review/


That's a power estimator. If you want a power meter, you have to
pay.


For a start, the iBike Newton measures a number of inputs and
*calculates* the input power according to the laws of physics. It does
measure power, just not in the same way. And there are others like it,
mentioned here;
https://www.dcrainmaker.com/2016/09/...-edition..html

The Stages device, like most other power meters, uses strain gauges
(resistors) that detect elastic deformation in a part of the drive train
that is expected to have a predictable force/strain relationship, and
from that and other measured inputs like crank position, for example,
can calculate the input power.

Make no mistake, the Stages device does not measure power directly, and
is not infallible. The electronics are not perfect and even if the
device is calibrated in the beginning, it will not remain calibrated for
ever. For this reason people often compare power meters to try to find
the most accurate and stable.

https://www.cyclingweekly.com/news/l...-meters-330322

I don't care about power, but those who do get power meters.


I don't care what other people do. Tom wondered about measuring the
external forces and such to calculate the input power. That is
precisely what the iBike Newton device does, which is why I included a
link to it, for his interest.

My son works for Stages, so I'm plugging that product.


I knew that already.

Good enough
for Chris Froome.
https://www.roadbikereview.com/revie.../froome-stages


Again, I don't care what other people do. I make up my own mind on
these sorts of things.

And Chris Froome is the least of my interests - except that I hope the
TdF organisers would listen to me and include a dozen stages where it is
fairly flat and there is a good chance of strong cross winds. That
would fix the spider and others like him!
https://michaelvalenti.com/the-oddit...-chris-froome/

There is a particular inaccuracy between pedal power meters and things like the PowerPod - the strain gauges measure the actual input power and not that used to actually move the bike forward. There are significant losses (by significant I mean measurable and not large) in pressing the pedals, flexing all of the components from frame to chain rings and cogs and the chain angle. This might add up in general to perhaps 5 watts so who cares as long as you get one power meter and stick with it.
  #155  
Old December 21st 18, 08:18 PM posted to rec.bicycles.tech
[email protected]
external usenet poster
 
Posts: 1,261
Default Power on hills.

On Thursday, December 20, 2018 at 9:29:43 AM UTC-8, jbeattie wrote:
On Wednesday, December 19, 2018 at 10:04:58 PM UTC-8, James wrote:
On 20/12/18 4:19 pm, jbeattie wrote:
On Wednesday, December 19, 2018 at 6:42:50 PM UTC-8, James wrote:


The iBike Newton does the maths & physics of what you theorise, but
it isn't perfect according to the reviews I've read.

https://cyclingtips.com/2013/09/ibik...-meter-review/

That's a power estimator. If you want a power meter, you have to
pay.


For a start, the iBike Newton measures a number of inputs and
*calculates* the input power according to the laws of physics. It does
measure power, just not in the same way. And there are others like it,
mentioned here;
https://www.dcrainmaker.com/2016/09/...6-edition.html

The Stages device, like most other power meters, uses strain gauges
(resistors) that detect elastic deformation in a part of the drive train
that is expected to have a predictable force/strain relationship, and
from that and other measured inputs like crank position, for example,
can calculate the input power.

Make no mistake, the Stages device does not measure power directly, and
is not infallible. The electronics are not perfect and even if the
device is calibrated in the beginning, it will not remain calibrated for
ever. For this reason people often compare power meters to try to find
the most accurate and stable.

https://www.cyclingweekly.com/news/l...-meters-330322


Yes, but a meter calculating power based on wind speed and gradient is useless for indoor training and unreliable outdoors. https://www.bikeradar.com/us/road/ge...wton-15-49649/ You get what you pay for. Personally, I wouldn't pay for any of them since I'm not training for anything.

-- Jay Beattie.


Absolutely it cannot be used indoors on a trainer. But it is completely reliable under normal riding conditions. And precise power isn't necessary. Repeatability is with sufficiently accurate data - you don't need to know if you are generating 300 watts today. You need to know if you made any improvements or losses.
  #156  
Old December 21st 18, 09:25 PM posted to rec.bicycles.tech
[email protected]
external usenet poster
 
Posts: 1,261
Default Power on hills.

At low speeds - those below 100 mph of so, aerodynamic drag really isn't a large loss unless you're playing for real small power such as that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ρ A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
ρ = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

..5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
..5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200 lbs is huge. KE = ½mv²



  #157  
Old December 22nd 18, 12:52 AM posted to rec.bicycles.tech
Andre Jute[_2_]
external usenet poster
 
Posts: 10,422
Default Power on hills.

On Friday, December 21, 2018 at 8:25:37 PM UTC, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really isn't a large loss unless you're playing for real small power such as that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ρ A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
ρ = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200 lbs is huge. KE = ½mv²


Thanks for putting the numbers to my argument, Tom.

Andre Jute
DESIGNING AND BUILDING SPECIAL CARS; Batsford, London; Bentley, Boston
  #158  
Old December 22nd 18, 02:00 AM posted to rec.bicycles.tech
James[_8_]
external usenet poster
 
Posts: 6,153
Default Power on hills.

On 22/12/18 7:25 am, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really isn't a large loss unless you're playing for real small power such as that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ρ A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
ρ = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200 lbs is huge. KE = ½mv²




Ignoring the mixture of units, (you used 30 sq. feet without converting
to sq. metres), you also forgot to square the velocity. The difference
in force between those two speeds is huge.

30 sq. ft = 2.8 sq. m

0.5 x 0.5 x 1.2 x 2.8 x 27 x 27 = 612 N
0.5 x 0.5 x 1.2 x 2.8 x 45 x 45 = 1701 N

The difference in power to maintain those speeds is even greater
(proportional to V^3);

612 x 27 = 16.5 kW
1701 x 45 = 76.5 kW

You also talk about power to accelerate, but give an energy equation,
and include an imperial mass value in an otherwise metric discussion.

364.5 kJ to accelerate to 27 m/s
1012.5 kJ to accelerate to 45 m/s

(ignoring aerodynamic drag)

22 seconds travelling at 27 m/s will require the same energy as the
automobile has in kinetic energy.

13 seconds travelling at 45 m/s will require the same energy as the
automobile has in kinetic energy.


Now how huge is the power required to accelerate compared with the power
required to maintain such speeds?

--
JS

  #159  
Old December 22nd 18, 02:08 AM posted to rec.bicycles.tech
Ralph Barone[_4_]
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Posts: 853
Default Power on hills.

Andre Jute wrote:
On Friday, December 21, 2018 at 8:25:37 PM UTC, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really isn't
a large loss unless you're playing for real small power such as that
developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ρ A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
ρ = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200
lbs is huge. KE = ½mv²


Thanks for putting the numbers to my argument, Tom.

Andre Jute
DESIGNING AND BUILDING SPECIAL CARS; Batsford, London; Bentley, Boston


You both missed the V^2 in the equation. To go 1.6 times as fast requires
2.56 times as much force and around 4 times as much power.

PS: you also missed converting the 30 sqft to square metres.
  #160  
Old December 22nd 18, 02:14 AM posted to rec.bicycles.tech
JBeattie
external usenet poster
 
Posts: 5,870
Default Power on hills.

On Friday, December 21, 2018 at 5:00:57 PM UTC-8, James wrote:
On 22/12/18 7:25 am, wrote:
At low speeds - those below 100 mph of so, aerodynamic drag really isn't a large loss unless you're playing for real small power such as that developed by a human over relatively long periods of time.

Just as a demonstration.

30 square feet of frontal area
coefficient of drag of 0.5
This is similar to a family car

F = 0.5 C ρ A V^2

A = Reference area as (see figures above), m2.
C = Drag coefficient (see figures above), unitless.
F = Drag force, N.
V = Velocity, m/s.
ρ = Density of fluid (liquid or gas), kg/m3. (dry air at 70 degrees F ~ 1.2)

.5 x .5 x 1.2 x 30 x 27 m/s (60 mph) = 243 N
.5 x .5 x 1.2 x 30 x 45 m/s (100 mph) = 337 N


Whereas the power to accelerate the mass of a car which is about 2200 lbs is huge. KE = ½mv²




Ignoring the mixture of units, (you used 30 sq. feet without converting
to sq. metres), you also forgot to square the velocity. The difference
in force between those two speeds is huge.

30 sq. ft = 2.8 sq. m

0.5 x 0.5 x 1.2 x 2.8 x 27 x 27 = 612 N
0.5 x 0.5 x 1.2 x 2.8 x 45 x 45 = 1701 N

The difference in power to maintain those speeds is even greater
(proportional to V^3);

612 x 27 = 16.5 kW
1701 x 45 = 76.5 kW

You also talk about power to accelerate, but give an energy equation,
and include an imperial mass value in an otherwise metric discussion.

364.5 kJ to accelerate to 27 m/s
1012.5 kJ to accelerate to 45 m/s

(ignoring aerodynamic drag)

22 seconds travelling at 27 m/s will require the same energy as the
automobile has in kinetic energy.

13 seconds travelling at 45 m/s will require the same energy as the
automobile has in kinetic energy.


Now how huge is the power required to accelerate compared with the power
required to maintain such speeds?



Exactly 12. No, 13. The speed of light? Wait . . . now its coming to me. Blue.

God bless Nikki Terpstra (love him or hate him)
https://www.velonews.com/2018/04/new...lassics_462590

-- Jay Beattie.


 




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