I am interested in measuring the work (energy, measured in Joules, or
"watt-hours") used when riding a bike, in a certain time interval.
What I need to monitor is the tension in the chain (the stretching force
in the chain) and chain speed (see below).
Chain speed can be measured optically, with a photo-cell for example.
How can tension in the chain be measured? I can imagine an arm with a
spring and wheel that presses down on the chain; the angle of the arm is
proportional to the tension in the chain. Sounds impractical.
-- Are there more practical ways of measuring or estimating the amount
of work spent?
Thanks very much.
PS: "Proof".
Work = Force * displacement = Tension in chain * chain displacement
(assuming force = constant)
The instantaneous power (at moment t) is:
Power(t) = Force(t) * chain_velocity(t)
In article <xJ2ob.6546$[Only registered and activated users can see links. ].prodigy.com >, [Only registered and activated users can see links. ] says...
> I am interested in measuring the work (energy, measured in Joules, or
> "watt-hours") used when riding a bike, in a certain time interval.
>
> What I need to monitor is the tension in the chain (the stretching force
> in the chain) and chain speed (see below).
>
> Chain speed can be measured optically, with a photo-cell for example.
>
> How can tension in the chain be measured? I can imagine an arm with a
> spring and wheel that presses down on the chain; the angle of the arm is
> proportional to the tension in the chain. Sounds impractical.
>
> -- Are there more practical ways of measuring or estimating the amount
> of work spent?
Power is also torque * rpm (with appropriate unit conversions), so:
Speed can also be measured by wheel or pedal rpm, and you could
measure crank strain (bend) with a strain gauge as a measure of
applied torque. Then with a bit of math and a couple of calibration
checks you can calculate power.
.....
--
Dave Kerber
Fight spam: remove the ns_ from the return address before replying!
>> I am interested in measuring the work (energy, measured in Joules,
>> or "watt-hours") used when riding a bike, in a certain time
>> interval. What I need to monitor is the tension in the chain (the
>> stretching force in the chain) and chain speed (see below).
>> Chain speed can be measured optically, with a photo-cell for example.
>> How can tension in the chain be measured? I can imagine an arm with
>> a spring and wheel that presses down on the chain; the angle of the
>> arm is proportional to the tension in the chain. Sounds
>> impractical.
>> -- Are there more practical ways of measuring or estimating the
>> amount of work spent?
> Power is also torque * rpm (with appropriate unit conversions), so:
> Speed can also be measured by wheel or pedal rpm, and you could
> measure crank strain (bend) with a strain gauge as a measure of
> applied torque. Then with a bit of math and a couple of calibration
> checks you can calculate power.
Others have had similar interests and for that purpose a system was
developed for measuring power expended while riding a bicycle. See:
[Only registered and activated users can see links. ]
Jobst Brandt [Only registered and activated users can see links. ]
223rem wrote:
> I am interested in measuring the work (energy, measured in Joules, or
> "watt-hours") used when riding a bike, in a certain time interval.
>
> What I need to monitor is the tension in the chain (the stretching force
> in the chain) and chain speed (see below).
>
> Chain speed can be measured optically, with a photo-cell for example.
>
> How can tension in the chain be measured? I can imagine an arm with a
> spring and wheel that presses down on the chain; the angle of the arm is
> proportional to the tension in the chain. Sounds impractical.
>
> -- Are there more practical ways of measuring or estimating the amount
> of work spent?
>
> Thanks very much.
>
>
>
> PS: "Proof".
>
> Work = Force * displacement = Tension in chain * chain displacement
> (assuming force = constant)
>
> The instantaneous power (at moment t) is:
> Power(t) = Force(t) * chain_velocity(t)
>
> Then total work from t1 to t2 is
>
> W12 = integral of Power(t) from t=t1 to t2.
>
Polar already does this with the power measurement option on
their high-end heart monitor (720?). They use a magnetic pickup
on the upper run (the tensioned side) of the chain. Normal road
roughness will make the chain vibrate. The chain's resonant
frequencies depend on its unsupported length, its linear density,
and its tension. The length is almost constant, and the linear
density is fixed if you ignore dirt buildup and wear. Polar
measures the speed of one of the jockey pulleys, IIRC, to measure
chain speed. I don't think they make any attempt to subtract the
tension in the lower chain run from that in the upper, but it
shouldn't make much difference.
> Normal road
> roughness will make the chain vibrate. The chain's resonant
> frequencies depend on its unsupported length, its linear density,
> and its tension. The length is almost constant, and the linear
> density is fixed if you ignore dirt buildup and wear.
I've estimated power (and energy) using the output from a GPS. The GPS
records position and altitude as a function of time. My power output can be
estimated knowing my weight, the bike's weight, speed and slope (I assume
wind speed is zero).
In article <bnugb3$kbi$[Only registered and activated users can see links. ].pol.co.uk>, [Only registered and activated users can see links. ] says...
> I've estimated power (and energy) using the output from a GPS. The GPS
> records position and altitude as a function of time. My power output can be
> estimated knowing my weight, the bike's weight, speed and slope (I assume
> wind speed is zero).
>
> The results are disappointingly small numbers.
Unfortunately, this doesn't take into account the power expended
against air resistance and frictional losses, so your actual output
will be greater than what this calculates for you, often significantly
greater when on relatively less-steep hills.
--
Dave Kerber
Fight spam: remove the ns_ from the return address before replying!
>>
>
> Polar already does this with the power measurement option on
> their high-end heart monitor (720?). They use a magnetic pickup
> on the upper run (the tensioned side) of the chain. Normal road
> roughness will make the chain vibrate. The chain's resonant
> frequencies depend on its unsupported length, its linear density,
> and its tension. The length is almost constant, and the linear
> density is fixed if you ignore dirt buildup and wear. Polar
> measures the speed of one of the jockey pulleys, IIRC, to measure
> chain speed. I don't think they make any attempt to subtract the
> tension in the lower chain run from that in the upper, but it
> shouldn't make much difference.
>
> Dave Lehnen
Do a clean chain and a dirty chain have the same resonance behavior
for the same tension ?
In article <[Only registered and activated users can see links. ]>, [Only registered and activated users can see links. ] says...
> >>
> >
> > Polar already does this with the power measurement option on
> > their high-end heart monitor (720?). They use a magnetic pickup
> > on the upper run (the tensioned side) of the chain. Normal road
> > roughness will make the chain vibrate. The chain's resonant
> > frequencies depend on its unsupported length, its linear density,
> > and its tension. The length is almost constant, and the linear
> > density is fixed if you ignore dirt buildup and wear. Polar
> > measures the speed of one of the jockey pulleys, IIRC, to measure
> > chain speed. I don't think they make any attempt to subtract the
> > tension in the lower chain run from that in the upper, but it
> > shouldn't make much difference.
> >
> > Dave Lehnen
>
> Do a clean chain and a dirty chain have the same resonance behavior
> for the same tension ?
No. Dirt will add to the mass and to the frictional losses, which
will reduce the resonant frequency and increase the damping
respectively (the latter may be negligible, but I doubt the former
would be). That's why he said you have to ignore the dirt buildup.
--
Dave Kerber
Fight spam: remove the ns_ from the return address before replying!
David Kerber wrote:
> In article <bnugb3$kbi$[Only registered and activated users can see links. ].pol.co.uk>,
> [Only registered and activated users can see links. ] says...
>
>>I've estimated power (and energy) using the output from a GPS. The GPS
>>records position and altitude as a function of time. My power output can be
>>estimated knowing my weight, the bike's weight, speed and slope (I assume
>>wind speed is zero).
>>
>>The results are disappointingly small numbers.
>
>
> Unfortunately, this doesn't take into account the power expended
> against air resistance and frictional losses, so your actual output
> will be greater than what this calculates for you, often significantly
> greater when on relatively less-steep hills.
Right. But there are another source of error.
I guess the OP used:
Power(t) = Mass * Acceleration(t) * Velocity(t)
Acceleration(t) can be derived from Velocity(t).
You also have to account for gravity.
When you go downhill, gravity does part of the work,
so you have to subtract g*sin(slope) from Acceleration. You
add that when you go uphill.
When you brake, Accelaration(t) comes from the brakes, not
from your legs, and this will contaminate your results.