How many watts to rear-brake a fixie?
carl...@comcast.net
on a fixie led me to wonder just how much effort is involved
in slowing a fixie down by just fighting the pedals (no hand
brakes).
This dreadfully important question does not appear to have
crossed the minds of the authors of my three favorite
bicycle speed and power calculators, which all choke before
I can bring my imaginary fixie to a halt on 7% slope by
putting in negative watts. Using their defaults, here are
the results:
-7% 0% -7% 0%
0 watt watts limit on -x watts watts
coasting___needed negative coasting___needed
calculator mph on flat watts mph on flat
----- --- ---- ----- ---
Kreuzotter 33.4 952 -299 20.0 233
Austin 32.65 823 -294 19.65 202
Analytic* 39.82 916 -332 23.73 214
* converted from meters/second
Here are the calculators:
http://www.kreuzotter.de/english/espeed.htm
http://w3.iac.net/~curta/bp/velocity/velocity.html
http://www.analyticcycling.com/ForcesSpeed_Page.html
For programmers, the short Austin source code is here:
http://w3.iac.net/~curta/bp/velocity/Velocity.java
A few questions . . .
First, does it seem reasonable that the rider would have to
put out around 300 watts of resistance to slow his gravity
propelled descent from around 32-39 mph down to around 20-23
mph?
That seems like a lot, but the numbers don't seem to add up.
If it takes 952 watts to go 33.4 mph on the flats, and only
233 watts to go 20.0 mph on the flats, the difference would
seem to be around 720 watts, not the -299 that may be
fooling the calculator.
Next, why do all the calculators accept negative numbers up
to around the same point (294 to 332 watts) and then choke,
showing Nan, 0 mph, or confused exponential notation? Were
these programmers so blinded by anti-fixie prejudice that
they failed to consider this crucial situation?
Finally, how much anti-pedal effort should it really take to
slow a fixie from around 30 mph to around 20 mph on a steady
7% grade? Much less than the roughly 300 watts predicted by
these sadly abused calculators? About 300 watts? Much more?
Curiously,
Carl Fogel
lime...@gmail.com
It always seems to me that it takes a HELL of a lot of effort to slow
appreciably, let along stop, on my fixie. I don't have a rear brake on
it and make a conscientious effort to brake with my legs but I usually
have to include some front braking, especially if I'm on a hill.
Lewis.
*******************
Jim Smith
Just subtract the power needed to go 20mph on the flat from the rate
of change of gravitational potential energy when descending the grade,
which is the same as the power needed to climb the grade if the wind
resistance and rolling resistance are zero. or
mass(kg)*9.8(m/s^2)*grade*speed(m/s).
Dan
<carl...@comcast.net> wrote in message
news:v8tv91p2nuosmcs03...@4ax.com...
Carl-
Some time ago I think I started a thread concerning internal vs. external
work and bicycles. The part that most of us seem to forget is the internal
work required to move your legs. I am not talking about chemical energy,
just physical effort. Physical effort is what makes you tired.
The external work is easy to calculate since it is based on the fairly
simple Newtonian physics of potential/kinetic energy and losses such as air
and rolling resistance. The internal work is harder to quantify, it has much
to do with technique and how gracefully you operate.
In a biomechanics book, I read about an actual experiment where two cyclists
pedaled stationary bikes connected by a chain. One rider pedaled forward and
the other provided resistance by back-pedaling. The physical effort of the
two riders was measured and compared. Although both riders were doing the
same amount of external work, the resistance pedaler was doing less total
work. The forward pedaling cyclist had to overcome the internal losses of
both riders.
In the book they provide an example in which the internal work of the two
riders is 75 watts. If this is the case, the forward pedaling cyclist will
always be doing 150 watts more work than the resisting cyclist.
So, you ride up and down a hill on your track bike. You descend at the same
speed as you ascend so the rolling and aerodynamics losses are the same. In
fact, you go so slow that we can ignore the rolling and aero losses. The
external work in the climb is just the weight of you and the bike times the
climb rate. The external energy you need to expend in the descent is the
same. In both the climb and the descent you will burn up something like 75
watts of internal losses. The result is that climbing will take 150 watts
more physical effort than the descent.
Please see page 106 of: http://tinyurl.com/a9arx
search for freewheeling
Michael Wileman
>In a biomechanics book, I read about an actual experiment
>where two cyclists pedaled stationary bikes connected by a
>chain. One rider pedaled forward and the other provided
>resistance by back-pedaling. The physical effort of the two
>riders was measured and compared. Although both riders were
>doing the same amount of external work, the resistance
>pedaler was doing less total work. The forward pedaling
>cyclist had to overcome the internal losses of both riders.
I once heard a lecture by a researcher who had performed
these experiments. He said that a backward pedaler could
wear out three forward pedalers. But the next day the
backward pedaler could hardly walk. I'm surprised your
biomechanics book didn't mention this.
He didn't say (perhaps didn't know) whether this was a
temporary effect caused by using muscles not commonly used,
or an effect of using the joints at an intensity they were
not designed for under that loading.
Mike
carl...@comcast.net
Dear Jim,
I'm not sure what is subtracted from which and how it's
calculated.
214 pedal Watts for 22.73 mph (10.61 mph) 0% grade
916 coast Watts for 39.82 mph (17.80 m/s) 7% grade
75 kg Mass of bike and rider
-0.07 slope of 7%
10.61 Braked descending speed 23.73 mph = 10.61 m/s
17.80 Unbraked descending speed 39.82 mph = 17.80 m/s
75 * 9.8 * 0.07 * 17.8 = 915.8
(That's the 916 watts already given by the calculator.)
Subtract from this 916 watts the 214 watts needed to go 20
mph on the flat, and the rider must provide about 700 watts
of braking power, which was what I was asking about.
Are we saying that a rider must provide a steady 700 watts
of reverse pedal power to descend a 7% grade at around 23
mph instead of around 40 mph?
Or is that 700 watts just what's needed to slow a 40 mph
descent down to 23 mph, after which a lower reverse pedal
effort will serve to maintain a steady 23 mph? If so, what's
the steady-state power needed to stop the bicycle from
accelerating back up from 23 mph to 40 mph?
Of course, I may be missing something and using the wrong
terms, but I think that I followed your suggested equation.
Carl Fogel
carl...@comcast.net
Dear Lewis,
Yes, it does seem to be a lot of effort.
Where I'm puzzled is that the calculations seem to suggest
something like 700 watts to slow from 32-40 mph down to
20-23 mph by reverse pedal effort.
Maybe the calculations are wrong? (Not unlikely, given who
was playing with the calculator, the odd failure of three
calculators to work below about 20 mph with negative watts,
and the fact that the calculators probably weren't
programmed to take such a bizarre input.)
Maybe riders really can put out 700 watts of reverse pedal
effort steadily on a 7% descent? (If so, why doesn't someone
whomp Armstrong by riding backward up the Alp D'Huez?)
Maybe it's only 700 watts to slow down and then some smaller
steady effort to hold the speed to 23 mph? (I'll need
someone to explain this, since it sounds implausibly like
saying less effort is needed to maintain top speed once you
reach it.)
Maybe it really takes 700 watts, but no one does it? (Most
fixie riders would either use a front brake to slow to 23
mph on a steady 7% slope or else spin much faster--less
braking--and thus descend at much higher speeds.)
Carl Fogel
Jim Smith
>
> I'm not sure what is subtracted from which and how it's
> calculated.
>
> 214 pedal Watts for 22.73 mph (10.61 mph) 0% grade
Why 22.73mph? I thought you were interested in 20mph?
> 916 coast Watts for 39.82 mph (17.80 m/s) 7% grade
> 75 kg Mass of bike and rider
> -0.07 slope of 7%
> 10.61 Braked descending speed 23.73 mph = 10.61 m/s
> 17.80 Unbraked descending speed 39.82 mph = 17.80 m/s
>
> 75 * 9.8 * 0.07 * 17.8 = 915.8
>
> (That's the 916 watts already given by the calculator.)
You want to calculate this at the speed you are interested in, not at
17.8 m/s
At 22.73mph (10.61m/s) decending a 7% grade, gravitational potential
energy is changing at a rate of: 75 * 9.8 * 0.07 * 10.61 = 546
jules/second (watts) This is the power input to our bike. Of these
546 watts, 214 are going to overcoming wind and rolling resistance,
leaving 546 - 214 = 332 watts that need to go somewhere so we don't
speed up. You can either use these 332 watts to try and melt the glue
holding on your tubular tires by pushing rubber pads against the rim,
or you can use them to convert donuts into sexy man-flesh by resisting
on your fixie. The choice is yours.
carl...@comcast.net
Dear Jim,
So the subtraction is from the power of the final braked
speed, not from the initial terminal velocity?
The defaults for the Analytic calculator give a steady
terminal velocity of 39.82 mph (17.80 m/s) for coasting down
a 7% grade, reflecting 916 watts.
The question is how much reverse pedal effort will be needed
to slow a fixie down to a steady 22.73 mph (10.61 m/s),
which takes 214 watts on the flats. (The Analytic calculator
chokes if you try to reduce the speed on the 7% grade below
10.61 m/s by increasing the reverse watts--using -332 works,
but -333 fails.)
So if I'm following you, a steady 332 watts of reverse pedal
will slow a fixie coasting steadily at 40 mph down a 7%
grade down to a steady 23 mph.
Any idea why the calculators all fail when the coasting
speed on a 7% grade (32-40 mph, depending on defaults) is
reduced below around 20-23 mph)?
Carl Fogel
John Dacey
>A current thread debating the desirability of a rear brake
>on a fixie led me to wonder just how much effort is involved
>in slowing a fixie down by just fighting the pedals (no hand
>brakes).
<snip>
>A few questions . . .
>
>First, does it seem reasonable that the rider would have to
>put out around 300 watts of resistance to slow his gravity
>propelled descent from around 32-39 mph down to around 20-23
>mph?
>
>That seems like a lot, but the numbers don't seem to add up.
>
>If it takes 952 watts to go 33.4 mph on the flats, and only
>233 watts to go 20.0 mph on the flats, the difference would
>seem to be around 720 watts, not the -299 that may be
>fooling the calculator.
>
>Next, why do all the calculators accept negative numbers up
>to around the same point (294 to 332 watts) and then choke,
>showing Nan, 0 mph, or confused exponential notation? Were
>these programmers so blinded by anti-fixie prejudice that
>they failed to consider this crucial situation?
>
>Finally, how much anti-pedal effort should it really take to
>slow a fixie from around 30 mph to around 20 mph on a steady
>7% grade? Much less than the roughly 300 watts predicted by
>these sadly abused calculators? About 300 watts? Much more?
>
>Curiously,
>
>Carl Fogel
No! no arresting the vast wheel of time,
That round and round still turns with onward might,
Stern, dragging thousands to the dreaded night
Of an unknown hereafter.
- Charles Cowden Clarke
You seek an answer to a question that has no practical application.
Even with a 90 inch gear (which exceeds by far what would commonly be
used by fixed-gear road riders), at 39 mph the cranks would be
spinning at a rate such that no rider would be able to effectively
apply meaningful back-pressure to the pedals. The best for which one
could hope would be to sit up and let aero drag attenuate some speed.
Even at 30 mph, the pedal cadence will be such that, on a 7% grade,
the bike would probably regain as much speed in the intervals between
back-pedalling efforts as you could slow the bike with those efforts.
After fatigue sets in after a few futile attempts to arrest the bike,
you'd be back to careering downhill at a rate limited only by the
slope and aerodynamics.
-------------------------------
John Dacey
Business Cycles, Miami, Florida
Since 1983
Comprehensive catalogue of track equipment: online since 1996.
http://www.businesscycles.com
Peter Cole
It's probably a sign switch. If you look at a curve of power vs. speed
for your 7% grade, the (brake) power is 0 at 0 m/s, increases to a
maximum at around 10 m/s (~ -300W), then starts decreasing toward 0W
again at the terminal velocity. I'm guessing the computation gets
screwed up when you put in a power greater than the maximum because that
would reverse your velocity. (sorry about the sloppy min/max incr/decr
use with neg numbers, but it reads better).
It's an interesting analysis, because it shows exactly how maintaining a
moderate speed on a steep grade can put a huge thermal load on brakes,
going either faster or slower will put less.
Jim Smith
It sort of sounds like you have work/energy and power a little bit
confused, or maybe I am not quite understanding you. It doesn't make
sense to ask "how much power does it take to slow a bike from x mph to
y mph." It DOES make sense to ask: "how much energy does it take to
slow a bike from x mph to y mph," or equivilantly: "how much work must
be done to slow a bike from x mph to y mph." Work and energy are
measured in Jules, or foot pounds, or Calories, or killowatt hours. Power, which is the
rate of doing work, is measured in Watts, or horespower. With this in
mind, we can see that it DOES make sense to ask: "how much power do I
need to slow a bike from x mph to y mph in z seconds."
Anyways, the 332 watt figure doesn't involve slowing the bike from 40
mph to 23 mph. It is the amount of power it takes to keep the bike
which, unrestrained, would be going 40 mph only going 23 mph. It
doesn't matter how you arrive at this 23 mph on this hypothetical
bike, you can slowly accelerate there from a stop, or slow down from
40, 50, or 60 mph, but once you are going 23 mph on this particular
bike, on this particular grade, you have to do 332 joules per second
(watts) of work to keep it from going faster than 23 mph.
> Any idea why the calculators all fail when the coasting
> speed on a 7% grade (32-40 mph, depending on defaults) is
> reduced below around 20-23 mph)?
Well, this is just a hunch, but calculating this involves solving a
cubic equation, which is going to have three solutions, only one of
which we are interested in. The other solutions may be complex, or
negative, and not have a usefull interpretation. Kind like if you
ask: what size cube has a volume of 8? So, you want to solve: x^3=8,
which has three solutions: 2+0i, -1+squrt(3)i, and -1-sqrt(3)i it is
up to you to pick which one is the one you are interested in
considering the physical interpretation. I suspect something happens
around 20-23mph that makes it hard for the calculator to pick the
correct solution. Again, this is just a guess.
Zog The Undeniable
> Finally, how much anti-pedal effort should it really take to
> slow a fixie from around 30 mph to around 20 mph on a steady
> 7% grade? Much less than the roughly 300 watts predicted by
> these sadly abused calculators? About 300 watts? Much more?
To maintain a constant 20mph on that gradient you'd need to develop more
than 300W of back-pedalling power. Simply calculate the rate of change
in potential energy and subtract 200W for air resistance (I'm assuming
you're sitting bolt upright to maximise drag!).
If you were additionally trying to slow from a higher speed, you'd need
to know how long the deceleration took to do the calculation. Either
way, it's going to hurt.
Zog The Undeniable
> After fatigue sets in after a few futile attempts to arrest the bike,
> you'd be back to careering downhill at a rate limited only by the
> slope and aerodynamics.
Just stick yer feet on the fork crown and let her rip :-)
Seriously, I've smelt my front brake burning - that singed Bakelite
smell - on a long 16% descent using the fixie. Holding the speed at
30mph puts a terrifying amount of heat into the rim.
carl...@comcast.net
Dear Peter,
I'm struggling, but it's me, not your min-max and sign (I
had the same problem trying to say increasing reverse pedal
power from -290 to -295 watts).
If I'm following you, the rider at 40 mph is putting out 0
watts of pedal braking. (There, no damned sign!)
He starts fight the pedals and slows down until at around
300 watts he's braked to around 23 mph.
But after that speed is reached, he can continue to slow
down from 23 mph with less and less braking power until at 0
mph he's using 0 watts.
So the hardest speed to maintain, the speed that's hardest
on the brakes, is (very) roughly in the middle.
I'd been wondering about why it seemed unlikely that anyone
was putting out a huge effort to hold the bike back at a
walking speed, so this is a nice explanation--if I haven't
missed your model.
Thanks,
Carl Fogel
carl...@comcast.net
<3.141...@gmail.com> wrote:
Dear Jim,
I think that I'm following you and Peter Cole (he's right
next door now).
Peter's point (I think) is that the maximum braking power is
reached at the point where the calculators give up the
struggle and that they can't handle the reduction that
follows the peak /\ of the braking power curve.
40 mph 23 mph 0 mph
0 watts braking 332 watts max 0 watts braking
pure coasting braking complete stop
The peak braking power is (very) roughly in the middle of
the speed from pure coasting to a complete stop. Peter's
post explains what may confuse the calculator better than I
can.
Carl Fogel
Fritz M
> Finally, how much anti-pedal effort should it really take to
> slow a fixie from around 30 mph to around 20 mph on a steady
> 7% grade?
>From 30 mph I bleed the speed off in a runout. There is no way I can
apply enough backpressure to slow to 20 mph on a 7% grade.
RFM
Jim Smith
Peter Cole
> Dear Peter,
>
> I'm struggling, but it's me, not your min-max and sign (I
> had the same problem trying to say increasing reverse pedal
> power from -290 to -295 watts).
>
> If I'm following you, the rider at 40 mph is putting out 0
> watts of pedal braking. (There, no damned sign!)
>
> He starts fight the pedals and slows down until at around
> 300 watts he's braked to around 23 mph.
>
> But after that speed is reached, he can continue to slow
> down from 23 mph with less and less braking power until at 0
> mph he's using 0 watts.
>
> So the hardest speed to maintain, the speed that's hardest
> on the brakes, is (very) roughly in the middle.
>
> I'd been wondering about why it seemed unlikely that anyone
> was putting out a huge effort to hold the bike back at a
> walking speed, so this is a nice explanation--if I haven't
> missed your model.
I think it's Newton's model, actually. There's always a bit of confusion
over force vs. power, kind of like torque and horsepower in the auto
world. When a rider is stopped and resisting the force of gravity with
force on the pedals, they are generating no power. On the other hand,
when going very fast, even a small force generates lots of power, power
being proportional to force and speed. Since you asked the question of
how long it would take to decelerate from terminal speed to a moderate
speed with a constant power, you have to accept that you framed a rather
complex problem since the braking force is dependent on velocity
(decreases with decreasing velocity under constant power) and the drag
force is dependent on velocity, so the velocity/time function is
complex. I'm sure the on-line calculators are not up to solving
arbitrarily bounded problems.
If you use the analyticcyycling calculator, and go to the "power, given
speed" section, put in your -7% slope, and some arbitrary speed, like 15
m/s, run the model. You'll see a graph at the bottom left that shows the
complex power/speed curve, similar to the one Jim Smith did.
Dan
"Michael Wileman" <jwil...@panix.com> wrote in message
news:d7puvu$dmi$1...@reader1.panix.com...
I have wondered why they didn't just turn the backwards guy around and let
him use his normal cycling muscles, it seems to me that this would be a more
appropriate way to measure the internal work.
Mark Hickey
>In <11a0ssa...@corp.supernews.com> "Dan" <banquo...@yahoo.com> writes:
>
>>In a biomechanics book, I read about an actual experiment
>>where two cyclists pedaled stationary bikes connected by a
>>chain. One rider pedaled forward and the other provided
>>resistance by back-pedaling. The physical effort of the two
>>riders was measured and compared. Although both riders were
>>doing the same amount of external work, the resistance
>>pedaler was doing less total work. The forward pedaling
>>cyclist had to overcome the internal losses of both riders.
>
>I once heard a lecture by a researcher who had performed
>these experiments. He said that a backward pedaler could
>wear out three forward pedalers. But the next day the
>backward pedaler could hardly walk. I'm surprised your
>biomechanics book didn't mention this.
I've read that your muscles can "absorb" a lot more energy than they
can produce. I remember using some gym equipment that worked on this
basis - you'd push x pounds in a leg press, and the machine would push
back with x time y pounds on the rebound.
This only make sense, and you can do an experiment to prove it.
First, jump off a six foot wall. Notice how much energy you use to
stop at the bottom of the fall.
Now jump from the ground to the top of that six foot wall... ;-)
Mark Hickey
Habanero Cycles
http://www.habcycles.com
Home of the $695 ti frame
Zog The Undeniable
> Michael Wileman <jwil...@panix.com> wrote:
> I've read that your muscles can "absorb" a lot more energy than they
> can produce. I remember using some gym equipment that worked on this
> basis - you'd push x pounds in a leg press, and the machine would push
> back with x time y pounds on the rebound.
>
> This only make sense, and you can do an experiment to prove it.
>
> First, jump off a six foot wall. Notice how much energy you use to
> stop at the bottom of the fall.
>
> Now jump from the ground to the top of that six foot wall... ;-)
Tendon material is basically a spring which absorbs the peak energy from
the landing and releases it more slowly.
The Greeks used real tendon in their mangonels, which were a
considerably more efficient way to throw big rocks than the frankly
retarded trebuchet.
http://digitalclass.org/Tech%20Ed/Exploring%20Technology/Machines/SiegeEngines.htm