woodcraft

OK, I taped 5lb weights to the spin bike pedals.

With the resistance backed off all the way, I pedaled for a while before the weights, with the weights, & again without the weights.

Without the weights, I spun up to the point of bouncing in the saddle without really trying.

With the weights, the RPMs didn't go as high, & more effort, but not as much as I would have thought. Possible I held back for the risk of the weights flying through the plate glass.




We'll see how the 40k TT goes...

gregf83 

Hat's off for carrying out an experiment. A little tough to control what you have though since you're essentially trying to measure a very low force (zero + friction).

An alternative would be to remove the chain if possible and just spin the cranks with your finger. Ideally you'd have a cadence magnet attached so you could measure RPM. If the BB bearings are decent you should find it takes very little pressure from your finger to keep the cranks spinning at a particular RPM. If you're able to spin them up to 90 RPM and remove your finger you should find the weighted pedals take longer to come to rest. This is the opposite of what you'd expect if the force required to spin the heavier pedals was higher.

You should also notice it takes more force to accelerate the heavier pedals up to speed but the same low force to keep them at a particular speed.

Extra kudos if you do the above experiment with a Garmin recording the results and post them here.

kpotier16

So essentenially you are denying that centripetal acceleration is real and suggesting that gravity is unnecessary to make planets orbit......

kbarch

As far as one leg going down pushing up the other goes, the effect of the crank can be described simply by realizing that, for any given speed, it takes the same amount of total energy to turn the crank with one leg as it does to turn it with two, it merely allows us to split the effort between legs. At certain points around the circle we use the left to move both the left and the right, and vice versa, but we still have two shoes to move around, and circular motion doesn't make their mass evaporate. Let's not forget; on bikes with freewheeling hubs, the cranks don't turn unless we turn them by moving our shoes.

The real issue with the equation above is whether the energy required to move a pair of shoes in a circle is correctly defined and converted to the energy required to move them up stairs.

gregf83 

Not at all. I thought we were discussing the work required to keep an object rotating in the absence of friction. I'm saying the work required is zero.

kbarch

And the energy required to keep moving up stairs in the absence of gravity is zero, too.
The significant resistance in this equation is not friction and the rotation of the crank is not the issue. Bear in mind, on freewheeling bikes, the momentum of the crank is completely overcome and the cranks come to a stop the instant we quit swinging our shoe-burdened feet around.

Bunyanderman

If you didn't know, gravity doesn't do work on orbiting planets. Yes it puts a force on orbiting planets, but 0 joules is done.

2manybikes

The work from rotating the cranks is offset by the hidden motor in the seat tube.

Doge

Really good idea. Actually we have the stuff to test this. Got pedal PM and Hub PM. So if I get around to it...

woodcraft

Calculation #2

200g= .0027 of body weight

Over ride with 6k' climbing, 6,000'x.0027=16.2'

200g additional weight is equivalent to climbing 1.8 flights of stairs.

Unequipped to calculate the acceleration aspect, but some of the climbing would come from momentum

from previous accelerations.

kpotier16

Exactly my point that no work is done! This is because the leg is analogous to gravity, which applys a force to counteract the foots momentum and make it move in a circle. I realize most poeple will still not get this and this is the last ill say on this, at the very least, we all agree on the conclusion that the weight of the shoe is not much more important than any of the other component's weight

DaveLeeNC

This time of year I do a good bit of cycling indoors on a spinner bike (Lemond Revmaster Sport) with Vector Power Pedals. This thing is turning a 40 pound disk via a belt drive (no slipping) and the load is friction on the disk (not the disk weight). My max power output on this contraption vs. (same power meter) on a real bike is quite similar as is FTP measurements. Max cadence - same thing.

Not exactly the same thing as 10 pound shoes, but close.

dave

Bunyanderman

Now this makes more sense, not related to watts saved through the crank, but a decent start. But congrats on saving 0.27% energy with 200g dropped. That's is more than half a watt gained at 200 watts!

Igualmente

Somehow 6000 feet of climbing seems to have made it into this, so I will assume we are "hauling" the 200g up 6000 feet of elevation change (net upwards movement).

So, half a watt? Not on average, anyways. Hauling 200g up 6000 feet (1829 metres), ignoring additional friction from the 200g and power for accelerations, requires just under 3600 joules. If that is done over 5 hours, it will be about .2 watts average. Of course, instantaneous power required will be higher and lower. I believe 3600 joules of energy is slightly less than a kcal, and if a human body is only about 20% efficient, 5 kcal of food (1/12 of a cookie?) should just about to it.

Hauling 80kg (bike and rider and stuff) up the same 1829 metres, and ignoring all friction, accelerations and aero, is 1435000 Joules which over 5 hours requires an average 80W output.

woodcraft

The 6,000' in question at the moment

https://ridewithgps.com/routes/12199278

The last, slightly flatter, ride showed ~3,500kj on Strava (I'm closer to 85kg)

So on the order of 40% raising the weight, 60% everything else?

Igualmente

If I understand that ride correctly, you gained 6000 ft, but also dropped 6000 ft, during your ride, so your net elevation gain was basically zero. In that case, you put gravitational potential energy into your body and bike (and the 200g of extra shoe weight) when climbing, but gain it back when descending.

So, when you are climbing, you may use 170 watts to go up a 5% grade at 8mph (just an example, don't really know the real numbers that would apply to you). When you go down after climbing, you may be use 0 watts (coasting) and get to over 25mph on the same 5% grade (downhill). In effect, on the climb you are using something like 85% of your energy to raise your weight, and on the descent you are using no energy if you are just sitting there and are gaining all the stored gravitational potential energy back and using it to accelerate and overcome air resistance, tire losses and friction.

In other words, as a proportion of the overall ride, you used no net energy to raise your weight, because you retrieved that energy on the descents.

pacificaslim

Very funny! That would only be true if after going down that hill, one could now go back up utilizing that energy "gained" on the way down. But of course, we can't store that energy (other than the minor amount we can store as momentum) so each time we pedal we need to spend new energy.

Igualmente

Not sure what is funny. I said a cyclist stores gravitational potential energy going uphill. Then the cyclist loses it going downhill. That is why we can go downhill without pedalling.

pacificaslim

The idea that one used no net energy to raise our weight up a hill because he then proceed to coast back down that hill for "free" is indeed a funny way to look at things.

wphamilton

Potential Energy is a funny concept. What all this arguing is really about, is because gravity is a conservative force. It takes the same amount of energy to go from point A to point B, regardless of whether you go up the hill or around it. (in a vacuum that is )

But that's energy, and as cyclists we generally care more about the power requirements than energy. And how fast we get there, which is a big part of the sport or recreation as your inclination may be. That's a whole different story.

DaveLeeNC

FWIW, if you were to start at the top of a hill where (somehow) there was no air resistance (say in a vacuum somehow) and just coast down that hill, and then just coast back up an identical hill using only the momentum that you gained going down, you would come very close to getting back up to the same height. Friction losses (Crr and bike stuff) would prevent you from getting exactly the same height.

But since we don't ride in a vacuum and the opposing force of apparent wind varies as the square of velocity, things not in a vacuum don't work that way. I don't see it as an energy vs. power thing. It is simply that the energy lost to aerodynamic forces is less obvious.

dave

Igualmente

Sigh.

My comment at post 91 was a response to woodcraft asking if 40% of the work went into raising the weight. Since woodcraft started and finished at the same elevation, there was really no energy lost in "raising the weight". Lots of energy was lost to friction, tire rolling resistance, aerodynamic resistance, brake friction, etc. Those things are always lost, whether climbing or on the flats. But, between start and finish the rider and bike had virtually no change in gravitational potential energy because the rider and bike started and finished at the same elevation (actually, there was a 7' difference, but I'm ignoring that here).

wphamilton

What you're missing, and we're trying to explain gently, is that you'll lose energy, or expend more power over a period time (same thing said a different way), if you're trying to keep the same speed, due to the effect of air resistance and the different speeds involved. It was pretty clear that woodcraft hadn't considered that and didn't ask precisely, but in the reality of cycling it's the most important consequence. If you're keeping other things equal, and overall velocity is one of those things, then you do expend more energy going up and down rather than around.

And having more weight increases that effect, almost in proportion to the extra weight.

Doge

But lost at entirely different rates depending if you are going up, down or on the flats.
If you are using distance as the base, the bearing friction is higher going up than going down.

DaveLeeNC

Hmmm, that would seem to me to be something requiring additional explanation. Can you provide additional information/explanation. Thanks.

dave