Maths Question: Cadence/Gears/Speed/Power

MatthewBulmer

I don't have a powermeter and my bike computer has a front wheel sensor, so when I'm on my turbo I use cadance as a very rough proxy for power (I know all the drawbacks of this.)

My questions is this: what equation can I use to equate the relative power/speed generated by a specific cadence in different gears?

For instance, if I am at 95prm in 50/16, what would be the % increase in power if I changed to 95 rpm in 50/15, 50/14, etc?

Cheers,

Matt

amaferanga

Depends on the power curve of the turbo.

MatthewBulmer

Yes, hence why I only use it as a very rough proxy.

However, all other things being equal, there must be a mathematical relationship between rpm/gears/power that would allow for such a calculation to be made.

Al_38

Not really, it still depends on the turbo as to the actual power you are producing.

However power increases with increased angular velocity (w) of the back wheel, in some fashion, cant be bothered to work out the leading order: at a guess I would say 3rd.

Assuming this, P = k *w^3 (neglecting lower order terms), where k is some arbitrary constant for the turbo.

at Power 1: p1 = kw1^3
at Power 2: p2 = kw2^3
(sorry cant do subscripts...)

Power increase = ((kw2^3 - kw1^3) / kw1^3)*100
= ((w2/w1)^3 - 1)*100

So what you want to know is how fast your back wheel is rotating:

w = Crank RPM * ( front chainring teeth / rear cog teeth ), gives the back wheel rotation in RPM.

so in your case:
in 50/16 @95 RPM, w = 296.9 RPM
50/15 @95 RPM, w = 316.7 RPM
50/14 @95 RPM, w = 339.3 RPM

So assuming you hold the same rpm, power is a function of gear ratio.

so power increase from 50/16 to 50/15 is 21.4%
power increase from 50/15 to 50/14 is 23.0%

If it isnt a 3rd order equation for the power curve, or the smaller order terms are significant, this is all BS. But if you have the same wheel angular velocity for 2 different gear ratios and cadence, then you are producing the same power, which holds regardless of what the turbo is doing.

now back to my degree

wheeler585

Good skills AL_38 !!

MatthewBulmer

Very impressive.

Thank you!

amaferanga

But the above calculation assumes a given power curve (a simple cubic) which almost certainly isn't accurate.

I can't find it now, but I did have a graph showing the power curve of several trainers and the variation is large, particularly towards higher powers.

So using the simple cubic as above, the change in power could easily be out by a factor 2 or more. It's therefore a pretty pointless guess-timate.

IMO if you don't have an accurate means to measure power then trying to train using power is a waste of time.

Al_38

Here come the much more involved answer:

As Amaferanga pointed out, if the trainer has other components in its power curve then things look a bit different.

We need to look at where the power is dissipated in the turbo and then characterise these:
bearing loss
windage loss on rotors (all but the air fan trainers)
magnetic brake loss (if its a mag trainer or a motor one)
Fan loss (on air fan or fluid trainers)

Now to say what form these losses take:
bearing loss is a linear term: P = aw, for some a
windage loss on rotors is a 3rd order term, P = b*w^3, for some b
magnetic brake loss is a 2nd order term P = c*w^2, for some c
fan loss is a 3rd order term, P = d*w^3, for some d

From this, we can now make a general power curve equation:
P = z*w^3 + y*w^2 + x*w + v, for some z,y,x,v as combinations of a,b,c,d
we know that v = 0 as there is no power dissipation at no rotation.


Next we want to find out how fast the turbo is rotating:
Converting the angular velocity of the back wheel to radian /sec (SI units)

50/16 @95 rpm: w = 31.1 rad/s
50/15 @ 95 rpm: w = 33.2 rad/s
50/14 @ 95 rpm: w = 35.5 rad /s

The turbo flywheel rotates at this speed multiplied by the ratio of the diameters, assuming a 40mm diameter of the flywheel contact and 662 mm for the wheel (20mm tyre depth)

50/16: wf = 514.7 rad /s
50/15: wf = 549.5 rad /s
50/14: wf = 587.5 rad /s
This corresponds to around 5000 rpm ish, so quite sensible

Now we want the relative sizes of these terms, at the angular velocities calculated above:
w^3 is of magnitude 10^8
w^2 is of magnitude 10^5
w is of magnitude 10^2

This means that the factor in front of the linear terms needs to be enormous for it to have any comparison. It wont be and so these terms have a fairly negligible effect.
Leaving us with 2nd and 3rd order terms.

Hence for fluid or air fan trainers, we have an equation of form P = m*w^3 + n*w, and I have said that the linear term is ignorable. The analysis I gave above is fine.
For Mag/Motor trainers:
P = m*w^3 + n*w^2 + o*w, again the linear term is ignorable. From experience with these things, I would expect that the windage loss to be around 20-30 W, the bearing losses to be around 10W. Everything else some 200-300W going into the brake. So about 85% of the power. Replace The cubed bits in the analysis above with squares and you are good to go.

Why the power curves for different trainers looks different at large powers: There are some other terms that come in at both large and small powers, these are considerably harder to model and not really that important in the middle range of operation. Also the constants for each trainer will be different.

Is this a perfect substitution for power? Nope, but then the power meter on my turbo can be somewhat dubious at times too. It should be perfectly acceptable for saying if the change from one gear and cadence to another is a 5, 10 15 % change in power, trying to do anything more accurate is clutching at straws somewhat.

Al

Sources:
http://ntrs.nasa.gov/archive/nasa/casi. ... 027690.pdf
http://books.google.com/books?id=0XYE2p ... ss&f=false
http://en.wikipedia.org/wiki/Faradays_law
http://www.scribd.com/doc/16648848/Fans ... n-of-Power

Bronzie

amaferanga wrote:

IMO if you don't have an accurate means to measure power then trying to train using power is a waste of time.

Agree with this....................

..............but for a bit of fun, I had a go at this on the turbo last night and these were the results for my turbo (an ancient magnetic resistance unit) using a PowerTap wheel.

                 mph     %incr    Power  %incr
53/21 @ 95rpm   18.8              275W
53/19 @ 95rpm   20.8     +10.6    311W  +13.1%
53/17 @ 95rpm   23.0     +10.6    347W  +11.6%
53/15 @ 95rpm   25.7*    +11.7    388W  +11.8%
53/14 @ 95rpm   27.2*    + 5.8    404W  + 4.1%

* - average speed over 30 secs does not tie up with predicted speed given known gear and cadence - probably because at these resistance levels it was getting quite hard to hold a steady speed

So the relationship for my turbo is fairly linear BUT the speed/power relationship this time is very different to the last time I tried to "calibrate" my turbo last May - too many variables to make much sense of this I think.

Al_38

Woop, some data to play with...

Should hopefully fit roughly a square relation

chrisw12

I had a very very quick play with power and speed (when using the turbo) in excel. When I plotted the thousands of points for speed against cadence there was a very linear distribution of points (eyeball tested), obviously not un-expected. So I could use speed as a proxy for power if I could be bothered to do some testing and get the relationship. but tbh it's just easier to track speed and not worry about power if you haven't got a power meter.

So to the op, what you could do is borrow a power meter, pump the tyre up to 100 psi, ride for 2hours varying speed cadence gears, then plot a graph of speed vs power. Get an equation for your turbo and use as you want.

Bhima

What if you have no resistance? (rollers)

Would doubling the speed also double the power required, as the resistance is fixed?

chrisw12

Bhima wrote:

What if you have no resistance? (rollers)

Would doubling the speed also double the power required, as the resistance is fixed?

Bhima, where are these rollers that provide no resistance? Can I buy them off you, I think I may pick up the Nobel prize for physics with these.

Bronzie

chrisw12 wrote:

So to the op, what you could do is borrow a power meter, pump the tyre up to 100 psi, ride for 2hours varying speed cadence gears, then plot a graph of speed vs power. Get an equation for your turbo and use as you want.

But, like I said above, I tried this in May and I got totally different results to what I got this time (resistance higher at lower speeds this time).

- Tyre and tyre pressure is the same (although tyre a bit more worn obviously).
- Roller pressure on the tyre is the same (as far as it's possible to 'set').
- Ambient temperature is pretty different obviously, but I suspect the main difference is that my turbo has variable magnetic resistance, which I tend to have on "max" when using for intervals, although the resistance on this setting is obviously not consistent.

It may be possible with a more reliable turbo to plot a speed / resistance profile if you can recreate conditions really closely, but in reality there are quite a few variables that you can not be certain are identical.

chrisw12

Yes, I agree with you and I did try to emphasise this in my post, it's too problematic/pointless to do.

Saying that, I note that this came up on the wattage forum and one thing they mentioned that you missed is the mass of the system. Have you put on weight?

Just thinking about this though what does change on a day to day basis? I keep my trainer at a constant resistance setting so that problem should be eliminated. I pump the tyres up to the same pressure every time (don't know how accurate the pressure gauges are?) I suppose the big one for me is tyre temp. my turbo system definitely feels smother after a few minutes, less slip. This is something I couldn't control.