# Power output from Turbo

Something another thread got me thinking about. Jack Hughes asked about a power meter for a turbo. Not being particularly scientific (in fact not at all), how much power, roughly, is generated from an hours turbo session?

I realise that its not enough to power a house, but does anyone know how long it would say, run a light bulb for?

• Well, the gym machine just gave me a rating of 240w [I was doing a brick session after a hard 5,000m row - and that was all my legs would do].

This doesn't mean much, without a time - but I guess it is watts per hour - i.e. 0.25Kwh. Which isn't too bad - a bar on an electric fiire uses about 1Kwh.

Of course, I don't know what the units are... but you could probaby run a DVD player with that, even if you couldn't power the TV screen to display it on.
• Power desn't necessarily work like that. when it says 240W it means 240W in that instance, be it 1 second or whatever.....

A watt is a equal to 1 joule of energy per second. 1 Calorie = 4.2kilo-joules.

But, if you did it for an hour you would have generated 240W for an hour which would equal 240Watt-Hour or 0.24 KW-Hr which would be (60x60x240) joules.
• Jack Hughes wrote:

Well, the gym machine just gave me a rating of 240w [I was doing a brick session after a hard 5,000m row - and that was all my legs would do].

This doesn't mean much, without a time - but I guess it is watts per hour - i.e. 0.25Kwh. Which isn't too bad - a bar on an electric fiire uses about 1Kwh.

Of course, I don't know what the units are... but you could probaby run a DVD player with that, even if you couldn't power the TV screen to display it on.

No, W stands for watts. Watts are a unit of power. Electrical devices have their power rated in watts, not Wh or kWh (the latter would be a measurement of how much energy it consumes over a particular length of time; but what length of time is that?).

Your electric fire's power rating is 1kW, not 1kWh. Of course, if you ran it for an hour, you'd consume 1kWh of electrical energy. "Power" and "energy" are different physical concepts.

If you could actually convert all of your 240W effort into electrical power fairly efficiently, and if you exercised for an hour, and if you had some way of storing that energy, you might be able to run a 100W lightbulb for a couple of hours.
• Thanks for the correction Bob!

While your brains all fired up, how about this: What power is required to move a mass (i.e. bike + rider) of, say, 80kg at 25 miles per hour (feel free to convert to SI units [:)]), assuming no head wind, no rolling resistance. And more importantly, what's the formula! Then how do we factor head wind/wind resistance/rolling resistance in - I remember it being something to do with the surface area, a square, an velocity...
• Changing to SI Units 25 mph = 11.18 m/s

Assuming you start from rest and there are no external losses (i.e. no resistance at all) then the energy required to change to this speed is given by KE=0.5*M*V^2, which for an 80kg mass would be 5kJ.

To maintain this speed then you have to put as much energy into moving the mass as is being lost resisting movement. This resistance comes from numerous sources and so it quite hard to quantify.

Fluid Drag is the main one: D=0.5*C*p*A*v^2 (C is the drag coefficient, a function of both shape and speed, p is the air density, A is the effective cross sectional area) assuming C to be about 0.9 and cross-sectional area to be 0.5m^2 with an air density of 1.2kg/m^3 (20°C, 1010mbar) gives a drag force of 33.74N. The work done is given by W=force*distance moved, so to move 11.18m (i.e. 1 second of work) requires 377 joules, or put another way it requires 377 watts to overcome air resistance and maintain a speed of 25mph. This assumes there is no wind blowing, if there is then you add(subtract) the head(tail) wind speed.

Rolling Resistance: F=Cr*M*g (Cr is the coefficient of rolling resistance, M is mass, g is acceleration due to gravity) using Cr=0.004 and g=9.81 gives a force of 3.14N, converting to work again gives 35 joules (35W @ 25mph)

Gradient: If your on the flat then there is no additional work required. Otherwise it is PE=M*g*h (h is height gained or lost). A +1% (0.1118m gained per 11.18m forward)gradient would require an extra 87.74 watts at 25mph and a -1% gradient requires 87.74 watts less.

Other resistance: I don't know how to quantify the other sources such as hub friction but they should be quite small really.

The values used are reasonable ones found googling, if you have a flash TT bike that has been properly fitted then the fluid drag may well be a lot less.
• Oooh! I'm going to set a spreadsheet up, so that when my turbo arrives, I can get cracking on a plan!
• Yeah, what he said.
• Yeah, what he said.

Actually, I have read jibby's post in detail and I'd just like to say that I disagree.

• Bopomofo wrote:
Yeah, what he said.

Actually, I have read jibby's post in detail and I'd just like to say that I disagree.

About which part? If there is a flaw in any reasoning/equations or i've overlooked something please let me know and i'll re-evaluate it.
• something tells me bopo is kidding

we could get that nice lady from the other thread to write a dissertation on the subject
• jibby26 wrote:

Changing to SI Units 25 mph = 11.18 m/s

Assuming you start from rest and there are no external losses (i.e. no resistance at all) then the energy required to change to this speed is given by KE=0.5*M*V^2, which for an 80kg mass would be 5kJ.

To maintain this speed then you have to put as much energy into moving the mass as is being lost resisting movement. This resistance comes from numerous sources and so it quite hard to quantify.

Fluid Drag is the main one: D=0.5*C*p*A*v^2 (C is the drag coefficient, a function of both shape and speed, p is the air density, A is the effective cross sectional area) assuming C to be about 0.9 and cross-sectional area to be 0.5m^2 with an air density of 1.2kg/m^3 (20°C, 1010mbar) gives a drag force of 33.74N. The work done is given by W=force*distance moved, so to move 11.18m (i.e. 1 second of work) requires 377 joules, or put another way it requires 377 watts to overcome air resistance and maintain a speed of 25mph. This assumes there is no wind blowing, if there is then you add(subtract) the head(tail) wind speed.

Rolling Resistance: F=Cr*M*g (Cr is the coefficient of rolling resistance, M is mass, g is acceleration due to gravity) using Cr=0.004 and g=9.81 gives a force of 3.14N, converting to work again gives 35 joules (35W @ 25mph)

Gradient: If your on the flat then there is no additional work required. Otherwise it is PE=M*g*h (h is height gained or lost). A +1% (0.1118m gained per 11.18m forward)gradient would require an extra 87.74 watts at 25mph and a -1% gradient requires 87.74 watts less.

Other resistance: I don't know how to quantify the other sources such as hub friction but they should be quite small really.

The values used are reasonable ones found googling, if you have a flash TT bike that has been properly fitted then the fluid drag may well be a lot less.

I believe that this is sceintific proof that red is faster...
• Joking, mate.

Does this all mean that my output power as measured on the bikes in the gym is possibly quite accurate? After all, the easiest way I can think of to put a variable brake on a bike would be a generator and resistor set-up. While my simple P=IV is no match for your half mv squared (in terms of looking good) it should be pretty accurate given that current and voltage are easy to measure accurately? I'd guess that the friction losses are probably low...

So, if the gym bike says I'm putting out 250W then I probably am, right? (Subject to possibly laughable calibration).

Not that I care, really. I just wanted to say an equation.
• AndI think you guys should spend your timeless talking, more training... [:D]
• At least I don't need to worry about that bit of paper with my physics degree on becoming worthless then ... [:)]

The power meters on the gym bikes have the potential to be accurate, depending on the quality of the components and the calibration.

As for having better things to do, it was a rest day, how better to spend it than looking through all those expensive physics books working out the best way to cycle faster (or with less effort)?
• After half a bottle of wine (must post on other thread) I wish I'd never bothered asking the question.....

Jibby, fantastic answer, just dont have a clue. But I can explain the offside rule.........
• In answer to your original question, my turbo reckons it can generate 400W @ 27km/h. In an hour that would be 1440kJ (or 0.4 kW/h).

This is enough to:

run 60W light bulb for 6 hours 40 mins

boil a kettle 2.5 times

power a 42" plasma tv for 1.5 hours

or more importantly keep the beer for after a hard day's training cold for 9.5 hours in an A rated fridge
• Jibby. Don't know where you got the 24kW bit from.... its 400W, always 400w. if generated for an hour its, as you said, 0.4kwh.
• How much water is in the kettle?
• sfuller wrote:

Jibby. Don't know where you got the 24kW bit from.... its 400W, always 400w. if generated for an hour its, as you said, 0.4kwh.

quite right, stupid mistake, I meant 1,440 kJ.