Supermono 1300cc racer project.

[Propellor]

Look forward to that. Rest assured the I pad will be cranked to volume 11.

All I was thinking is that if you knew the turbine tyre speed, then we could easily know the absorbed power, assuming the torque arm is on the input to turbine and not the housing. If it is the housing then wouldn't we also need to know losses (slip) inside the water turbine itself? In that case the actual bike rear wheel torque figure goes up. Maybe I'm wrong with that bit. I'm not confident, but I instinctively think that turbine losses would affect the calculation if the torque arm was on the housing. Also, I'm struggling to imagine where else it would be. I'll have to study the photo!

As it is we know the crankshaft rpm but torque at the turbine, either input or output, I'm not sure. Am I right there?

Working out the crankshaft torque (and therefore crankshaft power) would seem to be tricky if there is unknown loss at the tyres and/or turbine, assuming the torque arm is on the turbine output.

Cheers anyway.

[Propellor]

Torque arm looks to be attached to the input shaft? The same shaft as the tyre. Tricky to get ones head around!

[guest1306]

the crank stroke is 86mm
18" bike back tyre
13" car tyre

this is a paddle wheel turbine, but the car wheel axle is only attached to the paddle wheel. the inner casing of the outer independently upphanged bit which is attached to the 1 meter long arm is internally raffled , which makes the water " grip" the whole kit and kaboodle

there will be a lot of friction between the two wheels.

i must turn the crank and measure how many crank turns in top gear to do one bike wheel rev.

then measure the bike wheel outer diameter, and car wheel outer diameter - i guess

[guest1306]

it all gets a bit too complicated i think.

so all i do is use the max bike revs (at the moment) 7460, and multiply by the kilos 12

7460x12=89520

and the result 89520 is what i have to aim to beat the next time i run the bike after slight tuning.
If the result is worse, then ive done something wrong.

This engine should run to 10000 rpm, but perhaps not with all that friction going on.

[Propellor]

Apologies, I think I was just getting my knickers in a knot. The torque reaction at the housing is equal to the input shaft torque to the paddle, regardless of the fact that there will be significant losses (I imagine) within the turbine.

If that statement is true, then the torque arm tells us the torque at the turbine tyre. Phew! I got there in the end.

The rest is easy maths if we know the transmission ratios ie primary drive, gearbox and rear chain. Also the respective tyre rolling diameters, as you say. Maybe tricky to measure due to deformation? How about measuring the radius of each tyre to contact point, once preloaded with your straps? Maybe they deform even more when power is applied?

I'd be surprised if the overall ratio was a nice easy round number, therefore difficult to account for with just one tyre revolution?

What power are you reckoning on/hoping for?

[guest1306]

over 45 hp would be good, and when the best power is sorted, then its time to back off the advance a couple of degrees and tune her with the laughing gas. that should take her up to 60 hp.

[Propellor]

it all gets a bit too complicated i think.

so all i do is use the max bike revs (at the moment) 7460, and multiply by the kilos 12

7460x12=89520

and the result 89520 is what i have to aim to beat the next time i run the bike after slight tuning.
If the result is worse, then ive done something wrong.

This engine should run to 10000 rpm, but perhaps not with all that friction going on.

Ha ha. That is incredible. Stand well back mate. Behind a wall, and get a "friend" to operate it.

If my last proposal re turbine shaft/housing is correct then I reckon we could work out a power figure.

As you say, at least the data you have is a datum, regardless.

I wish you good luck. What you are doing is brilliant.

[Propellor]

Once we know the turbine speed and allowing a small amount for losses between the two tyres, don't  we have a reasonably accurate bike rwhp figure?

That's what I was initially intimating. Sorry for going around the houses. I enjoyed it!

[guest1306]

Once we know the turbine speed and allowing a small amount for losses between the two tyres, don't  we have a reasonably accurate bike rwhp figure?

That's what I was initially intimating. Sorry for going around the houses. I enjoyed it!

right Propellor, ive been and measured up in the garage. tonights brain teaser for you, cos my brain is shot.

4.5 crank revs turn the bike back wheel 1 rev
bike tyre outer circumferance is 1.965 meters
brakebench car tyre outer circumferance is 1.832 meters
if the bike is running 7460 rpm
what revs is the car wheel doing?
what speed is the bike doing , if it was on the road?
the 1 meter arm is presing 12 kilos

the inner rotor of the turbine is approx 0.3 m diameter
the stroke of the engine is 86mm, if it helps with any future calculations

here is a similar water turbine on Wiki  http://en.wikipedia.org/wiki/Water_brake

[Propellor]

Once we know the turbine speed and allowing a small amount for losses between the two tyres, don't  we have a reasonably accurate bike rwhp figure?

That's what I was initially intimating. Sorry for going around the houses. I enjoyed it!

right Propellor, ive been and measured up in the garage. tonights brain teaser for you, cos my brain is shot.

4.5 crank revs turn the bike back wheel 1 rev
bike tyre outer circumferance is 1.965 meters
brakebench car tyre outer circumferance is 1.832 meters
if the bike is running 7460 rpm
what revs is the car wheel doing?
what speed is the bike doing , if it was on the road?

the stroke of the engine is 86mm, if it helps with any future calculations

Evenin.

We can treat the tyre circumferences (or diameters, doesn't matter) as a ratio to multiply the ratio from crank to bike wheel, although the ratio between the tyres is speed increasing, whereas the ratio from crank to rear wheel is speed decreasing.

Torque transfer works the opposite way. Speed decreasing ratios multiply torque by that amount, always less a percentage of loss, even if the transmission is positive by gears etc. There is no loss of speed if the transmission is positive but there will be a loss if friction is required ie the two tyres. It may be small but it'll be there, you can bet.

7460/4.5=1657 rpm (rear wheel)

1.832/1.965=0.932 (ratio between tyres)

1657/0.932=1777 rpm  (turbine, pre slip allowance)

Guessing 2% loss 1777x0.98=1741rpm at turbine.

12kg at 1 metre radius = 12 kgm that's about 120 Nm

120x1741/9550=22kW that's about 30hp.

If I've not made a mistake then we can put back the two percent loss for a representative power at the rear wheel.

So 31hp ( being generous with rounding!)

That'd be about 34 hp at crankshaft.

Anyone spots a numpty mistake please yell!

I'll get back to you on the road speed shortly, once I've got my head around how many inches in a mile etc!

Cheers

[Propellor]

1.965m travelled per rev of rear wheel. Wheel speed is 1657 rpm.

1.965x1657= 3256 metres in a minute.

3256/1000= 3.256 km in a minute

3.256 x60 = 195 km an hour

195 x 0.6 = 117 mph.

Hopefully I've not made a mistake there!

[guest1306]

Geez Propellor, thats pretty clever. I didnt understand where you got the 9550 from to work out the 22 kilowatts . if the bike is only 31 hp then ive got a lot of work to do.

[Propellor]

The 9550 is just a constant in the mechanical power formula when you want to know kW. The 9550 is derived from the basic formula for watts which has 2 pi radians in there.

kW=Nm x rpm/9550.

[Propellor]

........... if the bike is only 31 hp then ive got a lot of work to do.

Do you believe it?

[guest1306]

You were spot on Propellor, I just got word from Tony Foale. 30 hp

Well done.