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Stock Phazer Dyno'd - HP/Torque.
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Sled Dog
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PostPosted: Tue May 09, 2006 10:57 pm    Post subject: Reply with quote

I read somewhere also that 10 lbs reduction = 1 hp you can add 10 hp to the equation from where you will feel it in the ride and acceleration.
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LazyBastard
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PostPosted: Tue May 09, 2006 11:59 pm    Post subject: Reply with quote

rs-yam wrote:
BV1 wrote:
Perhaps, but it was accounted for mathematically then.

There is no way that engine is only making 22 ft/lbs of torque (37 after gear reduction).

-Steve

Shouldn't the gear ratio be applied to both the hp and torque?

If 80hp/37t is used in the equation, I would think both hp and torque numbers would be proportional to the gear reduction - not just the torque. The above equation is correct but the motor output before the gear reduction is about 53hp/24t. (assuming a 1.5 gear ratio.) (Tq = 53*5252/11500 = 24)


No. You are confusing power to be the same as torque. Power is NOT affected by gear ratio. It is theoretically exactly the same on the crankshaft as on the track (and would be if not for our biggest enemy... friction), that is the reason why power is used as a measure of what an engine can do more so than torque.

In fact, the formula shows this quite clearly...

Torque = Power * (5252 / RPM).
Lets say we have two engines capable of making 100 hp, one makes its power at 5000 rpm, the other at 10000.

Fast engine:
Torque = 100 * (5252 / 5000) = 105.04

Slow engine:
Torque = 100 * (5252 / 10000) = 52.52

Now lets take the fast engine and gear it down 2:1
Torque = 100 * (5252 / (10000 / 2)) = 105.04 <-- identical to slow engine.


And yes, from this it follows that torque is a useless number for judging a vehicle's performance. A higher torque engine spins slower and makes the same power. The same results can always be achieved by changing the gear ratio.


Where torque makes a difference is mainly in vehicles with distinct gear ratios - particularly boats, which have only ONE gear ratio. Marine engines need to put out big torque to accelerate the boat up to its high end. A snowmobile doesn't need to, it can wind its engine out to peak power nearly instantly, and progressively shift in order to maintain absolute maximum possible torque on the track, which, funny thing is, is NEVER achieved at the point of maximum ENGINE torque.


Take for example this phazer engine;
At any particular speed, lets say W mph, the driveshaft (thats the one that goes through the front of the track) is spinning at X rpm.

At peak torque, 37 (crankshaft), we get;
37 * 9800 / 5252 = 69 hp
At 69 hp and X rpm, we get;
362388 / X footpounds of torque

At peak power 78 hp, we get;
409656 / X footpounds of torque.

That means that we get 13% more driveshaft torque (acceleration) with the engine spinning at peak power speed than spinning at peak torque speed.

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Sled Dog
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PostPosted: Wed May 10, 2006 1:13 am    Post subject: Reply with quote

What LB is getting at you can gear down (add torque or leverage) to a 1/2 hp motor to do tremendous work like a hydraulic press or any winch they are geared down lots. A small amount of torque can do lots of work if it is geared correctly. A 20 lb weight could lift 200lbs on a fulcrum with the focal point close to the heavier weight giving lots of leverage to the 20lb weight.
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BV1
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PostPosted: Wed May 10, 2006 7:41 am    Post subject: Reply with quote

LazyBastard wrote:

No. You are confusing power to be the same as torque. Power is NOT affected by gear ratio. It is theoretically exactly the same on the crankshaft as on the track (and would be if not for our biggest enemy... friction), that is the reason why power is used as a measure of what an engine can do more so than torque.

In fact, the formula shows this quite clearly...

Torque = Power * (5252 / RPM).
Lets say we have two engines capable of making 100 hp, one makes its power at 5000 rpm, the other at 10000.

Fast engine:
Torque = 100 * (5252 / 5000) = 105.04

Slow engine:
Torque = 100 * (5252 / 10000) = 52.52

Now lets take the fast engine and gear it down 2:1
Torque = 100 * (5252 / (10000 / 2)) = 105.04 <-- identical to slow engine.


And yes, from this it follows that torque is a useless number for judging a vehicle's performance. A higher torque engine spins slower and makes the same power. The same results can always be achieved by changing the gear ratio.


Where torque makes a difference is mainly in vehicles with distinct gear ratios - particularly boats, which have only ONE gear ratio. Marine engines need to put out big torque to accelerate the boat up to its high end. A snowmobile doesn't need to, it can wind its engine out to peak power nearly instantly, and progressively shift in order to maintain absolute maximum possible torque on the track, which, funny thing is, is NEVER achieved at the point of maximum ENGINE torque.


Take for example this phazer engine;
At any particular speed, lets say W mph, the driveshaft (thats the one that goes through the front of the track) is spinning at X rpm.

At peak torque, 37 (crankshaft), we get;
37 * 9800 / 5252 = 69 hp
At 69 hp and X rpm, we get;
362388 / X footpounds of torque

At peak power 78 hp, we get;
409656 / X footpounds of torque.

That means that we get 13% more driveshaft torque (acceleration) with the engine spinning at peak power speed than spinning at peak torque speed.


Agreed, simply put power is energy and cannot disappear into nothing or come from nothing.

If the engine is outputting 80 HP then after all the the gears and reductions etc there still better be pretty close to 80 HP or something is absorbing alot of engergy and gonna catch fire.

IE 1 HP = 746 watts, so

if the engine is outputting 80 HP and after the gear reduction the driveshaft is only outputting 75 HP, well then you have 5 HP lost some where.

5HP = 5*746 Watts = 3730 watts.

We all know how hot a 1500 watt heater is, so if your losing 3700 watts due to a gear change, some poor gear is generating 3700 watts worth of heat. Ouch.

So yes even after all the gear ratios your still going to be very close to 80 HP minus some small frictional losses.

-Steve
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rs-yam
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PostPosted: Wed May 10, 2006 11:29 am    Post subject: Reply with quote

Sled Dog wrote:
What LB is getting at you can gear down (add torque or leverage) to a 1/2 hp motor to do tremendous work like a hydraulic press or any winch they are geared down lots. A small amount of torque can do lots of work if it is geared correctly. A 20 lb weight could lift 200lbs on a fulcrum with the focal point close to the heavier weight giving lots of leverage to the 20lb weight.


That makes sense. How does the math make up for the gear reduction? It's obvious that the hp/torque values were taken with the gear reduction. If the hp isn't different with or without the gear reduction, then is it an 80hp/24t motor without the gear reduction and an 80hp/37t motor with it? -as measured by the dyno.

(I'm just trying to understand all the math that's been posted - some of which seems conflicting.) Dunno

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Sled Dog
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PostPosted: Wed May 10, 2006 12:02 pm    Post subject: Reply with quote

I would say it is with the reduction gear so it is an 80 hp sled as Yamaha states.
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PowerValve-700
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PostPosted: Wed May 10, 2006 3:04 pm    Post subject: Reply with quote

So let me get this straight, please correct me if I'm wrong:

A gear reduction will increase torque but will decrease horsepower.

Origional numbers: (these are crankshaft numbers):

37 lbft torque @ 9800 rpm and 78 hp @ 11500 rpm

Gear reduction: 1.57

They didn't spin the PTO shaft 11500 rpm, this would mean the crank shaft was spinning at 18055 rpm! So if the dyno was connected to the PTO shaft, how did they get these numbers?

Either the dyno calculates for the gear reduction or threw manual calculations you can figure out the numbers at the crankshaft:

Dyno results @ the PTO shaft were probably something like:

58 lbft torque @ 6250 rpm and 69 hp @ 7350 (only my estimate, derived threw calculations and rounded to the nearest 10)

Output torque "PTO shaft" = input torque "crankshaft" x gear ratio
so working backwards:
Input torque = output torque / gear ratio
Input torque = 58 / 1.57
Input torque = 36.9

@ what rpm?

Output rpm "PTO shaft" = input rpm "crankshaft" / gear ratio
so working backwards:
Input rpm = output rpm x gear ratio
Input rpm = 6250 x 1.57
Input rpm = 9812.5

For horsepower:

Input horsepower "crankshaft" = input torque x input rpm / 5252
(Input rpm = output rpm x gear ratio)
(Input rpm = 7350 x 1.57)
(Input rpm = 11539.5)
Input horsepower = 36.9 x (7350 x 1.57) / 5252
Input horsepower = 36.9 x 11539.5 / 5252
Input horsepower = 425807.55 / 5252
Input horsepower = 81
Input horsepower = 81 (@ 11539.5 rpm)

So horsepower and torque at the input shaft "crankshaft" is:

36.9 lbft torque @ 9812.5 and 81 hp @ 11539.5 rpm

That's pretty close to the origional numbers:

37 lbft torque @ 9800 rpm and 78 hp @ 11500 rpm

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monker
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PostPosted: Wed May 10, 2006 3:51 pm    Post subject: Reply with quote

you guys are getting yourselves all confused.

Measurements are taken at the pto. 80hp and 37ft-lb were measured where the clutch attaches. the 11500 rpm is engine rpm, which is greater than clutch rpm.

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AndersY
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PostPosted: Wed May 10, 2006 7:10 pm    Post subject: Reply with quote

PowerValve-700 wrote:
So let me get this straight, please correct me if I'm wrong:

A gear reduction will increase torque but will decrease horsepower.

No, horsepower is always constant, except for friction losses.

So just measure power against engine rpm, doesn't matter with of without gear reduction, and then calculate the torque.

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Yammerhead
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PostPosted: Wed May 10, 2006 11:30 pm    Post subject: Reply with quote

monker wrote:
you guys are getting yourselves all confused.

Measurements are taken at the pto. 80hp and 37ft-lb were measured where the clutch attaches. the 11500 rpm is engine rpm, which is greater than clutch rpm.


Sorry Monker, I have to disagree with you. The dyno was attached to the PTO shaft and measured the torque available there. Horsepower was then calculated by the dyno. On the graph, they plot the horsepower and torque along the same x-axis using ENGINE RPM. This means the torque has to be adjusted to crankshaft torque, otherwise the graph is erroneous.


On another note, did MC-Xpress edit their PDF? I could have sworn it used to have a stock dyno graph? I don't even see the stock peak HP and torque figures any more. If so, think Yammie made them pull it? Hmmmmm.......


Alf, you out there?? LOL!!

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rs-yam
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PostPosted: Thu May 11, 2006 3:27 pm    Post subject: Reply with quote

Yammerhead wrote:

On another note, did MC-Xpress edit their PDF? I could have sworn it used to have a stock dyno graph? I don't even see the stock peak HP and torque figures any more. If so, think Yammie made them pull it? Hmmmmm.......


Yes they did change it. It was there. Big Brother must have called and censored them.

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BV1
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PostPosted: Thu May 11, 2006 3:59 pm    Post subject: Reply with quote

Someone must still have it! Check your internet caches! :)

-Steve
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PostPosted: Thu May 11, 2006 10:56 pm    Post subject: Reply with quote

rs-yam wrote:
Yammerhead wrote:

On another note, did MC-Xpress edit their PDF? I could have sworn it used to have a stock dyno graph? I don't even see the stock peak HP and torque figures any more. If so, think Yammie made them pull it? Hmmmmm.......

Yes they did change it. It was there. Big Brother must have called and censored them.

Its definitely NOT something that Yam would want out just yet. As such a fast spinning engine, the torque numbers will seem very low, which will be misread by the misinformed to mean that the engine is weaker than it actually is. This kind of information would be much better AFTER the sleds have been bought and paid for, so that real world testing will prove that they actually run really well for their class.

The next year, it will be OK since there will be real performance results to prove that its good.

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PostPosted: Tue May 16, 2006 11:00 pm    Post subject: Reply with quote

the phazer has alot of throttle response and it pulls hard right from the start, it has plenty of torque. i am very impressed with the stock phazer, it is already running close to 12's in the 1/4 mile at 91+ mph and we have only had it to the track three times so far. jeff
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phaz4
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PostPosted: Wed May 17, 2006 10:03 am    Post subject: Reply with quote

Srxspec wrote:
55/35 reduction so it'd be a 1.57 reduction roughly.


is the official/actual/verified gear reduction between the motor and pto? it looks right.

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