| Author |
Message |
    
Rocketman
| | Posted on Thursday, April 05, 2001 - 04:07 pm: | 
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Aaron :
I'm interested in how Rogers 100 horses compares to my 100 horses. I went along with your example because it is relative (in theory) to the 100 horse question. However, what you have explained is how the dyno sees the two bikes in your example (identical) but the dyno hasn't explained how those same bikes will perform and they will not perform the same.
I'm done.
Rocket in England |
Jmartz
| | Posted on Thursday, April 05, 2001 - 04:22 pm: |
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Vrooom |
Aaron
| | Posted on Thursday, April 05, 2001 - 04:34 pm: |
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"... they will not perform the same"
Sure they will, their performance is identical in every respect! What makes you think it's different? Didn't both bikes accelerate from 50mph to 100mph exactly the same? |
Axtell
| | Posted on Thursday, April 05, 2001 - 05:34 pm: |
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Evidently when you guys say your "done" you have your fingers crossed :-) |
Court
| | Posted on Thursday, April 05, 2001 - 09:07 pm: |
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Sit down Dickey.........let'em go.
I, who am NEVER done, am well on my way to organizing the first annual inter-cultural Buell mud wrestling championship wherein a bearded bloke (that's what they call eacah other) from the UK and Redbeard, the emerging Elf-Dyno-ius, both have rudimentary engineering text books duct taped to their oil soaked (no wait....that would start a mineral vs. synthetic thread) anti-freeze soaked....(no wait....that would start a new Buell motor thread)....well just hose'em down and let'em wrestle in the salt. Afterward we'll let the 17 hungry Schauzer's chase them and lick the salt off.
I'm ordindarily not like this but current developments in the TEAM ELVES camp and a 45 minute call from THE WOZ today have me so excited that if I had your pacemaker I'd have to reverse the polarity and put myself in the microwave.
What time you have a free moment tomorrow? Drop me a note, I wanna talk.
I'm still pissed that the name RONCO had been taken just as your fame skyrockets.
Court |
Hoser
| | Posted on Thursday, April 05, 2001 - 09:27 pm: |
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MikeJ:
Re: Mtn bike on the dyno , I couldn't find a spot to connect the inductive pickup so my graph does not show Torque ( I'm sure some of you could offer suggestions ) but what I ended up with was 2 HP (after 4 beers ) . I will find that graph and post it for a laugh.
Jeff |
S2no1
| | Posted on Thursday, April 05, 2001 - 10:51 pm: |
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Aaron, Axtell,
Just a suggestion. Using the graphs from each dyno run in every gear, integrate the power curve to determine the maximum energy available from the rear wheel. We'll need to make a few assumptions:
1) That the transfer of energy to the drum is representative to the real world (i.e. same friction.
2.) That the horsepower is also the same.
Now what you'll probably see is that a Buell has more energy below, say 40 MPH, but after 40 MPH, (assuming that we pull to redline) a 600 will have more available above the 40MPH.
Now you can convert this to acceleration with basic physics.
And here I didn't think I had anything to add to this discussion.
Court,
In order to be totally fair you would need to put Rocket in the "rider conditioning thread" for a month or two and make sure he leaves his his stealth brick and launcher in England. Alternatively Aaron could bring those "American" Horses" (I'm partial to Quarter Horses myself).
By the way, you aren't the only one who is never done.
Arvel
who should be working all night but is too tired |
Aaron
| | Posted on Thursday, April 05, 2001 - 11:47 pm: |
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I just thought of a simpler way to prove my point ...
If you accept the hp = (tq*rpm)/5252, then you will quickly realize that the back wheels of two different motorcycles cannot POSSIBLY have the same ground speed (rpm), the same horsepower, and different amounts of torque. It violates that formula.
AW |
Blake
| | Posted on Friday, April 06, 2001 - 12:02 am: |
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Ain't this stuff great! I love it! Almost as much as riding.
BHR:
"...'splain to me why a 100 hp 600 with 45 lbs of torque will spank a 100 hp Buell with 90 lbs of torque." I know you know this, but you did ask... The 600 likely has significantly aerodynamic drag and significantly less mass (weight). Rider weight also matters too, so in your/our case... 
One corrolary... the dyno drum does not see the HP required to spin up (accelerate) the bike's rotating/circulating parts (drivetrain, rear wheel/tire, oil...), however, the power used to spin up the rotating parts varies with acceleration, so wouldn't a 5th gear run (longer duration, slower engine acceleration) always show higher dyno HP versus lower gear dyno runs (shorter duration) especially a 1st or second gear run. The high gear run uses less of the engines HP to spin up the bike's rotating parts.
Ponder this: The power required to accelerate a rotating mass increases in proportion to the angular velocity (rpm). For example, considering only the engine's rotating inertia, it takes twice as much power to accelerate (at a constant rate) our engine between 3000 and 5000 rpm as it does between 1000 and 3000 rpm.
But, the rotational inertia is proportional to the rotating mass and the square of its radius.
So an engine with half the rotating mass that rotates at half the radius would have 1/8th the rotational inertia... (1/2)*(1/2)^2 = 1/8.
So lets do virtual compare of a 600 and our Buell to see which migh have the advantage. Assume the 600 hase half the rotating mass at half the radius of the 1200 (this is a huge whopping assumption solely for the sake of argument, but it is probably conservative).
Let's say our 1200cc Buell is running from 5800 to 6800 rpm, a 1000 rpm range. The 600's gonna run double the rpms over double the rev range (say from 11600 to 13500 rpm, a 2000 rpm range) with 1/8th the rotational inertia in the same amount of time.
Compared to the Buell the 600's Relative Power Lost = 2 * 2 * 1/8 = 1/2 of the power that the Buell would lose due to engine inertia.
Or in pertinent terms, if the 600 has the same hp as the Buell then it will spin up twice as fast even though it has to cover double the range at double the rpm.
Since the aforementioned radius (radius of gyration for nerd engineer types) is closely related to the engine's stroke, it may be likely that the inertia of a 600 is significantly less than even 1/8th that of the Buell.
Aaron, the discussion at the torque/HP link you posted started out okay, but lost integrity when the guy stated that "Torque is the only thing that a driver feels, and horsepower is just sort of an esoteric measurement in that context. 300 foot pounds of torque will accelerate you just as hard at 2000 rpm as it would if you were making that torque at 4000 rpm in the same gear, yet, per the formula, the horsepower would be *double* at 4000 rpm." The "in the same gear" comment apparently hints that he is comparing one vehicle to itself. I guarantee that a Corvette in second gear spinning 4000 rpm with 300 FT-LB engine torque output will toast a diesel in second gear (same gearing) spinning 2000 rpm with 300 FT-LB engine torque output.
Like you said... (another Aaron quote to save) Power is Power and the one with the most average power across the band for the most time wins. Huh? wuzzat? |
Aaron
| | Posted on Friday, April 06, 2001 - 12:49 am: |
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Blake: Hmmm ... so what you're saying is that when you accelerate something more slowly, ultimately you use less power to get it up to speed.
BUT ... in top gear, don't you have to accelerate it farther? Sure, the engine will be spun up the same amount, but the thing on the other side of the gearing has to be accelerated much farther than it does in a low gear. Doesn't that make the total power used a zero-sum game?
I'll have to go ponder that.
With respect to short strokes and corresponding reductions in inertia, yes sir, I can definitely see that. Square of the radius is a powerful thing. That's why I'm really wondering what effect Rocket's new wheels will have on his dyno runs. He's taken a bunch of weight off the outside of the wheel, where it matters most.
With respect to the 600's, or any other short stroke design, it'd be interesting to actually measure their powertrain losses and compare them to our typical 15% or so.
Let me ask you a question. The dyno readout we get is after these drivetrain inertia losses, but what about the front wheel? As long as it's on the ground, isn't it mechanically connected to the rear wheel, and the motor has to spin it up too, overcoming it's inertia as well?
Seems like a dyno design that spun it up at the same rate the bike was spinning up the rear wheel would give a more useful, real world result.
Y'know, at our Bonneville run, I left those front brakes on. Two honkin big rotors. Hmm ... maybe those should go. What do you think?
AW |
Brentx1
| | Posted on Friday, April 06, 2001 - 01:07 am: |
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O.K. I've waited long enough to try this. I have a '99 X1 with Nallin stage 2 head work, Nallin pistons, total seal rings, race ecm, V&H muffler, Force intake. I will be doing some cam work later. I did all the work myself, and I didn't know anything about motors. Thank the lord for service manuals. I'm not too sure about reading dyno's so if someone could tell me if this seems normal for these modifications, I would appreciate it. Thanks. |
Airborne
| | Posted on Friday, April 06, 2001 - 01:53 am: |
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Now that you have the equipment. Try dialing in the tuning for more HP. My 99 X-1 only has a V&H slip on and a home garage modified air intake with K&N filter the engine is bone stock and I'm making 88 RWHP on a good day with fresh service, i.e. correct tire pressure, belt tension, fresh air filter, fresh plugs. Keep up the work though you are headed in the right direction. I don't have a garage worthy of rebuilding a motor or I would do the same. |
Rocketman
| | Posted on Friday, April 06, 2001 - 06:01 am: |
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OK I'm not done .........yet 
So we have two identical bikes making identical charts on the dyno except one does it up to 5K the other up to 10K, not that it matters to the dyno where the RPM's are because the tach pick up is not connected right.
Aaron reckons if we measure the 50 to 100 acceleration of both bikes they will still be identical so they will ride the same. Surely that's a mistake my wizard friend ?
If these bikes perform the same in the real world what does the one bike do with the potential spare 5K it has left over and what did the other bike do with the first 5K it wasted ?
Now me being a brick head and at least one brick short of a bungalow, please explain how these two bikes will ride the same because I just ain't getting it
Rocket in England |
José_Quiñones
| | Posted on Friday, April 06, 2001 - 08:03 am: |
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Rocket,
Which one is lighter (rider + Bike)?
Everybody seems to be forgetting the power to WEIGHT ratio!
Blake,
If that Diesel Truck and that Corvette had the same power to weight ratio, they would accelerate the same under your conditions. But they are not the same, which is why the Vette is faster.
Why is a Blast!, with 26-28 hp, faster than traffic? Because it's only pushing 400 lbs + 150 lbs rider (19.6 lb/hp) vs. a car (PT cruiser) making 150 hp that weighs 3184+ pounds (21.2 lb/hp).
That's all I have to say about that...... |
Court
| | Posted on Friday, April 06, 2001 - 09:06 am: |
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>>>That's all I have to say about that......
Yeah...right....hehehehe.
I've put in for vacation time to allow me to monitor this discussion. This is more fun than when by biology teacher electrocuted the frog.
Court |
Axtell
| | Posted on Friday, April 06, 2001 - 09:51 am: |
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Great posts Blake
He knows too much
Court..Make him set down too |
Aaron
| | Posted on Friday, April 06, 2001 - 10:01 am: |
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I'm still pondering Blake's point about rate of acceleration versus power. He may well be right. Damn, I was never very good at this physics stuff. I got through the classes by memorizing formulas.
"Aaron reckons if we measure the 50 to 100 acceleration of both bikes they will still be identical so they will ride the same. Surely that's a mistake my wizard friend ?"
Arrrghhh! You're just not going to let me out of this, are you?
I don't know what you mean by "ride the same".
Accelerate the same? Absolutely. Same torque reaching the rear wheel, at least over those ranges I described.
"If these bikes perform the same in the real world what does the one bike do with the potential spare 5K it has left over and what did the other
bike do with the first 5K it wasted ?"
I think you're trying to read too much into the example. I was just trying to show that it doesn't matter what rpm the hp is made at, if the powerbands are the same and the gear reduction is changed to make the ground speeds match. But okay. Let's assume 5K and 10K rev limiters. And let's assume the low revving bike isn't usable below 2500rpm and the high revving bike isn't usable below 5000rpm, they'll stall.
Both powerbands are the exact same size, you can double your speed over the course of the powerband. Both make an identical horsepower curve over the course of that powerband, 47.6 to 95.2hp. If the only difference in the gearing is the 2X reduction in the high revving bike to make the ground speeds the same, and both bikes weigh the same, I contend that the performance will be the same, yes.
The ONLY difference is that the motor in one of the bikes is spinning twice as fast as the motor in the other bike at any given point in the powerbands. But it's also making half the torque and geared twice as short. So it looks the same to the back wheel, and the forward thrust on the bike is identical. At least that's my story, and I'm sticking to it until someone can disprove it!!!
Now, in the real world, high revving 600cc sportbikes making the same rear wheel power as our Buells tend to have a wider powerband. They also are equipped with 6-speed gearboxes, allowing them to operate over a narrower rpm range (in relative terms) and therefore maximize the torque reaching the rear wheel as you run through the gears. And they tend to be lighter and have better aerodynamics. So, yeah, they "feel" different and they're probably quicker.
I think the horse is dead 
No mas!
AW |
José_Quiñones
| | Posted on Friday, April 06, 2001 - 10:19 am: |
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From the article Aaron referred to earlier:
The Only Thing You Really Need to Know:
Repeat after me. "It is better to make torque at high rpm than at low rpm, because you can take advantage of *gearing*." :-) |
Al_Lighton
| | Posted on Friday, April 06, 2001 - 10:25 am: |
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Blake,
you wrote:
"it takes twice as much power to accelerate (at a constant rate) our engine between 3000 and 5000 rpm as it does between 1000 and 3000 rpm."
Unless I'm interpreting you wrong, this makes no sense to me. I'm a linear kinda guy, don't run many rotational mass calcs. But the linear version of that statement would be:
"it takes twice as much power to accelerate (at a constant rate) a mass from 30 to 50 ft/sec as it does from 10 to 30 ft/sec". Now, neglecting wind resistance and such, I know this not to be true. So why should it be true for rotational calculations?
I'm not saying you're wrong, I'm saying I'm lazy and haven't bothered to look it up to verify. But it sure doesn't make sense to me intuitively. Can you please 'splain it some more?
Al |
Al_Lighton
| | Posted on Friday, April 06, 2001 - 10:30 am: |
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This one has almost reached contact patch proportions, eh? Aaron, I don't believe you about the not very good at this physics stuff. You may not have all the theory memorized, but your grasp of the practical aspects is generally right on.
Al |
Mikej
| | Posted on Friday, April 06, 2001 - 10:38 am: |
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Some stuff just can't be explained with numbers and formulas, sometimes you just got to sit your seat down and feeeeeeeeeeeeel the difference.
2+2=4, usually.
That's all the math I'm going to do today. |
Aaron
| | Posted on Friday, April 06, 2001 - 11:07 am: |
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Brent: Look at that dyno sheet, it's really obvious what's happening. It noses over when it hits the stock ECM's high rpm retard. Race ECM + careful setting of the timing, by the book, it's critical on the injected bikes.
Jose': YES ... but notice he said torque, not hp. Hp is a number that combines the torque and rpm and that's why we use it.
I find the torque curve to be useful because it tells me what kind of a cylinder fill I'm getting at a given rpm. Tuning the LSR bike, I basically ignored the hp curve, I looked at the torque peak and then did my damndest to maximize it's size and move it to the right, knowing that the hp would follow. By watching the rpm of the torque peak I was able to see the effects of changes long before I moved it enough to affect peak hp, telling me whether or not I was going in the right direction. This was all done with the exhaust system, BTW. It has a huge effect on cylinder fill & the rpm where you get it. People always think in terms of cams driving your powerband, but the exhaust is a huge factor. One of many things I've learned from Mr. Wizard.
Al: I was thinking the same thing, re: contact patch. Ahh, how I long for those days
AW |
José_Quiñones
| | Posted on Friday, April 06, 2001 - 12:52 pm: |
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Aaron,
I think we are saying the same thing. Torque is what can actually be measured (Foot/lb of force).
HP is mathematically derived. You can't measure HP directly. You have to measure the torque and rpm and calculate HP out of the formula
hp = (tq*rpm)/5252.
So like you said, the higher up the rpm band that you can make your peak torque will determine how much HP you make. Take care of the Torque and the HP will take care of itself. |
Aaron
| | Posted on Friday, April 06, 2001 - 01:01 pm: |
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Jose': Yup, that's the way I understand it, on all counts. |
Ralph
| | Posted on Friday, April 06, 2001 - 01:41 pm: |
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Nope, as interesting as the contact patch arguement was (not!) this is riveting. Keep going. I'm not ever going to be done.
bighairyralph |
Axtell
| | Posted on Friday, April 06, 2001 - 02:20 pm: |
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Aaron, on your 10:01am post I think you made a small error
" I think this horse is dead"
you mean horsepower don"t you????
Heheheheheh |
Rocketman
| | Posted on Friday, April 06, 2001 - 02:46 pm: |
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The ONLY difference is that the motor in one of the bikes is spinning twice as fast as the motor in the other bike at any given point in the powerbands. But it's also making half the torque and geared twice as short. So it looks the same to the back wheel, and the forward thrust on the bike is identical. At least that's my story and I'm sticking to it until someone can disprove it!!!
Well actually, his name is Einstein but he's dead now. He did say though, when I last saw him at the drag strip, that given the choice of which bike, the torque monster or the rev nutter bastardio, he'd have chosen the torque monster because it would use its RPM range more efficiently where as the screamer motor would need to launch at 5K to be on a par with the torque motor, and even Einstein who never rode Top Fuel recognised the contact patch theory, or lack of it, in the screamers case.
However Aaron, I've confused myself now because I've still got both wheels spinning at the same rate. Moving from standstill or from 50 MPH ? They simply must ride different.
Mmmm. Do you know what a salt and pepper flavoured cap tastes like ? Thought not. Me neither and I'd like to keep it that way.
Rocket in England nearly done |
Blake
| | Posted on Friday, April 06, 2001 - 08:01 pm: |
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Aaron: As far as the engine spin-up goes, conservation of energy dictates that a top gear run and a 1st gear run between the same rev points would entail the exact same amount of work done to the rotating engine parts. This work would be the integral of torque versus revs (not revs/min, just revs, or degrees if that makes more sense). Same amount of work in less time means that the 1st gear run hadda use up more power spinning up the rotating engine parts.
However, consider the rear wheel, which in top gear gets spun up to a mucho higher rate albiet at a mucho lower acceleration, we have done more work. I'd have to take some time to work out the equations to be sure, but I'm guessing that the HP used to spin up the rear wheel will be identical for each gear. I'm thinking the greater amount of work for a high gear run would be offset by the greater amount of time required for the run.
Really if the rear wheel had enough inertia, we wouldn't even need the dyno drum would we. There ya go... a poor man's dyno... stick on a thousand wheel weights and let her rip LOL!
Front wheel... you are correct. It would be a simple task to measure the front wheel/tire assembly's inertia, spreadsheet the equation and use it to adjust your dyno curves accordingly. You could do the same thing with the bike/rider mass too. The aerodynamic drag would be the tricky thing. But there are fairly good approximations that can be had for that too. Remember my dyno runs versus throttle position request?
Not really too difficult to come up with a pretty good real world simulation on your dyno.
Wish I were in CO; I'd be calling, laptop and instrumentation in hand, asking if I could come over and play.
Jose: Load the vette with bricks (non-CF bricks please ) so it's the same mass/weight as the diesel. If they both have the same overall drive ratios/gearing, and have the same engine torque, but the vette is pulling that torque at twice the rpm range as the diesel, the vette will finish the quarter mile in half the time. The vette will have twice the HP.
HP is THE benchmark that relates engine torque to the rear wheels. In linear motion terms, Power (P) equals force (F) times distance (d) divided by time (t)... P = F*d/t.
For our bikes F is the kick in the pants we feel when accelerating.. it's the thrust that accelerates the rider/bike and overcomes inertial and drag loading. Distance (d) and time (t) are simply that, the distance traveled and the time it took to do so.
So for Rocket's 1/4 mile run we have...
d = 1/4 mile = 5280FT/4 = 1320 FT = 402.3m
t = 11.25 seconds
Useful Equations:
F = ma where "m" is mass, and "a" is acceleration.
a = (V2-V1)/t where starting velocity (V1) = 0.00 (like in a drag race), so let's just use "V" for final velocity and drop the "V2".
a = V/2t (for constant or avg acceleration)
V = 2d/t (again for constant acceleration)
Assuming constant acceleration...
V = 2*1320 FT / 11.25 s
V = 234.7 FT/s = 160.0 mph = 71.53 m/s
a = 234.7 FT/s / 11.25 s
a = 20.86 (FT/s)/s = 14.22 mph/s = 6.358 (m/s)/s
Thus Rocket's average acceleration for the entire 1/4 mile run was 14.22 mph/s (speed increasing 14.22 mph each second after he left the line).
Now of course we know that his acceleration was not constant. Acceleration drops off with each upshift and with increased velocity. So it is not surprising and makes perfect sense that Rocket does not actually attain a trap speed of 160 mph. But since we are only trying to calculate HP required to get Rocket and his bike down the 1/4 mile in 11.25 seconds, the details in between really don't matter.
Here goes...
I'm gonna cheat and use conservation of energy instead of actually calculating based on the givens above. The result should be exactly the same though based on our assumptions and neglecting aerodynamic drag and front wheel inertia and rear tire slippage...
Assumptions:
Mass (m) of Rocket and his bike...
m = 300 Kg = 660 LBm = 20.51 Slugs
Conservation of energy says that the sum of all energies will remain constant within a closed system or something like that. In our case the "system" is Rocket+Bike so we can simplify and say that the potential energy (PE) plus kinetic energy (KE) will remain constant between the start and finish line (recall that we're using constant acceleration (a)).
PE = mad
KE = 1/2mV^2
KE at starting line = 0.00 and
PE at finish line = 0.00
so PE at starting line must = KE at finish line or
mad = 1/2mV^2 simplifying we can even say ad = 1/2V^2
PE = mad = 300 Kg * 6.358 (m/s)/s * 402.3 m
PE = 767,300 (Kg-m/s^2)m = 767,300 N-m
check KE...
KE = 1/2 * 300 Kg * (71.53 m/s)^2
KE = 767,500 N-m (close enough, I'm sticking to 4 significant figures.)
I'll give Rocket the benefit of the doubt and use the KE.
Well, since we started with KE = 0.000, we had to use all our own energy to achieve the required acceleration. How much work (W) did it take to reach the KE above? Neglecting all the losses discussed previously, it takes exactly 767,500 N-m.
Power (P) = W/t
P = 767,500 N-m/11.25s = 68,220 N-m/s (aka Watts)
HP = 68,220 W * 0.001341 HP/W = 91.5 HP
Rocket, considering HP losses due to clutch slippage, aerodynamic drag, and rear tire slippage, and simply being off the peak HP for much of the run, I'd say that either 300 Kg (660 LB) is too high of an estimate for rider+bike, or your dyno numbers may be very conservative.
Hope y'all enjoy the above.
oao,
Blake (nerd weenie web dyno dragrace simulator)
PS: Someone please get me the drag coefficient for a brickhead flying on an S1! |
Blake
| | Posted on Friday, April 06, 2001 - 08:34 pm: |
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Al:Unless I'm interpreting you wrong, this makes no sense to me. I'm a linear kinda guy, don't run many rotational mass calcs. But the linear version of that statement would be:
"it takes twice as much power to accelerate (at a constant rate) a mass from 30 to 50 ft/sec as it does from 10 to 30 ft/sec". Now, neglecting wind resistance and such, I know this not to be true. So why should it be true for rotational calculations? Sorry, but I'm right. Nya nya!
Think of kinetic energy. KE = 1/2mV^2
Since the "1/2m" portion is constant, simply compare relative changes in KE as
50^2-30^2 = 1600
versus
30^2-10^2 = 800
Thus, it takes twice as much energy to acclerate a mass from 30 to 50 mph as it does from 10 to 30 mph. If we accelerate at the same rate (over the same time (t)) we'd have to have twice the power.
And Ralph, since that wonderful contact patch thread got lost... If your contact patch area times your pressure ain't close to the axle loading, something is wrong. So if total bike/rider weight is 660 LB and tire pressure is 36 psi, total contact patch area is ~18 sq inches.
Did I mention how much I love this stuff?
Blake |
Fastback69
| | Posted on Friday, April 06, 2001 - 09:07 pm: |
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Right! |
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