Torque or HP?

[LVjetboy]

"Please, just show me"
Kojac, allow me show you by addressing each point in your first two post in a bit more detail...
"Higher horsepower will naturally correspond to the higher torque...the rpm an engine makes when torque falls off more than horsepower gains is the rpm that the pump stops pulling."
Actually, the pump stops pulling when it reaches the engine-impeller match point, which is typically calculated with engine and impeller power curves. While it's true at any given engine rpm, higher torque means higher power, that does not mean the magnitude of torque applied to the impeller (or how quickly engine torque falls off) is a true measure or predictor of jet thrust and performance. Example: matching impeller to peak torque instead of peak power. Matching to peak torque makes sense if you believe torque drives performance...right? But assuming no pump losses, your peak torque jet will generate less thrust, accelerate and run slower than the jet dude with the same identical engine but impeller matched to peak power. Trust me. This is why it's all about power.
Focusing on torque instead of power can lead to the wrong conclusions.
Another example: Jetboat Joe says he's putting 500 ft-lbs of torque to his pump and Jetboat John says he'll keep up 'cause he's putting 500 ft-lbs to his pump too and they have the same jets. Is it a dead heat, if not, who wins? You can't know because torque to the pump means nothing without rpm...in other words, power. So you search for more information... Jetboat Joe turns a Berkeley b/c to 6300 rpm absorbing 615 hp. Jetboat John turns an American Turbine 9.5 to 4300 absorbing 415 hp or 200 hp less. By knowing power, you now can estimate performance. Even though John's running the same torque as Joe, he's nearly 15 mph slower...loosing by a long shot. But wait you say, maybe that 9.5 is more efficient than the small b/c cut and makes up for the difference? From the data I've seen which I'd be happy to show you, not likely...although there will be some loss effects, and John may only be 10 mph slower than Joe.
In the real world there certainly are pump losses, some related to impeller cut (matching) and some to impeller rpm. These losses are expressed in terms of power btw, not torque...
Side note: There's a good reason for losses defined as a power ratio with no mention of torque...goes back to why performance is based on power not torque.
When real world drive train power loss is accounted for, the best performance match will be short of peak power. Oh Geez!! So does that mean performance is driven by peak torque instead power? No. Or some as yet unknown relationship between torque slope quickly dropping off and power gain? No. Here's the key...
The best jet boat performance engine-impeller match applies maximum engine power to the impeller up to the point where matching that impeller to peak engine power incurres losses exceeding engine power gain.
Have you heard or read that anywhere? I think not because I just wrote it.
If you know the relationship of cut size to efficiency, along with engine power curves, you can predict the optimum performance point. Just because we don't have impeller or pump efficiency curves doesn't mean we should base the match point on some bogus relationship of torque crossing power or how quickly torque drops off. If you want a rule-of-thumb accounting for losses, I'd suggest something relating power gain to rpm change...or the slope of the power curve. Although without measured pump efficiency curves, it'd be like any ROT, a wag. At least it'd be based on the true driver of performance...power.
"I have seen a small block on the dyno make 720 hp at 8200 rpm's. Put it in a jet boat with an a/b impeller and it wouldn't pull more than 5600 rpm's. Why?"
That's easy. For lack of better information let's assume you're talking a Berkeley a/b, then at 5600 (on the lake not the dyno), the small block only puts out 510 hp and his a/b impeller absorbs 510 hp. His impeller won't spin faster because at that rpm it needs more engine power (more than 510) to spin faster. 5600 rpm is the engine-impeller match point. To turn higher with the same engine he'd need to cut the impeller. To figure how much performance he'd gain by cutting to a B, C or D, he'd need to know how pump efficiency changes with impeller cut size and the shape of the engine power curve. In other words, how much power he'd gain from the engine by spinning it faster adjusted for how much power he'd loose from the pump by cutting the impeller smaller. There will be a point of diminishing returns. Notice you don't need to know how quickly the torque curve drops off or even the value of torque to figure a match point or to predict performance. In fact, focusing on the magnitude of torque applied or the behavior of the torque curve can lead to wrong conclusions.
"I had a buddy with a 468 Chevy 13:1 compression...turned a pump to 6100 doing 108. He built a 588 and turned the same rpm's and same speed even though the big motor made more power and torque...he just couldn't understand it...all that money to accomplish the same thing."
If the big engine turned the same rpm full throttle that engine developed the same power as the 468 at the same rpm. No mystery. Even though the bigger engine may have more peak power, at 6100 rpm it had the same power as the 468. You mentioned possibly 250 more peak hp? That seems a bit high for two engines matching power at 6100, but if true then impeller matching for the big engine would regain some of that. You mentioned he then changed cam profile and did some head work for more velocity and ran 128 at 7200 rpm. If he kept the same impeller cut, that'd be about 570 hp (depending on impeller make and size) gained from cam and head work. Seems also a bit high and I'd guess he may've cut the impeller too? Either way, once you start focusing on power, performance becomes less mysterious.
jer

[LVjetboy]

SteelComp posted: "...what is it that you are calling peak power? You keep saying power- power- power. What is power? Where is peak power achieved and if we don't use q as a factor, then what do we use?"
Peak power is the maximum power an engine produces through the rpm range of the application. Power is a combination of both torque and rpm as the equation posted time and again shows. Peak power is reached when the product of torque and rpm (not just torque) is maximum. Yes, torque is a factor in power, but so also is rpm. And you need BOTH torque and rpm to describe performance. The term "power" combines both torque and rpm in a way that can be used to relate performance. When I say, it's all about power, that doesn't mean I'm saying torque is not used to calculate power. Or that by knowing both torque and how it varies with rpm you couldn't calculate power and then figure performance.
But there IS a difference in focus.
If you read my somewhat long reply above to Kojac, I think you'll understand how some misunderstanding and even wrong conclusions can be drawn from focusing on torque instead of power. Joe and John concluded the wrong thing about performance by focusing on torque applied to the pump. I'm sure there's a racer or two who've matched impeller to peak engine torque and found less than best performance. Yet some insist it's all the same thing. I'm not saying you don't understand the difference Steel. But from responses to this and other power vs torque threads, you can see how some really don't understand. I'm not sure if I answered your question...they seemed a bit tongue-in-cheek? But if you're asking if I understand how torque is used to calculate power or the relationship between torque, power and performance...then yes I do.
One last somewhat related thought. Some claim torque is more important because after all, "that's what a dyno measures." Another myth. Does anyone know what a dyno actually "measures"? You'll also hear, "Power is calculated, Torque is measured." But torque is also calculated by the equation, Torque = Force x Distance, does that make force a better predictor of how that engine will perform in a jet boat? By above logic, force must be a more basic and therefor a more important quantity?
jer

[LVjetboy]

VictorFB posted: "If I remove the 270 hp 454 from my duelly, and install nissans 320 hp V6, will I be able to pull the 5th wheel easier? If hp is a true measure of power, then it should pull much harder with the added 50 hp. Hp is hp isn't it?"
Hp is certainly one true measure of power, not the only one. And yes, hp is hp. Neglecting corrected power (variations in test conditions) and standards of course. Corrections and standards do make a difference. And often those differences account for much of the misunderstanding of the data. You know, how can this engine with this reported power run better than that engine with that reported power? But wait, how do you know that engine has that power? And if it has that power on the dyno, how much at the lake? What standard rated at?
Back to your question, the difference is those two engines can't deliver their peak power at all towing rpm’s. Unlike a jet boat where you mash the throttle and you go to max rpm's (and maximum thrust) quickly, with a truck towing a load you'll spend a lot more time short of peak engine power rpm. And the more weight you tow, the longer time at lower power slowly accelerating through the gears. For that application, a broad fat power curve down low rules. Not peaky up high. Gears help too but there's limits and a wide low end power curve give the best compromise for the full throttle rpm’s your truck towing engine typically sees. Does that make sense?
"...many seem to forget if you change the equation and not have the 5252 meeting point you will not get the correct hp."
For English units 5252 rpm is the crossing point. For metric units the torque/power crossing point is quite a bit higher because metric units require a different constant relating torque and power. Does unit choice change engine or drive train performance? NO. So also the torque vs power crossing point has nothing to do with jet boat performance. Sorry if I'm stepping on someones learning from their favorite guru or preconceived notions here.
Naaa...come to think I could give a sh*t.
"A turbo charged mazda rotary can achieve 400 hp but they lack the torque for any heavy loads. You think that rotary will push a 19' jet easier and faster than a 300 hp big block? NOPE..."
Nope? Actually yes, I think it's possible. Depending on what rpm that turbo rotary gets to 400 hp. Unlike a tow truck, full throttle on the mazda hooked to a jet pump will go quickly to the engine-impeller match point…for the most part unaffected by boat weight. And if you don't give away more than 100 hp by cutting the impeller down to get that peak 400 hp (pump efficiency hit), then yes the rotary can put more power to the impeller and that extra power will be converted to more thrust. More thrust = pushing the jet faster. As an extra bonus, less rotary engine block weight = pushing the jet easier. But that’s not important here.

jer

[LVjetboy]

Well said Lv. Oh wait, what you say not what my engine guru says. You're freakin' bogus Lv!

[Floored]

Kind of a shame to find out all these years later that Keith Black, Wes Cerney, and (Famous)Amos Satterlee didn't know shit and taught me wrong. Now what can I believe in. :argue:

[LVjetboy]

You could believe beyond here there be sea monsters.

[kojac]

SteelComp posted: "...what is it that you are calling peak power? You keep saying power- power- power. What is power? Where is peak power achieved and if we don't use q as a factor, then what do we use?"
Peak power is the maximum power an engine produces through the rpm range of the application. Power is a combination of both torque and rpm as the equation posted time and again shows. Peak power is reached when the product of torque and rpm (not just torque) is maximum. Yes, torque is a factor in power, but so also is rpm. And you need BOTH torque and rpm to describe performance. The term "power" combines both torque and rpm in a way that can be used to relate performance. When I say, it's all about power, that doesn't mean I'm saying torque is not used to calculate power. Or that by knowing both torque and how it varies with rpm you couldn't calculate power and then figure performance.
But there IS a difference in focus.
If you read my somewhat long reply above to Kojac, I think you'll understand how some misunderstanding and even wrong conclusions can be drawn from focusing on torque instead of power. Joe and John concluded the wrong thing about performance by focusing on torque applied to the pump. I'm sure there's a racer or two who've matched impeller to peak engine torque and found less than best performance. Yet some insist it's all the same thing. I'm not saying you don't understand the difference Steel. But from responses to this and other power vs torque threads, you can see how some really don't understand. I'm not sure if I answered your question...they seemed a bit tongue-in-cheek? But if you're asking if I understand how torque is used to calculate power or the relationship between torque, power and performance...then yes I do.
One last somewhat related thought. Some claim torque is more important because after all, "that's what a dyno measures." Another myth. Does anyone know what a dyno actually "measures"? You'll also hear, "Power is calculated, Torque is measured." But torque is also calculated by the equation, Torque = Force x Distance, does that make force a better predictor of how that engine will perform in a jet boat? By above logic, force must be a more basic and therefor a more important quantity?
jer
Jer,
Thanks for the input. I believe if I understand you right in your quote to Steel that "Peak power is reached when the product of torque [b] and rpm[b] (not just torque) is maximum." is what I am observing when I see the motors that I am fortunate enough to see dynoed and then placed in a jetboat and the pump impellers and are paired close to their peak power combinations.
I remember in one of the post regarding Duane's testing of impellers and your data with dyno sheets that he tested three impellars with different cuts on the same boat with the same engine. As I recall the one with the smallest cut pulled the highest rpm's (also near highest dynoed horsepower output)but didn't pull the boat as fast as the one with the larger cut with less rpm's. And the biggest impeller slowed the boat down even more with less rpm's. (sorry I don't recall the size's of impellers off the top of my head or the rpm's I'll search for them and try to put them here.)
The conclusion I came to was according to the dyno sheet the motor was pulling the impeller that ran the boat the fastest was closer to the point where the torque started to drop off more than the horsepower gained. Is this what you mean when you when you refer to peak power. I think this is the point where the peak efficiency of the motor and impeller work best.
These are only my observations over the years with my limited experience with motors that I have seen dynoed or dyno sheets with posted rpm/impeller cuts.
I'm still giving thought to your answers to my post.
"While it's true at any given engine rpm, higher torque means higher power, that does not mean the magnitude of torque applied to the impeller (or how quickly engine torque falls off) is a true measure or predictor of jet thrust and performance. Example: matching impeller to peak torque instead of peak power. Matching to peak torque makes sense if you believe torque drives performance...right? But assuming no pump losses, your peak torque jet will generate less thrust, accelerate and run slower than the jet dude with the same identical engine but impeller matched to peak power. Trust me. This is why it's all about power"
I agree that matching impeller to PEAK torque is not the goal. What I am saying is that at the point ('RPM") when your engine starts to lose MORE torque than it is GENERATING horsepower is where you should be matching your impellar to get the best efficiency and speed.(That point is usually is higher in the rpm range than Peak torque)
I also agree that the motor actually pulls to the engine impeller-match point.
If in my small block engine example it would be able to pull up to 8200 rpm's if you theoretically cut the impeller to a much much smaller size. I don't know if it would make the boat go correspondingly faster unless you had a real small ultra light jet boat. (I have a Seadoo jetski that pulls up to 8000 rpm's but only goes 70+mph bad example of course much different animal.)
However in one of my previous post I talked about an engine I had in my heavy marlin daycruiser jet (468 cubes 10-1 comp , small roller, oval port heads, single barry grant 1000 cfm carb.) horsepower unknown not dynoed. It pulled a berkley stock "a" impeller 5600 rpm's to 78mph.
Installed that same engine in my sterling 19 ft tunnel with a don bowers prepped berkley race pump with a stainless "B"impeller and pulled it to 5600 rpm's to 95mph.
Took it out and installed in my brother's light mantra with a stock berkley pump with a "b"impellar and pulled it to 5600 rpm's to a speed of 89.9 mph.
same motor, Three different boats, three different pumps. same rpm's. three different speeds. I later installed an "a impellar in my brothers boat and still got 5600 rpm's??
I also agree on the conclusion of your examples of jetboat john and jetboat joe. It is important to know what rpm's the motor is making it's best torque/horsepower to get the best impeller combination.
Quote"The best jet boat performance engine-impeller match applies maximum "engine power to the impeller up to the point where matching that impeller to peak engine power incurres losses exceeding engine power gain."
If I understand the above quote correctly I think you are saying what I think I have been saying.
I would still be interested to see what Diggler's motor pulls the a/b impeller. My guess not over 5800 rpm's
Kojac

[victorfb]

not to start an arguement, but the mazda rotary was tried and failed misserably with a small hull and berk pump. final data proved that the mazda rotary lacked the torque to compete with other engines in that application. kinda like riding a ten speed bike in tenth gear. yes you finally get to your desired speed but you will need to use alot of time and energy to get there. your legs dont have the power to to turn the crank with that much force off the line.

[steelcomp]

Kind of a shame to find out all these years later that Keith Black, Wes Cerney, and (Famous)Amos Satterlee didn't know shit and taught me wrong. Now what can I believe in. :argue:
So how many in depth conversations did you have with any one of these guys about HP vs. Q and the theory of power?
Just wondering. This is OLD stuff here....nothing new. THAT, you can believe in.

[steelcomp]

Thanks for the posts Jer. Really good reding.
LV posted:Peak power is the maximum power an engine produces through the rpm range of the application. Power is a combination of both torque and rpm as the equation posted time and again shows. Peak power is reached when the product of torque and rpm (not just torque) is maximum. Yes, torque is a factor in power, but so also is rpm. And you need BOTH torque and rpm to describe performance. The term "power" combines both torque and rpm in a way that can be used to relate performance. When I say, it's all about power, that doesn't mean I'm saying torque is not used to calculate power. Or that by knowing both torque and how it varies with rpm you couldn't calculate power and then figure performance.
That makes more sense to me than what you had said before. I misunderstood your intention as saying that q was a non factor...a "myth" as you said. I understand now that what you're saying that it isn't THE factor, as many believe, and that without relating it to RPM, it's only half the story.
Good explaination. Thanks again.