| Author | Message | ||
     Aaron |
"Is the realtionship between engine torque and driveline reduction linear?" Yes, as I understand it, engine torque to rear wheel torque changes linearly with gear reduction ... for example, if you have 100 ft/lbs at the engine (i.e. what a Dynojet shows) and you have a 5:1 overall gear reduction through the primary/tranny/final drive, then you'll have 500 ft/lbs at the rear wheel, i.e. 500 lbs. of force at a 1 foot radius from the hub. Your rpm at the rear wheel will be 1/5 of the engine's rpm. Horsepower will be the same at either the engine or rear wheel (ignoring losses), although the rpm & torque makeup of that horsepower will be different. If one were to increase a Buell rear wheel perfromance say to 120 HP and 100 ft. lbs. of torque leaving the driveline reduction stock would get you to the top of the gears quicker but retaining the same top speed" Assuming the stock engine could reach the same rpm in top gear as the modified engine (a big assumption), yes, the top speed will be about the same. Calculating acceleration is trickier (although I'm SURE Blake could do it), you have to look at the area under the curve, but I would expect a 120hp/100ftlbs bike to have quite a bit more area under the curve. "How would one go about raising the final drive ratio to say 2.00 (2.26 is stock) in order to put that extra energy into forward motion while accelerating?" Chain drive, just pick two sprockets that equal 2 when you divide the rear tooth count by the front tooth count. You'll lose rear wheel torque. But if the engine can pull it, you'll gain top speed. AW | ||
| Blake |
 | ||
| Craigster |
Jmarts, To add to Aarron's post, every gear interface results in a 2% reduction in calculated torque due to friction/inefficiency losses (typical - depends on a lot of factors). That's why a 5th gear pull on a sporty or Buell tranny results in slightly higher numbers on a Dynojet dyno. Thus actual torque at the crank will be slightly HIGHER than the results back calculated through Dynojets ignition pulse-gearing math. There's also slightly less inertia in 5th gear, but it's impact is minimal compared to gear mesh interface. | ||
| Moperfserv |
Jimidan, the key to the phenomenal torque figure is the brake dyno. The torque at the dyno shaft is probably 136 lb-ft. if that torque is the torque at the quoted HP engine RPM, the dyno shaft RPM is 4,400 RPM. The torque at the crankshaft would then be 108 lb-ft., a much more believable value. Roland and I have always looked at torque/liter. and a value of 80lb-ft/liter is very good for a normally aspirated engine. The only way to get more than that is by compression (and that only gets a little) or some kind of external enhancement, super-charge or nitrous-oxide. | ||
| Jmartz |
Aaron: When I asked you how to go about changing the trasfer ratio I meant in selecting an adequate value to take advantage of incresed egine output. There is a misconception out there that comes from the HD engine design philosohy that we must preseve torque when we hop up a Buell motor. Going fast depends on the total energy deposited to the ground per unit time. That is the reason as you pointed out above that 600 cc in line 4's will outrun a Buell eventhough they produce less peak torque. These bikes end up putting more to the ground by virtue of their increased rate of deposition. One could build a Buell up to make 100 ft. torque and 120 HP. If the driveline is left stock It would seem to me that this bike would not show much improvement over, say my bike (85/100), in a drag race. | ||
| Aaron |
Are you asking how to know what top speed a given horsepower level will support? The way I do it is start with a data point, i.e. "X" hp will support "Y" mph when geared properly (a taller or shorter gear would make it run slower). Once you have this, just apply the cubed relationship of hp required versus speed increase. For example, if on a particular bike, 100hp supports 150mph, and now you make a change and have 120hp ... 120/100 = 1.2 (20% more horsepower) cube root of 1.2 = 1.063 (will support 6.3% more speed) 1.063 * 150 = 159.4mph Gear it to go that speed on the power peak. Conversely, if you have a target speed of say 165, and you want to know the hp needed to get you there ... 165/150 = 1.1 (10% speed increase) 1.1 cubed = 1.331 (requires 33.1% more power) 1.331 * 100 = 133.1hp required "These bikes end up putting more to the ground by virtue of their increased rate of deposition." You hit the nail right on the head! "One could build a Buell up to make 100 ft. torque and 120 HP. If the driveline is left stock It would seem to me that this bike would not show much improvement over, say my bike (85/100), in a drag race." You lost me here. 120hp will accelerate a bunch faster than 100hp. Rate of deposition (love that term) is reflected in horsepower. AW | ||
| Thunderbolt |
JMartz, I couldn't agree more with the statement that there is a misconception about torque numbers when it comes to engines in general, and even more than average in the Buell/HD community. What Aaron stated above is 100% true and indisputable. Your engine's HP & bike's speed, at any given time, and mass of the moving system (you & the bike) determine how fast you are accelerating at any given time (neglecting ugly things like friction, wheel spin, wind drag, etc.). So why do people even bother talking about an engine's maximum torque value? Not really sure, myself. But, I think that it comes from the fact that a relatively high max torque combined with a relatively low max hp tends to mean that the bike is making, relatively, a lot of LOW END hp. This is the case of a typical 1200cc Buell compared with a typical Japanese 600cc in-line four. The Buell has a higher max torque, but less max hp. these factors tend to indicate that the Buell is going to have more LOW END power. Interestingly, this is usually what people describe as a 'torquey' motor. One that makes a lot of low end power. Another misconception, which I happen to think is even more prevalent than the torque/hp thing, is the 'one number' reference to an engine's performance. A 100hp engine. A killer engine with 120rwhp. We all see these references, all the time. Hell, I talk like that too. The fact that an engine makes 100rwhp at 6500rpm only matters when...the eninge is at 6500 rpm. That's it, the only time it matters at all. I personally feel that engine power at 80% of redline is more important than power at redline. The exception to this would be for top speed runs. But for most riding, you are at or near 80% or redline more often than redline itself, I would bet. Why do we (yes, me included) talk like this then? I think it's because there is no convenient method to relay an engine's REAL performance numbers. I think an ET is the closest thing, and that introduces a whole new set of variables (like the rider's ability). As Aaron was talking about, it is the integrated area under the hp/rpm curve which determines rear wheel torque, and, in turn, acceleration. Gearing also plays a big role, because it determines what portion of the power curve you are going to have to use. What speeds you're going to be accelerating to and from make a difference as well. In first gear, every bike has to go through the 'bottom' of the power curve. This is why Buell's compare favorably in stop light racing-0-10mph racing takes an in-line four out of its element (high rpm power), and pulls it closer to a long stroke v-twin's element (low rpm power). Now, talking about your 85/100 bike vs this 100/120 bike...ASSuming the bikes (and riders) are equal in weight (and ability), and ASSuming that the 100/120 bike makes 20% more power over the complete rev range (BIG ASSumption), and ASSuming that we are neglecting wind drag, and using your ASSumption that the gearing would remain the same, the 100/120 bike, with its 20% more powerful engine, will accelerate...break out the calculator...20% faster! ASSuming the 85/100 bike can make it to redline in top gear, the top speed would remain the same for both bikes. OK, fingers hurt. | ||
| Jmartz |
Aaron and T-bolt: Great thoughts... This idea of an enhanced power motor needing adequate gearing comes from my experiences drag racing (friendly side by side acceleration) with guys who have modified their motors especially with larger bore and its subsequent increase in torque. These seem to wind up their rpm range quickly (1st especially) but lack the ability to pull away from me. Granted there is rider technique involved. I try to extract from my motor all it can give w/o concern for its wellbeing allowing it to reach 7500 before shifting and barely letting off the throttle to do so. Since our motor is so rpm limited, in order to reach higher speeds through the gears a higher driveline is required when there is more torque to drive it. This seems to be the trend in Buell with the new destroked and repackaged motor. 7200 is still not going to do it much good especially in light of its smaller displacement and lower torque values. Aaron: That enegy deposition term is from my days in p-chem | ||
| Blake |
Thunderbolt said... "...it is the integrated area under the hp/rpm curve which determines rear wheel torque, and, in turn, acceleration." Integrating the hp versus rpm curve would give you hp x rpm. I have no idea what meaning that would have; it certainly has little to do with the rear wheel or the torque there. The torque reported by a dyno is, like the hp, an engine performance characteristic that is referenced to the rear wheel (accounts for losses between crankshaft and rear tire surface). Integrating the torque/rpm curve might give some form of a power result, but hp would require some serious intermediate conversions to arrive at the desired units of hp (550 ft-lb/sec). Revolutions/min would need to be converted to angular velocity in radians/min; then there's the 550 factor. Acceleration can be estimated from rear wheel torque, speed, and mass. Velocity is change in displacement versus time; acceleration is change in velocity versus time. Didn't we lay all this out at least three times before? Stupid archives.  Blake (I love this stuff) | ||
| Rocketman |
T Bolt: The fact that an engine makes 100rwhp at 6500rpm only matters when.....the engine is at 6500rpm. That's it, the only time it matters at all. If you mean 100rwhp is only achieved at 6500rpm, then obviously that's the only place you're ever going to see 100rwhp but your comment is misleading because it suggests there is no power increase before full rpm therefore making a 100rwhp motor not worthwhile unless you intend to use it flat out all of the time. Horsepower at any rpm is relative to torque. If you increase HP at any point in the rpm range, accordingly torque will increase too. As I've said before, even when torque is falling off, the rpm's will still give you more power providing they are rising faster than torque is falling. When this occurs the rpm's take the place of torque and keep the motor producing more power. This is a clear way to understand why we build a high horsepower motor AND lighten the resiprocating parts accordingly. We're aiming to fill the cylinder as quickly as possible and in the most efficient way, then exhaust it efficiently, in an effort to make the power we want all along the rev range. In Buell terms, I'd say 100 rwhp is a great target for a fast road bike, and take it from me, there's a big difference between stock rwhp and 100rwhp all along the rpm range and not just at the top end. It is a relative relationship which means 100rwhp is fantastic fun all the way to flat out, and in every gear Rocket in England | ||
| Thunderbolt |
OK. Here goes. First, Rocket, I didn't say that getting 100 rwhp out of a Buell wasn't a good thing. But, I have to really disagree with your notion that if you increase power at redline say 20%, then you've bumped it roughly that amount throughout the rev range. Hopefully, if the mods are done correctly with a lot of fore thought and careful consideration, you will have achieved something close to that. But, it certainly doesn't happen automatically. A perfect example of this is changing the cams in a Buell. It's real easy to go to a more aggressive grind and pick up hp, sometimes mucho hp. But, it's a lot more difficult to do that and even maintain low end power. All I'm saying is that "100rwhp" doesn't tell the whole story, not even close. Example: Rocket spends 3 yrs and $5k+ on mods to his Buell, and ends up with a 100rwhp bike. On a side note, this bike has power everywhere, even down low. Meanwhile, T-Bolt dumps a new set of cams (with a super aggresive grind) in his bike, puts a pipe on (tuned for peak, peak, power) and gets the 'same' results: 100rwhp. Impossible to tell from "100rwhp", but T-Bolt's bike hasn't got any balls below 4,000rpm. I really think we need new numbers to talk with. How about two numbers, to start. The first number would be an 'acceleration factor' for first gear (called "AF1"). It would 'somehow' take into account a bike's power through the entire rev range--from idle to the point where you'd shift to second gear for max acceleration (redline on most Buells). The second number would be an acceleration factor for all other gears (called "AF2"). It would take into account a bike's power through only a portion of the rev range. This portion would start at the lowest 'second gear' rpm where you shifted from first, and end the same place AF1's did. AF1 would be probably be relatively high for a Buell when compared to a 600cc Japanese in-line four, but I'm afraid AF2 would not compare favorably to the same bike. Would this be a perfect way to describe a bike's engine performance? No way. BUT, I think it would be MUCH, MUCH better than "100rwhp". Now, Blake, let me clarify what I meant when I stated that whole 'integrated...blah, blah, blah'. What I meant was that total acceleration over a certain speed (in the same gear) is directly proportional to the area under the hp/rpm curve through the corresponding rev range of the engine through said speed range. Now, is that correct? I'm pretty sure it is, but not 100%. I have to think about it a little more. Now, you say that "Acceleration can be estimated from rear wheel torque, speed, and mass." I don't think it can, not by knowing only those things. Rear wheel torque doesn't matter (oh boy, here we go). You want to increase your rear wheel torque real easy like? NO engine mods NO nitrous NO extra wear and tear on the ole pistons...increase the diameter of your rear wheel and change the final drive ratio accordingly. You'll have a bigger wheel turning at a slower speed but with more torque. now, of course the POWER will remain the same, and, likewise, so will acceleration. I would say that acceleration can be estimated from rear wheel POWER, speed (of the bike), and mass. Indeed, a=f/m where m=mass, and f = the tangential force at the tire's contact patch. f, mathematically, = T/r, where T = rear wheel torque and r = the radius of the rear wheel. Now, we can also say that: T=P/w, where P is power at the rear wheel and w is angular velocity of the rear wheel Therefore, f=P/(w*r) And, we can easily see that (w*r)=V, where V is velocity Therefore, f=P/V Replacing (f) with (P/V) in a=f/m, a=P/(V*m) Which, in words, is: "acceleration depends on rear wheel power, velocity (of the bike), and mass" NOT rear wheel torque. Again, we are ignoring those ugly, ugly things such as friction, etc. Now, I'm going to the corner to think about hp/rpm curves and how they relate to acceleration. | ||
| Thunderbolt |
Just a quick PS. you can also see that acceleration could be described as: a = T/(r*m) And, therefore say that acceleration does depend on rear wheel torque. And, when taking into account r (rear wheel radius), it does. But, I was specifically addressing Blake's (T, m, V) statement. Now I GOTTA get to work! | ||
| Sybren |
Hypothetical question: what would it take to make a Buell reliably run at ... lets say... 9000 RPM? Mathematically speaking this would be the easiest way to gain HP. Syb | ||
| Buelliedan |
Sybren, The only way you could run a Buell reliably at 9000rpms would be to eliminate the pushrods and go to a DOHC and gears, belt or chain set-up. Pushrod engines just aren't cut out for sustained rpm blasts. Thats not to say you can't run one at 9,000 rpms, it's just not going to be very reliable. Just watch the in car diagnostics on NASCAR. They always have them peak out right at 8,500 rpms. same stuff here. | ||
| Thunderbolt |
Well, while you don't normally see pushrod engines run at that high an rpm, I agree, I would offer that is only one of the factors which limits a Buell's rpm. The biggest obstacle is the piston speed. The long stroke of the motor also dictates that the piston is moving REAL fast at 6,000 rpm when compared to most other high performance motorcyle engines. Average piston speed, Ps is equal to: Ps=2*S*rpm where S is the engine's stroke. You can see that when you double the stroke, the average piston speed doubles for the same rpm. That said, if you shortened the stroke such that piston speed was at an acceptable level at 9000 rpm, you would certainly have to make changes to the stock valve train as Buelliedan said. The firebolt has a shorter stroke--and, not coincidentally, a higher redline. I thought I read that they did make some changes to the valve train to improve performance/reliability at the higher rpms too. (Although it's still a pushrod engine). | ||
| Jmartz |
bewlydan: NASCAR engines (4 x 3.5) spin up to 9000. They use a hardened crank and rods and titanium valves with Fireblast like springs and collars. Given the nature the Sportster's bottom end I think we can only aspire to an 8000 rpm redline with a similar valve train and lightened reciprocating assembly. Got 15K? I can give you the recipie and sources for what you'll need to put this motor together, or you can go and buy a Ducati and achieve the same results (maybe even better). | ||
| Aaron |
Another big obstacle: breathing. A typical Buell runs out of breath at about 6500rpm, i.e. the torque begins falling faster than the rpm is rising, hence your hp starts dropping. This 6500 number can be pushed upward somewhat with appropriate breathing mods, but all the way to 9000 would be a hell of a challenge. Particularly in an undersquare configuration, which allows less room for valves for a given displacement. Another reason short strokes are better for high rpm. But it's been done, even with a 3-13/16 stroke. A very special set of dual-carb heads. 153rwhp. It costs money, and it's not conducive to the kind of longevity expected from a production bike. You know the old saying, the candle that burns twice as bright burns half as long! AW | ||
| Rocketman |
Thunderlord : My point was, if you target 100rwhp as your tuning goal, and there are many ways to succeed (and fail), a bi-product of this feat would be to see a relative increase in power all along the rest of the rev range. Your comment seemed to imply that by doing this you only gained an advantage by hitting the 100 horses at full RPM. If that were the case, either you intended it that way or the mechanic needs shooting Rocket in England | ||
| Blake |
T-bolt, You are correct; I failed to state my assumption that we knew the radius of the rear wheel/tire. Unfortunately those "ugly, ugly things such as friction" (I assume you are including aerodynamic drag in that description) are what makes the whole subject of analytically predicting acceleration and performance from a dyno plot kinda tricky. At anything approaching highway speeds, it becomes necessary to include the effects of aerodynamic drag in any analytical description of acceleration. Take a look at my post on 08/29/01 and the discussions here and here and here for more fun with math. | ||
| Blake |
What we would really need is the engine power plotted versus velocity (transform gearing and rpm to velocity at the rear wheel) along with the power required to overcome aerodynamic drag also plotted versus velocity. The area between those two curves will give you a value that could be used to compare the acceleration performance of two bikes. | ||
| Tripper |
You guys are boring. Go twist the throttle and relax. Tripper - stuck in Wichita without my bike. | ||
| Jmartz |
Tripper: What is wrong with your bike. I feel your pain, mine was down 3 weeks. Jose | ||
| Thunderbolt |
Rocket, I can't understand where you're coming from. To think that all bikes with 100rwhp would have the same power curve through the entire rev range just doesn't make any sense to me. I guess we'll just have to agree to disagree. Blake, YES, I think HP versus velocity would be good. But, keep in mind that rpm and velocity are directly proportional. Which, is where I draw my HP vs RPM curve thingy being proportional to acceleration. Yes, it does work either way, I think. I really don't see the need in talking about wind drag--yes, it plays a big role in acceleration at h-way speeds, but (1)I would think that all production bikes would have similar enough drag coefficients to make the difference between the various drags of various bike simlar enough at speeds less than 100 mph or so to not influence acceleration on the same order of magnitude as does engine performance...why do I feel like Aaron is about to correct me? and, (2)I am specifically talking about a way to describe engine performance. We already have a GREAT way of talking about a vehicle's aerodynamic performance, and it's even a one number stat, the kind we all like to throw around. | ||
| José_Quiñones |
This is all kind of over my head, but I found something that might contribute to this, from Don Tilley, by Kevin Cameron of Cycle World:
| ||
| Rocketman |
Thundergod : Twice you've put words into my mouth that I never said. I suggest you take up some reading study just incase you misinterperit me again. This time I'll put it down to my Yorkshire accent. Next time I'll sue. Incidentally, my point was and still is, that horsepower is proportional to torque all along the rev range. If you build a power motor to produce 100 rwhp @ 6500 rpm, accordingly the torque will increase too. If you build a torque motor to produce 100 ft\lbs at 5000 rpm, accordingly the horsepower will increase too. What is it you don't understand about that ? Rocket in England | ||
| Blake |
T-bolt: Well let's see for ourselves; I'm not sure. At 100 mph we would be traveling at 100 MI/HR * 5280 FT/MI / 3600 s/HR = 147 FT/s. So, at 100 mph, every 10 LB of aerodynamic drag would require... 10 LB * 147 FT/s / 550 FT-LB/s/HP = 2.67 HP. Assuming that, including an average rider, coefficients of drag vary between 0.4 for an RR/Hayabusa-like machine and 1.0 for a cruiser, and assuming the incident areas are around 7 sq FT for sportbikes and 8 sq FT for cruisers... The aerodynamic drag can be calculated from the following equation: fd=A*Cd*Q where A is the incident area, Cd is the drag coefficient, and Q is the dynamic pressure coefficient for air at standard temperature and pressure (STP). Q is calculated as V2/391 with V being in MPH and the resulting Q units being PSF. The "/391" is a constant derived from properties of air, such as density and kinematic viscosity at STP) So, at 100 MPH and STP we have a dynamic air pressure (Q) of 1002/391= 25.6 PSF. For the cruiser, the drag calculates as... fd=A*Cd*Q = 8FT2*1.0*25.6 LB/FT2 = 205 LB That 205 LB at 100 mph requires about 55 HP; seems reasonable for some of today's big cruisers. For an RR or Busa, the drag will be closer to 71 LB; at 100 mph that only requires 19 HP. Doubt that result? We all know that HP required increases with V3, so the Busa's theoretical HP required to run 200 mph would be (200/100)3*19HP=8*19HP=152 RWHP. Darn near right on the money for an ungoverned Busa running near 200 mph (per the various magazine tests). So based on some crude assumptions for a cruiser's drag coefficient it seems the big cruiser needs an additional 36 HP at 100 mph versus an aerodynamic RR/Busa-like sportbike. Seems like that would affect the acceleration between the two bikes. The late model Buells will fall somewhere in the middle of those two examples with a Cd around 0.7. Still think the difference is negligible? Good stuff eh? Blake | ||
| Rocketman |
Blakey : Get lost Rocket in England | ||
| Sybren |
There is a guy here in Holland who has build a turbine engine into his bike. The turbine came from a starter engine for airplanes. The thing redlines at 60.000 RPM! That's right: 60.000 RPM on 2 wheels!! I'll post a picture later. Buelliiidan, Thunderbooolt, Jmartzzz, Aaaaron, thanks for your comments. I'll take my girlfriends Ducati when I want to rev it up.  Facinating stuff. Syb | ||
| Thunderbolt |
Blake, yes, that's a bigger difference than I thought. I still think it would be worthwile to have AFs which refer only to an engine's power output. Maybe it would be useful to have another set--total acceleration factors, TAFs, which would take into account the bike's mass (probably be a good idea to add the mass of an average rider, too), its engine's AFs, and its wind drag characteristics. It would probably be useful to have several TAFs. One for low speed (0-20mph?), TAF1, one for average back road speeds (20-60mph?), TAF2, and one for highway speeds (60-120mph?), TAF3. Jose, awesome, awesome quote. Where'd you dig that up? That's what I was TRYING to say! The best point made in that statement is that it would be SUPER easy with the data already being in digital form on the dyno. So why don't we see it? All we would need to get this ball rolling is someaarone with a dyno... | ||
| Jimidan |
Moperfserv, snip The torque at the dyno shaft is probably 136 lb-ft. if that torque is the torque at the quoted HP engine RPM, the dyno shaft RPM is 4,400 RPM. The torque at the crankshaft would then be 108 lb-ft., a much more believable value. snip Actually, I talked to Dan @ Hot Shot Motorworks about your theory of loss, and he said his water brake dyno already figures in an 18% loss thru the chains and jackshafts before it drives teh water brake. He is reading actual numbers off his dyno. He is not sure where you are getting the assumption that his shaft is turning only 4,400 rpm as all dynos are different. I wish I understood it better so I could make a more intellegent response. |
