| Author | Message | ||
     Blake |
Bill: Reducing mass and/or rotational inertias in the wheels or drivetrain will not effect true peak RWHP. It will however improve the overall HP/LB ratio and thus acceleration. It takes power to spin up the wheels and all the rotating parts (engine/drivetrain included), but once spun up, all the true RWHP is then available to combat aerodynamic drag. So, for a 1/4 mile drag racer, yes reduce mass and rotational inertias as much as possible, but for say a LSR bike, concentrate on peak HP and reduced drag/increased traction (a heavy rear wheel is probably and asset). Note: A rotating drum dynomometer does not measure "true" RWHP, which is why we get higher RWHP for dyno runs in higher gears versus lower gears (in higher gears less HP is used to spin up the engine/drivetrain/rear-wheel assy). Blake (notoriousinternetbaitfish) | ||
| Aaron |
Say what? There's a very minor difference in indicated hp depending on the gear you're in. Of course there's a massive difference in torque applied to the drum depending on what gear you're in. But there's a corresponding difference in rpm. Hence the horsepower it shows stays about the same. 5th gear is a direct route through the gearbox, that's why it shows more power. Less loss. I look at drivetrain mass as power storage. The more of the engine's power you divert into storage, the less is available to accelerate the bike. That's why you have less power at the rear wheel. Yes you do. And the dyno does show it. AW | ||
| Aaron |
I've been pondering this some more ... We already know that an inertia dyno just has a sensor to detect each revolution of the drum, I posted a pic of the sensor on the dyno sheets page. It seems to me like every sample of the drum's rotation, an inertia dynamometer can figure out acceleration, by simply recording the speed and comparing it to the speed of the previous rotation. It knows the mass of the drum, too. Therefore it knows how much force was applied to the drum (f=ma). And it's also got the distance it moved, the circumference of the drum. So every sample, it has distance (feet) and it can figure out force (lbs), so voila, it has torque (ft/lbs). The sample also tells it the rpm of the drum. Now it's got all the ingredients for horsepower. (torque * rpm) / 5252 I submit that these relationships will hold true no matter what gear you're in. Accelerating a constant mass from one speed to another will always describe force accurately, no?. The circumference of the drum won't change depending on your gear. Yes, your torque versus your rpm will change, but they change the same amount in opposite directions. So how can you say it takes less horsepower to spin up the drum in a higher gear? I don't see it. The differences in a brake type dyno intrigue me, and I'm not sure I fully understand it, nor do I necessarily buy into the notion that more rotating mass won't hurt top speed. Here's my thinking. The engine has already spun up, the mass has been accelerated. A load is applied to hold the rpm constant and a force measurement is made directly, either by measuring eddy currents or using a strain gauge. Now you have force, distance (I would assume the moment arm), and rpm, all the ingredients. The mass in the drivetrain is having no effect, because it doesn't have to be accelerated, you're measuring the force directly. But in the real world, getting from one speed to another *does* require acceleration, and mass will affect it! I don't know, it seems like the inertia measurement is more representative of the real world. Now consider this. Aerodynamic drag goes up with the square of the speed increase, right? But horsepower required goes up with the CUBE of the speed increase. Why the difference? Because the mass has to be accelerated? Hmm ... AW | ||
| Reepicheep |
Well... as someone smart enough to understand most higher order math concepts, but dumb enough to generally miss something critical on the derivation... let me make a suggestion. Add half pound of water to rear tire and do dyno run. Repeat several times. Perform curve fit. It would be interesting to see theoretical quarter mile times as well (based on dyno), and see how they degrade versus rotational mass increase, then go back and calculate what horsepower change would be required to achieve the same quarter mile times with the extra pound or so of rotational mass. That is the figure I am trying to get at... I don't really care much about top speed, and I understand you don't really change peak power, though you would significantly change the area under the power curve, which is really the ultimate desire of anyone not trying to set landspeed records. Aaron... you got me. I'm stumped. Where does the extra multiplier come from to make it a from a square relationship to a cubed relationship. I can think of lots of linear losses, but nothing to increase it exponentially. So long as F=MA, then the acceleration of the mass still ought to be constant multiplier, not an exponential factor. | ||
| Chuck |
So . . . does all this mean, that if I steal Rocket's carbon fiber wheels, and fill the tires with helium, my bike will show a bigger horsepower number on Aaron's dyno? | ||
| Aaron |
Bill: good point. I want to understand the cube thing, too. Chuck: yes, I think that's true. And I wouldn't be surprised if the Firebolt shows us less difference between crank and rwhp than previous Buells. We'll see. AW | ||
| Hootowl |
I believe it has to do with the viscosity of the air. As speed increases, it gets harder and harder to push the bike through it. Think of the sound barrier. The air can't get out of its own way. Speed affects the effective density of the air. Mix corn starch and water. You can stir it with your finger easily enough, but smack it with your fist, and it doesn't move. (Try this, your kids will get a kick out of it.) | ||
| Vr1203 |
Hey, enough of the theories  I need turbo help. Anyone around Eastern IA give me a buzz. | ||
| Aaron |
Yes, hootowl, but why the difference between aerodynamic drag (squared law) and horsepower required (cube law)? | ||
| Hootowl |
Just a guess, but friction HP increases rapidly with speed, so you would need more brake HP to overcome this. I don't believe this is a second order equation, but rather a third. But it's just a guess. I read the answer to this in a motor mag a few years ago, but I can't remember what it was specifically. Friction HP had something to do with it though as well as the air viscosity. | ||
| Hootowl |
Interesting quote... Awhile back, a gentleman by the name of Watt (the same gent who did all that neat stuff with steam engines) made some observations, and concluded that the average horse of the time could lift a 550 pound weight one foot in one second, thereby performing work at the rate of 550 foot pounds per second, or 33,000 foot pounds per minute, for an eight hour shift, more or less. He then published those observations, and stated that 33,000 foot pounds per minute of work was equivalent to the power of one horse, or, one horsepower. | ||
| Al_Lighton |
Aaron, if the extra velocity term (cube vs square) was due to the F=mA relationship, it would imply that if you had a JATO rocket strapped to the RR to get it up to a higher speed, it would stay there. Is that the case? I don't think so. From Marks mech engineering handbook: "Aeodynamic drag FORCE is a function of a car's shape, size, and speed. Air resistance varies closely with the square of the car speed" "Drag force is defined by the equation D=C'd*(1/2 p*V^2)*A " (A = area,V=velocity,p=density,C'd=drag coeeficient) "The AERODYNAMIC portion of the road load power increases as a function of the cube of the car speed. The mech portion increases at a slower rate and , from about 25-60 MH is almost a direct function of car weight" The first was a description of force, the second a description of power. Power is energy over time, energy is force over distance, and velocity = distance over time. So if D/T=V F*D= E, and E/T=P, then (d/t)^2 (aero FORCE) * D (distance) = Energy All the above divided by time = Power, so the D/T is essentially the third velocity term in the equation. Stumped me for a bit.... Al | ||
| Hootowl |
Been out scouring. This is the closest answer I can find. It's about fans, not bikes, but it still has to do with pushing objects through the air be it bikes or fan blades. -quote- A fan is actually an air pump. As such, the theoretical performance of a fan follows certain basic laws of physics: Airflow rate (Q) varies directly with fan blade speed (rpm). For example, if fan speed is doubled, the amount of air moved will also double. If fan speed is reduced by one-half, the amount of air moved is also reduced by one-half. This is how airflow rate is changed with variable speed fans. Static pressure (s.p.) capability varies as the square of fan speed (rpm2). This means that if fan speed is doubled, the pressure or suction that the fan can develop or the resistance against which the fan can operate is four times greater (22 = 2 x 2 = 4). This is one reason many grain drying fans operate at high speeds. Conversely, if the speed of a fan is reduced by one-half, the static pressure capability of the fan is reduced to one-fourth of its original value (1/2 x 1/2 = 1/4). This relationship illustrates a problem that can occur, particularly with variable speed exhaust fans. Wind blowing at a fan creates additional static pressure that the fan must work against. If the fan is operating at less than full speed, static pressure capabilities are often not sufficient to overcome this increased resistance, and airflow through the system is greatly reduced and/or becomes very erratic. With strong winds, air can flow backwards through the fan, even though the blades continue to rotate. Power (hp) required varies as the cube of fan speed (rpm3). This means the power required to operate a fan is a function of the speed cubed or to the "third power." In practical terms, if the speed of a fan is doubled, the power required to operate the fan will be eight times greater (2^3= 2 x 2 x 2 = 8). Conversely, if fan speed is reduced by one-half, the power will be reduced to one-eighth of the original value (1/2 x 1/2 x 1/2 = 1/8). The relationship between speed and power is frequently overlooked, and is a common reason for motor failure when pulleys are changed and a different size pulley is installed. A smaller pulley on the fan or a larger pulley on the motor will cause the fan blades to turn faster which greatly increases power requirements, often resulting in motor overload. | ||
| Al_Lighton |
minor error..in my text I said energy = force OVER distance....meant force TIMES distance. Wrote it right in the equation, though. Ft*lbf is an energy term, 1 ft*lbf = 1.3558 joules =.001285 BTU. Torque is expressed in ft*lbf, so it is therefore an energy term. And torque over time is power. | ||
| Blake |
Aaron: Concerning reported dyno RWHP versus gear run; ponder this... How much power does it take to spin up the engine to redline rpm? Does it depend on the time required to spin up the parts? Since it takes more time to reach redline in a higher gear, is this "engine spin up power loss" then different for each gear? The dyno drum itself isn't causing the differnece it's the engine inertia I was referring to. Think what would happen if you slapped a 24" diameter 1000 lb flywheel onto the crankshaft, the rear wheel would see very slow acceleration and thus the inertail dyno would report very low HP. A brake dyno though would still show actual RWHP since it negates the effects of engine and drivetrain inertial power loss (energy storage as you said). Running an inertial dyno in a higher gear is getting you closer to the brake dyno, closer to a constant rpm measurement that is not losing power to the spin up of engine and drivetrain parts . You are correct, the inertial dyno will only see a small reported power difference for different gear runs, but there is a real difference, and as you also point out the differences are not solely due to inertial losses. As to the cubed power versus velocity thing, recall the basic equations... Power(P)=Work(W)/time(t) and W=Force(F)*Distance(d) (for the linear motion case) We can then also say equivalently that P=F*d/t and since Velocity(V)=d/t we find that P=F*V. For aerodynamic drag... FaeroaV2 or simply Faero=cV2 where "c" is a constant depending on incident area, ambient conditions (air density), and drag coefficient. Substituting cV2 for F in the preceding power equation we get... Paero=cV2*V = cV3 and there you have it. Power to overcome aerodynamic drag is proportional to velocity cubed. Ain't it beautiful! Or alternately substitute for F in the work equation and we get... Waero=cV2*d. We can write d in terms of V as d=V*t or simply d=Vt. Substitute "Vt" for "d" in the Work equation and we get Waero=cV2*Vt = cV3t The power equation is then Paero=cV3t/t = cV3 the same result as above, and thus we have irrefutably proven the proportionality of velocity cubed to the power required to overcome aerodynamic drag. FYI, The rotational work equation is... Work(W)=Torque(T)*Angular Displacement(q) where T=I*a where I is the mass moment of inertia about the axis of interest, and a is the rotational acceleration. All a dyno actually measures directly is rotational displacement (q) and time (t). Knowing the geometry and inertia of the drum, the rest is simple math. Fun stuff or what?  Blake | ||
| Blake |
Al: Clarification to your last statements concerning torque... In a rotational system... P=Tq/t and as with work (W) in my post above, energy (e) is also equal to T*q Those pesky radians (basic unit of rotational displacemnt) are pseudo-nondimensional. As far as kinematics are concerned, without displacement there is no work; without work; no power. | ||
| Tripper |
Sick stuff or what? | ||
| Axtell |
Ring--whaaaaaaaaaassss up? answer---nutton--hanging with my buds talking radians true true Tripper---this is sick | ||
| Aaron |
Damn, I was wondering when I'd get a reaction out of you. The answer to your question is no. Because in a higher gear, you have less Force applied (which is why it accelerates slower) for a given distance, but more rpm. In other words, torque goes down and rpm goes up. And they change the same amount. Power is torque times rpm, therefore it stays the same. All you've changed is the mechanical advantage of the motor. I understand the math that explains the cube thing, that's simple enough, there's already a V term in power and you have two more with the definition of force with respect to aerodynamic drag. But I have a hard time with it conceptually. It's definitely true, though, I have several data points that support it. AW | ||
| Al_Lighton |
Ya know blake, I was thinking 'bout that.. "As far as kinematics are concerned, without displacement there is no work; without work; no power." So if I'm standing there pushing as hard as I can against a wall four feet off the ground that isn't moving, there is no displacement, therefore, Energy = 0 = force * displacement (displacement =0). But I'm standing there sweating up a storm, converting lots of chemical energy into heat, doing no "work" (in the kinematic sense). If I do it long enough, I'll get hungry for more calories. Since energy is neither created nor destroyed, how can this be resolved? Well, within my body system, I'm pushing with my hand against the wall, and reacting that force thru all kinds of moment arms in my body, creating some torqe in my arms, torso, legs, etc, ultimately being reacted against the ground. Nothing is moving, but major energy is being expended, or at least it sure feels like it. I'm not converting to kinetic energy, nothing is moving. I'm not converting to potential energy, nothing is getting higher and the wall isn't deflecting significantly and storing any energy like a spring. Basically, I'm just making heat. Units is units, right? Lb-ft is lb-ft, which has to equal energy, right? Doesn't matter if me the human or a robot does this, the harder the push, the more energy it takes, right? Or not. Please 'splain, oh wise equation endowed one. I used to be an engineer. Then I learned to spell, so they had to make make me a manager :-) Al | ||
| Road_Thing |
Stop! Please! My ears are bleeding! r_t | ||
| Tripper |
r_t - There's no stopping him. Once he gets wound up it is best to just step aside and let him go. Al might have just tied him in a knot with that last one so he will explode like a super nova. Watch for a bright light from the south. Watching this has aleviated my depression from not being on the road to homecoming today. You go Big Blake! | ||
| Hans |
Paero=cV2*V = cV3 ????? Think not. One of thos V `s is already souped up in the equitation of the drag. Just squared sound me more then enough.  | ||
| Chuck |
you guys are nerds  | ||
| Reepicheep |
Darnnit Blake... you took the wrong bait. Thanks for the derivation though. Now back to my original question (which will be pretty knotty). What is the perceived horsepower gain for loosing rotational mass. For example (purely hypothetical ), if some visionary company was to produce a spiffy new bike with (hypothetically) three to five pounds less weight spinning in the wheels, how would that translate in terms of acceleration (say zero to 100 MPH). Would it feel like a 10 HP stronger motor? 20 hp? It will be an area under the curve thing... and a pretty tangled derivation I suspect. I am not sure you could even come up with a clean equation, as the location of the mass away from the center of rotation would likely be a major factor. The more I think about it, the more significant I think it may be. Think about how big an engine it would take to spin a 5 pound wheel up to 100 miles per hour in 7 seconds. I don't think my 5 HP lawnmower motor could not do it. I know my 125 HP motor saturn would. I'm guessing around 10 horsepower... which makes the Firebolt look significantly better, even in stock trim. And this is free out of the box, anywhere in the powerband, and does not tap into the significant latent potential lurking in that engine. And interestingly, it would be a significant factor that would not show up on (some) dynos. Makes me want to start ignoring power output figures and start looking only at zero to 100 times if I want to get a real feel for how "fast" a bike is... Bill | ||
| Aaron |
Yeah, it may be nerdy but I'll tell you what, you cannot believe how helpful it was to the LSR effort to understand these relationships between power and speed. It meant we had our •••• together when we rolled it off the trailer. This info was used in tuning the engine for the proper powerband (we would've gone slower with a total focus on top end power) and of course for choosing the gearing. Nerds can be pretty useful sometimes! AW | ||
| Aaron |
Bill, I submit that lightening rotating weight is not just a perceived horsepower gain, it's an actual one. The motor isn't making more power, but more of it is getting to the rear wheel, because less of it is being diverted into storage, which is what rotating weight does. An inertia dyno will show it, and inertia dynos are the common type that most shops have, just because brake dynos are a whole bunch more expensive. Seems to me like for looking at acceleration, the inertia dyno does a good job. For looking at steady state conditions, though, no question, a brake dyno is the only way to do it. A brake dyno can do sweep tests, too, which should show the effect of mass. The radius of the extra rotating weight will play a big role in it's effect on horsepower ... a squared function. Blake laid out those calcs once, sometime back, very straightforward. Once you know the mass, seems like calculating it's effect on hp should be easy ... I mean, we already laid out the calculations for an inertia dyno. Just add (or subtract) the new mass into the front end and recalculate the horsepower. I think. AW | ||
| Al_Lighton |
Bill, I wouldn't touch that problem with a 10 ft pole, but Blake is probably salivating at the thought... Go Blake! However, I don't think it's sufficient to ask "what's the effect of removing 5 lbs of rotating mass"....I think you have to specify WHERE the rotating mass is. If it's on the hub, the answer is different than if it's on the rim. But Blake will tell us. Or not. Al | ||
| Court |
>>>Nerds can be pretty useful sometimes! No sh** Sherlock! As the manager of one of the world's LARGEST COLLECTIONS OF NERDS, Lord knows I can attest to that! I have nerd Doctors, Engineers, Choreographers, Radiator technicians, EMT's, Grocer's, a lamed in the line of work cop and the entire mix stirred by a "Video Artiiiiist". That's the understatement of the year. Court | ||
| Vr1203 |
PLEASE!!! Help .... I have sent for a Mikuni flat slide to replace the CV carb. Is that going to help the turbo VR? |
