Yeah tell me about it. Mark and Mitch sure are a couple of clowns aren't they? Hahahaa!!!! I kid, I kid.
OK I’m buying it now. Lightening a flywheel has no effect on torque delivered to the wheels. We agreed that it had no effect on torque developed by the engine, but I thought that, while accelerating, the amount of torque available to the wheels would be more. Not so. It is the same. The drive train experiences greater acceleration because its moment of inertia has been lessened. What finally convinced me was pondering a simple mental model of an engine. This engine is a drum with radius R that is free to rotate on a long shaft through its center. It has a rope coiled around its circumference and a mass M with weight W hanging from the rope. T = R x W The torque T provided by this engine is determined solely by the radius of the drum multiplied by the weight of the mass hanging off that radius. You can hang as many flywheels of varying moments of inertia on the shaft as you wish and the torque will not change because the diameter of the drum and the weight of the mass hanging from the rope have not changed. Thus, torque is not altered by altering the moment of inertia of the drive train. All altering it does is change the acceleration of the drivetrain. When you lower the moment of inertia (i), acceleration increases even though torque stays constant. a = T/i You can expand this thought experiment to illustrate the measurement of torque at any point on the shaft. To measure torque, use another drum of the same diameter and hang the same weight from it, but wind the rope in the opposite direction. Now insert the drum at any position along the shaft. If the torque at that location differs from the torque at the engine, then the shaft will begin to rotate. I think you can imagine that it will not. The torques are identical so the shaft will not angularly accelerate. Thanks to those who persisted in trying to convince me. Your confidence kept me pondering. Great education here!
A+ ,, but, you owe me a GeDunk. Yes, the moment of inertia is less, with a lighter flywheel. I mentioned, "polar," moment of inertia, because it is easier to integrate the acceleration component, when talking about Torque, in a set of polar coordinates and not in a set like X,Y,Z Bonus question,,? What do you get,, when you integrate the acceleration component, lets say from 0 to 11 seconds, of the Torque equation.. (a = T/i) ??? Edwardo PhD boot strap U.
Hmmm.. trick question... lots of attention? traffic ticket? 250pi rad/sec (redline on a 348) flywheel angular velocity? 11 sec is too long to stay at 7500 rpm though, so I will assume the operator is shifting through the gears so the answer will not be "engine damage". Assuming petal to the metal and proper shifting.. In a 348 I'm guessing something approaching 100 mph? ?????????????
Yes shifting,, finishing at 1 to 1... final drive ratio,, guess,, 225-50-16" rear tires. What is the answer... in MPH. hint hint,, Enzo would be proud.. give me a beer, I think I got it..
Can I use my life line to call PAP348 to see what the final drive ratio is? If he doesn't answer I'll have to use the Internet...
Well PAP didn't answer so I had to go to the Internet... I found final drive ratio for a 348 TS stated to be 3.53. I will also make an assumption which I know is not possible, but it gives an impressive number. Let’s say a 348 can go from zero to redline in top gear in 11 seconds. Then, using an online tire calculator I find the circumference of a 225-50-16 tire to be 78.06 in. Ignoring the flat spot in the tire let’s say its circumference is 78 inches. Then my final answer is… [(7500 rev/min)(60 min/hr)(78 in/rev)] / [(5,280 ft/mi)(12 in/ft)(3.53)] = 157 mph Hmmm... I think Enzo stated 171 mph? A larger tire would help. Here’s a GeDunk. Does that count for something? Image Unavailable, Please Login
MMMmmmm chocolate. If you hit the,, 157 MPH mark in 1/4 mile that's pretty damn good, in 11 secs. (many fudge factors need to be explained,,many Dyno hours need to be logged,, ,,, and, , , , , many beers need to be consumed.) Enzo would be proud. Edwardo
everything is a trade off, lighten it too much, no torque. Make it heavier, no acceration. Check out your Newtonian math. The equation is a differential. You're acce;erating faster but think you are going faster.It's apples and oranges. Take enough off, and you will stall coming out of the corner but you will get to the corner faster. Try it and you might not like it. Get lucky and you will love it.
Lighter flywheel or any driveline component will equal more power at the tires as you are using less engine power to turn the driveline. (and yes obviously engine output remains the same). You reduce the moment of inertia which in this example is resistance to rotational acceleration. The best example is the 4000lb flywheel. You would use nearly all of the engines power to turn it leaving little left over to turn the tires. In lower gears, where engine acceleration is faster you will lose more power to inertia and then in higher gears where engine acceleration is slower you will lose less to inertia.
We went through this, its not more power, its simply more acceleration due to the reduction of effective mass. To say it is power related is to demonstrate you don't understand what is going on. The vehicle do not loose power, the vehicle has greater effective mass. Changing the gears changes the rotationial inertia, not the power or torque. It only looks like you lost power because of the built in assumptions of the rotating mass dynomometer. If the same vehcle was measured on a static (brake) dynomometer, those difference would not be present. Same car, same day, same atmospherics, same gasoline,...
LOL. Power= work/time. All things remaining equal, quicker acceleration of a fixed mass equals more horsepower. (accomplishing more work in less time) When you spin up a dyno drum quicker you have more measured horsepower at the wheels. To say that quicker acceleration of a fixed mass is not increased horsepower means you do not understand what horsepower is. You must realize that the parts in between the engine and wheels have a parasitic loss and to reduce that loss equals more power transferred to the tires. No one is saying that the engine itself produces more horsepower, only that less is lost on the way to the wheels. And let me ask you in a real life scenario where the engine is running in a static situation? That would be never. You may have something like, 1800rpm/sec 1st gear and 400rpm/sec in 4th gear and that's why reducing the moment of inertia saps less power from the engine therefore more power reaches the tires. When you have a quicker rate of acceleration (in lower gears) you lose more power to accelerating mass and you will net less at the tire. Go ahead and throw a 4000lb flywheel on an engine brake and see how much power you make. Even at a slow 200rpm/sec it would be an amazing loss, if it could even accelerate it that quickly. Lightweight reciprocating assembly does show an increase in horsepower on an engine dyno/water brake. The faster you sweep the engine the more gains it will show. Static HP doesn't really mean anything in the real world unless you have a CVT which has the engine locked at one specific RPM.
As I most times feel I can understand something better than explain it I did some searching online and found this article- http://www.stockcarracing.com/techarticles/scrp_0610_reducing_moi/index.html
hmmm...sounds like the hour 15 I spend twice a day going to and from hour no? This thread should die.....it's much to painful to read.
Throttle position aside, that's the point though isn't it? WOT throttla and (nearly) static conditions occur tfor 500 miles of every NACAR race, Indy, the end of every straight and the end of every drag strip, any boat running WOT. HP is not in any way related to flywheel weight. Acceleration is related to flywheel weight, that's how thw math works.
All of your examples are nowhere near static conditions. Power= work/time Whether that's achieved by better flowing heads or a lighter reciprocating assembly the end result is if you do more work in less time you have increased horsepower. That's how the math works.
Really? WOT with no change in rpm is not static/steady state? Do tell. A flywheel does not recipocate BTW, it rotates which requires no force and consumes no energy once in motion.
Let us return to the concept of the brake dyno. On a brake dyno nothing is accelerating, but plenty of work is being performed. On a brake dyno, the car with large rotating inertia will show the same power (and torque) as the car with the low rotating inertia. Yet (porperly ballisted so as to weight the same) the cars will perform differently. Thus, with the same power (measured) and different acceleration (also measured) the car with the low inertia driveline will accelerate faster with no change in power (or torque). No, this is where you knowledge goes off the cliff. You are performing the same amount of work. You are performing the same amount of work an a vehicle that is effectively lighter. Thus the vehicle accelerates faster because the same amount of power was applied BECAUSE it weighs LESS. {in this sense the weight is both the inertial mass and the rotational inertia} Sure: Engine powering a water pump to irrigate 1600 acres. Runs 5,000 RPMS for 4 hours. Or better yet, a coal fired power plant, ocean going ship motor making 15 knots. No, static HP is what counts more for almost any INDUSTRIAL activity. Power generation, Pumping, Air ventilation,... The only place where accelerative power is paramount is in racing. There is more money spent in ONE coal fired power plant than in all sanctioned forms of auto racing put together. Do you have any idea how many billions of dollars it costs to contructs a single 2GWatt power plant? and how much money it takes to run the plant continuously at 2GWatts? These things are called power plants because power can be converted from one form (rotational) to another (60 cy AC) to another (heat, rotational, pressure). They can be measured in HP if you desire as there are direct conversions from HP to Watts. Indeed, in the metric HP is measured AS Watts. But I digress. Acclerative power is maximized by making things light, lighter things accelerate faster from the same amount of power. It is they dyno operator lieing to the dyno that makes the result look like increased HP when it is an effective decrease in mass of the vehicle being accelerated.
