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Billy Golightly
05-25-2007, 10:44 AM
I'm not a physics guru by any means, and this question has had me wondering for a while now. Some crankshaft companies like Falicon, will actually add weight to a crankshaft because they claim it increases torque and allows you to run a larger tire for drag racing. You can also of course remove weight, which probably makes less wear and tear on bearings, and give you less low RPM torque, but what else I'm not exactly sure. Is there some sort of a forumla to calculate, if your vehicle/static weight is xx you need xx amount of engine inertia weight to carry it?

Theres also another factor, that on something that is rotating, the larger in diameter it is the more leverage it has on whatever is turning it. For example, a 3lb flywheel 8 inches in diameter will be much much harder for the engine to turn then one that is 3lbs and 3inches in diameter. Or at least thats what I've learned so far.

Could someone enlighten me?

ccdhowell
05-25-2007, 02:38 PM
Don't have much to add except this. A few weeks ago I got out the mower to do the seasonal maintenance. I like to drain the oil while it is hot, but I had already taken the blade off. Well, I thought I'd just crank it and let it idle for a while to warm the oil a bit. I pulled and pulled and the little Briggs & Staton 6hp engine wouldn't even try to crank. I sharpened the blade and put it back on and it cranked on the first pull. I was surprised to say the least. I figured that it needed the extra weight to help the crankshaft turn through enough revolutions to fire the mower. I know their are different engines, but the idea would be the same. A heavier rotationg mass would tend to continue to rotate and be more difficult to stop, whereas a lighter rotating mass would be easier. Although a heavier mass would take more energy to start it rotating. The question I have is, once rotating at say the same speed, what is the difference in energy required to keep the heavier mass moving? Or put another way, are you using the energy gained from the heavier mass up in keeping it moving? Where is the tip point? Too much weight added would certainly take too much energy away from the system and would be a burdon.

My 2 cents, Chris

Billy Golightly
05-25-2007, 04:57 PM
Thats pretty much what I would like to know too, where the line is. I know, somewhere, there is a formula for flywheel weight to full machine weight with the rider but I don't have a clue what it is or how to calculate it. Finding that would be a good place to start I bet.

MyMistress86R
05-27-2007, 09:36 AM
Not sure if this will help your cause any or not Billy, but it a tad bit of info I found in regards to an explanation of your "example".

The info is based on a car platform, but the principle is sound for any motorized vehicle..."torque is proportional to force AND the distance from the epicenter of the rotating mass. 250 lbs at 1 ft gives you 250 ft-lbs. 250 lbs at 2 ft gives you 500 ft-lbs. Increasing either gives you more torque."

Ask.com and it's "algorithm" is a great search tool for issues like this.

sandpuppi101
05-30-2007, 09:58 AM
Something similar also to enlighten the subject,we have the old John Deere 2 cylinder tractors,we have them set up for tractor pulls,stroked,bored ,shaved cylinders etc.Anyways as you know some of the tractors had to be hand cranked via the flywheel ,which is external ,you grabbed it and spun it by hand,Ours have the eletric start,on one particular tractor we had an issue with starter drag,it would get the starter hot to simply start the tractor.Long story short ,we rebuilt the motor ,and put on a substancialy larger flywheel ,after doing so we had no more starter issues and had a gain in horsepower also,it makes sense,it creates more energy to get it rolling but once that energy is put into motion ,it takes alot to slow it down ,creating high torque.

Chevy200s
05-30-2007, 10:40 AM
I'm not sure if this helps or not, but torque is just the cross product of a force and the distance from that force to the fulcrum or point at which the system turns. Force is the mass of something multiplied by its acceleration. So the mass or indirectly the weight of the object directly effects the torque as well as the distance from the point turning. So the weight of an engine crank along with its radius if I've got this right.

Here's a website that gives a pretty good idea of the basics behind it
http://www.comfsm.fm/~dleeling//physics/torque.html

Theres a lot of other factors that go into this when determining the torque produced by a motor or any other "real" system, such as the magnitude of the force in the x, y, and z direction, force of friction would also have to be taken into account. But I'm not a physics major, just took a few classes for fun. Interesting stuff. Sorry if this doesnt help or wasnt what you were asking

NOS_350X
05-30-2007, 11:37 PM
Less rotational mass in an engine the faster reving it will be, less torque and easyer to stall. More is going to have more low rpm grunt. But becomes a turd when wanting the high revs fast.

pickleweasel_00
05-31-2007, 04:00 PM
I've always wondered about the heavier flywheels for cars, and how much it would affect "revving up" the engine? I guess it would make sense because if you have a heavy flywheel vs. a lightweight one, imagine taking off from a dead stop. The heavy one will have more force behind it and the lighter one won't have as much thereby probably bogging down a bit as the clutch is applied. Not sure if that's how it works, but it is an interesting idea. I always thought that a heavier flywheel woulg give slower acceleration though? now i'm confusing myself :wondering

Billy Golightly
05-31-2007, 08:17 PM
My question though is, at what point, does the additional weight help you, or hurt you? Its not secret that a lot of drag racers like to use stock ignitions because they feel the lightened flywheels of rotor type ignitions don't give you as much wheel turning torque off the line. My question is, for example, if I'm currently turning a 20inch tire to fast, and want to upgrade to a larger diameter but the engine is not capable of turning it as easy, is it a viable solution to add rotating weight/mass to gain additional inertia to turn the bigger tires? Or, would it be better if I lightened the existing flywheel and crankshaft to gain more RPM and high engine speeds?

cr480r
05-31-2007, 09:08 PM
I dont think the added weight would help in a drag situation besides maybe adding traction... the weight doesnt create any torque output... only increasing the cylinder pressure and/or stroke would... The heavier flywheel just increases momentary inertia... helping the initial takeoff only... after the drivetrain catches up with the engine the added weight would just be a burden to the engines acceleration IMO...

cr480r
05-31-2007, 09:13 PM
if I'm currently turning a 20inch tire to fast, and want to upgrade to a larger diameter but the engine is not capable of turning it as easy, is it a viable solution to add rotating weight/mass to gain additional inertia to turn the bigger tires?

Why are the 20's spinning to fast? Not enough contact patch? or too short of gearing?

Billy Golightly
05-31-2007, 09:23 PM
For intent and purposes of the overall question :) they are spinning because theres not enough contact patch with the ground.

Daddio
05-31-2007, 09:50 PM
Could it be as simple as changing air pressure?
Ron

jenndnn3
06-04-2007, 07:56 PM
Billy some info for you to check into....

http://en.wikipedia.org/wiki/Flywheel_energy_storage
http://en.wikipedia.org/wiki/Kinetic_energy#Kinetic_energy_of_rigid_bodies

All have some fun links that you can really get confused on....

I did a question this along time ago adding weight to flywheel it should be somewhere on the boards...

Sargon2112
08-06-2007, 07:17 PM
Here's another explanation that I hope might help. Think of torque as momentum. Compare pushing your trike to pushing a car... the trike is much easier to get moving, and easier to stop as well compared to the car because the trike is obviously lighter (less mass). But coast them both down a long hill, and see which one will coast farthest after it reaches level ground at the bottom of the hill... the car will because it has more momentum due to it's mass being so huge compared to the trike. Heavy vs. light on the flywheel is the same. The lighter one will rev quicker (engine rpms speed up easily), but since it is "lighter", it will also be easier to affect the rpms when a load is placed on it (stalls easier because there is less momentum keeping the engine turning). So you can see how a heavy flywheel could help in a drag, it might rev slower, but once it's tached up, dropping the clutch has less of an effect on the rpms because the heavy flywheel's torque (or momentum) keeps the engine turning. Torque = angular (or rotational) momentum. It boils down to this: how much force is needed to change an object's state of motion, whether the coasting car or the rotating mass of the flywheel. If it doesn't take much to speed something up (rpms), it's generally not going to take much to slow it down either. If it's hard to start moving, it's likely going to be hard to stop it.

I took a stab at it! Hope I helped!

David

Pistonhead
08-12-2007, 11:47 AM
I have an explanation based upon the physics concept of impulse:

force(f) x change in time = change in momentum (mass x change in velocity)

You would have more torque from the engine because, since f = (m x change in velocity) / change in time;

you have a larger numerator product with more mass on the crankshaft (more mass, presumably for drag racing a big change in velocity) divided by (if your machine is quick) a very small change in time.

Because with more weight you have big change in momentum divided by small change in time; you get more force, translating into more torque to the back wheels.


This explanation also shows why you want an elastic bungee cord for bungee jumping, to reduce the change in time of your fall such that the force exerted on your legs isnt enough to tear your feet from them.

Billy Golightly
08-12-2007, 01:24 PM
I don't understand the formula exactly, but I think I get what its saying is that the torque generated by the additional mass/weight easily overrides the loss in "time" or in our application "rpm" ?

Pistonhead
08-12-2007, 03:02 PM
I don't understand the formula exactly, but I think I get what its saying is that the torque generated by the additional mass/weight easily overrides the loss in "time" or in our application "rpm" ?

Yes and no, actually a loss in time is what you would be going for (less time to get the crank going at high rpm's) to make the big torque but that depends more on the power produced by the engine you are using. Where the additional mass comes in is in the product of the mass and change in velocity. If you are adding more mass to the crankshaft without sacrificing (to a certain degree) the rate at which the rpm's increase in a given time (ie, you still have a decent increase in velocity of the crankshaft in that given time) you will then get the big torque.

What it is saying is that the increased mass may slow the crankshaft to a certain extent, but it overrides this loss because you then gain more force (torque) to the rear wheels.

Rustytinhorn
01-26-2008, 07:05 AM
Some crankshaft companies like Falicon, will actually add weight to a crankshaft because they claim it increases torque and allows you to run a larger tire for drag racing.

Comparing lighter vs heavy flywheel on 4 stokes:
I think that with the heavier flywheel, once it is up to speed it takes more force to slow it down than a lighter flywheel. This way you can "pop" the clutch with big tires, instead of bogging down the engine or feathering the clutch. Also with big tires, the vehicle is going to want to slow down after every compression stroke. With the heavier flywheel, it will "push" the tires, valves, camshaft, and the engine through the other strokes until the engine fires again.
For two stokes:
I think the reason 2 strokes produce less torque is because resistance is met on every stoke, compared to a 4 stoke, where in all reality, the only hard resistance met is on the compression stroke.
Its like spinning a bicycle tire by hand, and then spinning a truck tire by hand. It takes a lot longer to get the truck tire up to speed, but once you do its really hard to stop. I think thats why the add weight to the flywheel in the tractor-pulls. As long as they get their engine revved up before the light, they can use that weight to pull them through. As for a formula for it, even though the 2 stoke flywheel may weigh 3 pounds, and a 4 stroke 6 pounds, I would think the 3 lb flywheel would be spinning fast enough to be equivilant to the slower spinning 6lb flywheel. My own numbers just for an example would be say like a 2 stroke spinning a 3lb weight at 50rpm wich equals 150. Or for a 4 stroke spinning a 6lb weight at 25rpm equals 150 also.
So you would have to spin the bicycle tire say about twice as fast as the truck tire to make it so it would take the same amount of force to stop each tire.
Just my .02.

Anyway, I didn't intend to answer anything, but more or less had a question myself.
How big of a difference does getting rid of rolling mass really make on a 250r rear axle? I know it will help improve acceleration since we just discussed flywheel vs tire weights on a 2 stroke. I did some weighing on my R the other day, and I can drop about 24lbs of rotating mass off my rear axle.
Thanks.

Rustytinhorn
01-26-2008, 07:21 AM
I always thought that a heavier flywheel woulg give slower acceleration though? now i'm confusing myself :wondering

It probably would give slower acceleration, except if you already had your engine revved all the way up when you poped the clutch, then your engine would continue to run at that same high rpm, and you wouldn't have to worry about accellerating cuz your already still it full bore. No bogging during take-off. I guess that is until you have to shift:rolleyes:

sandman829
02-21-2008, 06:18 PM
In stockcar raceing a litter rotating mass gave a faster throtle responce, but let the engine wind down faster. What I was told is that the flywheel, wheel/tire pacage, and now power producing parts needed to be as lite as posable. The more it takes to turn it the more it takes to stop it. I will talk to my engine builder to see if he has a formula for what you want.

Billy Golightly
02-21-2008, 08:45 PM
That'd be great, I'd love to hear if a formula for optimal rotating weight is around or not. Thanks!

sandman829
02-22-2008, 01:28 PM
Bad news, I have talked to two differante engine builders and they both say it is a matter of knowing the parts guys and what works together. Cameron said years of trial and aer. As a rule of thum he trys to not go heavier the 1/2. For every puond of crank weight no more than 1/2 pound added. Prefers light weight rods and piston. Hope that hel.ps

The Goat
02-22-2008, 02:18 PM
damn, i thought i had typed a response to this billy, my comp must have frozen and not sent it.

seems like you would have to factor in A LOT of things to get it right.

suspended weight of the trike, unsprung weight on the trike, rider weight, the amount of traction you have, and...i know I'm forgetting something.

each variable would change proportionately/exponentially with the others.

now it's been a few years since i've taken a physics or calculas class...but buddy, that would be one HELL of a formula to figure out. and at some point you'd need to test the trike on a skid pad to see what kind of G's you're pulling. :crazy:

Now I realize that was likely way more variables than you wanted...but that's what I came up with. :w00t:

TeamGeek6
02-22-2008, 04:31 PM
There is a maximum between "too light" and "too heavy" Its called "just right." Its already there because thats where the engineers that designed it put it.

This gets into UGLY math (calculus) and is mostly thrown out the window when cylinder conditions are brought in. Yes, a heavier crank mass (within reason) will absorb more power to then transfer it to the transmission. A lighter crank will turn faster and have a higher dT.
But if too much mass is present and it cannot absorb all the cylinder energy, then that extra weight is a waste of time and bearing life.

Youd think a lighter crank is easier on bearings, until the minimum bearing load is exceeded and vibrations start in, which wipe the bearings out. It has to be "just right"

These are "dT", not delta-T. Delta T only applies to linear situations, and this one isnt.

There is no such thing as an "action without a reaction" and changing the crank has an effect on cylinder conditions - how the fuel burns. Its all related. A much heavier or lighter crank also requires a heavier or lighter piston to balance, and the losses on the upstroke are important, especially in a 4 stroke.

The big problem is that not even the automotive engineers know exactly what the cylinder is doing, and never have. The chemistry inside is far too violent and unpredictable. So, they reduce it to a "constant pressure" system (which it isnt) and take the pressure from there, at the degree angles the piston, rod and crank are going through, and convert the cylinder pressure to mechanical force. Its fairly easy from there.

Untiil you get to the thing about the mower that wont start without a blade. Then the whole thing goes out the window. It defies logic that it wouldnt start, but it doesnt.

To actualy calculate this is a chain of ugly calculations, based on dP /dT(true change in pressure). Integral calculus gives total cylinder pressure, then thats sent to the piston mass, then to the crank. This is not practical since the cylinder conditions are so difficult to measure, and then they change with the weather.

I can do the math, but I have better things to do with 2 months of my life.

The best thing is to leave the crank weight alone and work on optimizing everything else, especially ignition.

The power output of my 250R is absolutely brutal, and its a stock bore and stroke. Pipe, reeds and radically modified fuel and ignition system and the power is unusable. The crank is totally unmodified.

Ignition is where the power is, not messing with the engine internals.

Think of the 20 degree span of crank rotation that power has to be made in. The max is from about 10 - 30 degrees ATDC. If the burn is too slow, pressure doesnt turn the crank. Speed the burn up to get the pressure in that degree span and power goes up. Speed it up too much (detontaion) and parts start breaking.

If you want to talk about dP, which is more pressure per time, increase the ignition rate and watch what happens. The torque goes insane.

Heres an example:

http://gl1200harness.tripod.com/gl1200_photo_album.html

MONTHS of massaging carbs, ignition and a little grinding in the cylinder domes to remove ridges, and the back tire went bald in 5000 miles. There were no hi performance parts available for this machine. Its all about a very hot ignition and the proper AFR.

RubberSalt
07-19-2016, 11:52 AM
Crank/flywheel weights are an interesting subject. I to would like to know where to focus on the math. Maybe I'll look further into this some day. From my readings over the years, Heavy cranks, like everyone is saying, does take some time to rev. But the torque produced is absurd. It isn't just when you start going, it holds through when changing gears.

If your on the pipe at 850 0rpm, you shift and now your at 7k RPM, below the pipe, that sucks. The engines RPM drive due to the load. If your at 8500 RPM and you shift and your at 7750 RPM. WOO! hold on. The inertia of the weights in action!

Think about 2 different flywheels. Both weigh in at 2lb. One has a 5in diameter and 75% of the mass is in the outside 1/2 of it. The next is 2in diameter and has 75% of its mass in the outter 1/2 of it (long skinny? lol). Both are spinning at 8500 RPM. The bigger one will have more energy. The inertia on the outside of the flywheel will create much much more torque that the skinny one of the same mass. How much? I'm not sure yet. I'll try to find some math.

I know my dads 64 1/2 Plymouth satellite runs an 80lb flywheel! 512 cubes in that monster.

RUNMEDOWN
07-19-2016, 12:20 PM
Plug in some #s and you are set (Use the link)
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

RamsesRibb
07-22-2016, 07:54 PM
[QUOTE=ccdhowell;498322]"Don't have much to add except this. A few weeks ago I got out the mower to do the seasonal maintenance. I like to drain the oil while it is hot, but I had already taken the blade off. Well, I thought I'd just crank it and let it idle for a while to warm the oil a bit. I pulled and pulled and the little Briggs & Staton 6hp engine wouldn't even try to crank. I sharpened the blade and put it back on and it cranked on the first pull. I was surprised to say the least. I figured that it needed the extra weight to help the crankshaft turn through enough revolutions to fire the mower."


Lawn mowers do not have a flywheel. The blade "is" the functionality of a flywheel. I have seen them run with no blade and believe me it is out of whack.

(Wow this thread is really old!!!)

xrider
07-25-2016, 12:49 PM
Your dipping your toe into a very big pool here. What you quoted (F=ma) is Newton's second law. In order to relate Newton's second law to torque you need to know the length of the lever arm. This is because torque is a product of force. Therefore to calculate the torque you need to take F x L, where L is the length of the lever arm and F is the force as calculated from Newton's second law. This properly factors in the rotational analogue associated with the law.

Also for clarification, L (lever arm) is the distance from the center of rotation to the force vector. This will have both an x and y component as the force is applied in different directions (there is no z component for obvious reasons). This further complicates the equation to Torque = yFx - xFy. It is a concept that seems simple on the outside until you begin factoring all of the variables and angles that effect the situation. If you really want to get into how numbers can be theoretically calculated for an Engine, I can point you in the right direction, but it would be difficult to explain correctly in a post. Note: My undergraduate degree is in Mechanical Engineering.

yellowoctupus
12-22-2016, 07:24 PM
To further clarify (if such clarity can be brought to this somewhat muddied up post):

Adding mass will not change horsepower or torque ratings of an engine whatsoever.

The only thing that adding rotational inertia to a mechanical system does is change the rate of engine speed change. This can be viewed in two different manners, as Macro or Micro.

Micro would be if you are viewing the engine speed as it completes a single (or two in the case of a four stroke) engine revolution. As the system's inertia approaches infinity (heavy flywheel) the engine speed would remain constant throughout the cycle. Conversely, as the system's inertia approaches 0, the engine speed would vary considerably, decelerating during compression and accelerating during the power stroke.

Macro is more along the lines of the 'stalling off the line' or 'fast revving engine' comments from previous posts. The more inertia an engine has, the 'easier' it will be to ride as user inputs (throttle, clutch) have a slower reaction time, in this case it would be harder to stall the engine, but the larger inertia will slow down the rate of acceleration so it will not rev as quickly. (Of course the opposite is true with a lightened flywheel.)

So, it all depends on application, whether you want a heavier or lighter system inertia.

And just to make note from my test engineering days; engineers designing engines DO have a pretty good idea on what inputs are seen to their system, even though combustion is a crazy/violent reaction. The instrumentation required to determine cylinder pressure is not terribly complex and even older DAQ (Data Acquisition) systems record at 20khz+. (roughly one measurement point every 2 at 7000rpm).

In the end though, there is still great value to 'seat of the pants' testing for overall system inertia to make a car or bike, trike, monowheel etc. 'feel' comfortable to the end user.