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Rotational Mass

  1. Nov 15, 2004 #1
    I need help figuring out how much HP/TQ is freed up by changing to a lighter flywheel in a car. Specifically the Mazda KLDE/KLO3 2.5L V6 found in the Ford Probe/Mazda MX6.

    I have listed all needed specifications (I hope). Basically I enjoy physics but am not smart enough to identify and apply the nessecary equations to get the desired result. I know that the benifit will be proportional to acceleration and gear ratio. I hope someone can help me identify the TQ/HP freed up!

    Stock flywheel is 23 lbs
    Aftermarket flywheel is 9lbs
    Flywheel diameter is 250mm

    Redline: 7000
    Peak Horsepower: 164@6000 rpm
    Peak Torque: 160@4800 rpm

    Gear Ratios
    1st: 3.307
    2nd: 1.833
    3rd: 1.310
    4th: 1.030
    5th: 0.795

    final: 4.388

    If you need more let me know and i will provide if i can!
    Last edited: Nov 15, 2004
  2. jcsd
  3. Nov 16, 2004 #2
    Even just telling me how much TQ or HP is needed to spin a 23 Vs. a 9lb flywheel at 7000 rpm would be helpful!
  4. Nov 16, 2004 #3


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    Changing Flywheels

    For a simple model of flywheel, there is no particular minimum torque or horsepower necessary.

    The flywheel essentially acts as a storage medium for rotational energy. As you are probably aware, an IC doesn't produce continuous torque, but generates torque spikes whenever a cylinder is fired. The flywheel acts to smooth out these spikes in torque.

    Switching to a lighter flywheel isn't going to free up or produce an appreciable amount of horsepower unless the existing flywheel system is extremely poorly designed. You might save some weight, but the 14 pounds you mention isn't going to make a huge difference by itself.

    Since I'm not a car buff, I'm not sure where the appropriate improvements for your car are, but, generally, to increase performance you'll want to improve fuel and air flow characteristics in and out of the cylinders. Changing air filters, for example, can have a noticable effect on performance and doesn't involve much effort.

    To determine how much energy is required to spin up flywheels, it's necessary to know the moment of inertia of the flywheel. A solid disk, for example, is easier to spin up than a disk with all of the mass on the outer rim. If the two flywheels have the same profile then it will take about [itex]\frac{9}{23}[/itex] of the energy to spin up the lighter flywheel.

    In summary:
    From a naive perspective, I would guess that changing to a lighter flywheel is not going to improve your engine performance in any significant fashion, and can lead to increased engine wear and a jerkier ride.
  5. Nov 16, 2004 #4
    Well I know for a fact that the flywheel helps tremendously. The power it frees up is phenomenal, and it cut .2 seconds off my 1/4 mile time. Consider it takes me about 2.3 seconds (thats my 60 foot time so it should be within .2) to get to redline in 1st, thats 7000 rpm in 2.3 second or 3000 RPM/second. That ALOT of angular acceleration, and dropping the weight makes traction in first tough to grab, however if you get traction my 164 HP probe and pull AWAY from a 240 HP Honda S2000 and they have a better 1st gear ratios, = torque, and more redline to use.

    As far a moment of inertia, the placement of the mass is = between the 2 flywheels. The difference is in the metal used. There for in my estimation, even if our moment of intertia is off, it will be off for both and thus proprotionally the figures should work out. So lets assume the weight is evenly distributed among the entire 250mm diameter of the disk. What units should i work in to calulate moment of inertia, and is this (I=mR^2) the correct equation to use assuming even weight distribution throughout the body.

    Knowing that the lighter flywheel is 9/23 easier to spin up is great but unless i know how much HP or TQ it takes ot begin with this dosent mean much to me.

    Can someone calulate how much tq is neede to HOLD the flywheels at 7000 rpm and tell me the difference there. I listed the diameter and weight, if you need more lemme know.

    Thanks for the input Nate TG, I REALLY wish i knew more, and want to take some classes cause so much could help me with clutch clamping, braking, weight distribution under dynamic loads (acceleration/braking) and other things i wish to understand better!
  6. Nov 16, 2004 #5


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    Nate, thats true at a constant speed, but a flywheel absorbs energy while the rpms are increasing, which affects acceleration. Since its a dynamic situation, that can be tough to calculate.

    pgt95, You need to know the moment of inertia of the flywheels - if it isn't a uniform disk, that can be tough. For a uniform disk, its [tex].5m r^2[/tex]. Multiply that by angular acceleration and you get torque. Your flywheels are a little under 6 inches in diameter, but probably have more mass out toward the edges, so we'll assume 6 inches. That gives a moment of inertia of 1.1 and 2.9lb-ft[tex]^2[/tex] (it would be nice if it were given and it seems like a logical thing to put in the specs for the flywheel, so check).

    Now, the angular acceleration can be tricky - and it'll be different for every gear. Say you start 2nd gear at 4,000rpm and switch to 3rd when you hit 7,000 and that takes 2 seconds. That's an increase of 1500 rpm/s or an angular acceleration of 157 radians per second per second.

    Applying f=ma (divide by 32 to convert pounds to mass) we get 2.9*157/32=14.2lb-ft and 1.1*157/32=5.4lb-ft or a savings of 8.8lb-ft.

    Obviously, you can tweak these numbers to fit the actual performance of the car.

    THIS may help with my explanation.

    edit: btw, welcome aboard!
    Last edited: Nov 16, 2004
  7. Nov 16, 2004 #6
    Thats what i needed Russ thanks alot!

    So to sum up,

    I need to have my diameter in inches.
    .5mr^2 (how do you guys do the cool numbers that look all text book?!?!) is my moment fo inertia.
    How do i translate a time and RPM into angluar acceleration, that is radians.

    Once I have my inertia, and my angular acceleration all I need to do is take them and apply them to F=MA/32 (the 32 to convert to mass) and then I can input varying masses and see how things come out?

    Now it is true that the mass is focused further out towards the edges, however, seeing this formula it seems that PROPORTIONALY the numbers should work, so while I am not accurate in absolute numbers, the difference should be failry accurate.

    Is there anyway to use the gear ratios, inertia, and peak TQ umbers to do something simmilar? I would love to figure out a set of equations where one could plug in flywheels mass, inertia, peak TQ @ RPM, and figure changes dependant upon gearing.

    You guys rule, I wish i would have found this place sooner!
  8. Nov 16, 2004 #7
    Speaking from experience and extensive no-how with cars, i can tell you that an lightweight flywheel wil give you know measurable horsepower gains and your torque will stay the same thru your torque bands. The only thing that wil happen changing to a lightweight flywheel will due is help improve your launch times. (REDLINE faster).
  9. Nov 16, 2004 #8
    Yes the motor makes the same power however there is now more availible to the wheels. Underdrive pullys do not MAKE power either but they certainly make your car faster.

    As well you state the car will redline faster. In order to accelerate the motor faster the car also has to accelerate faster, since it is the same car then alooking at F=MA it would seem that since M is the same and A is increased F would also have to increase.

    Please guys, I keep getting people trying to tell me my theory is wrong. From PLENTY of hands on experiance, the flywheel makes the car faster noticably in 2st and 2nd gear (Ihave time slips to back this up too), and to a much lesser degree in 3rd. I just want ot put a number to it or something close...
  10. Nov 17, 2004 #9
    No thoughts on using gear ratios?
  11. Nov 17, 2004 #10


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    Really, I don't think that the gear ratios are going to matter much in the calculations for how much power goes into the flywheel except that they affect how quickly the car accelerates.

    The amount of energy that goes into (or comes out of) the flywheel is [itex]K (\omega_f^2-\omega_0^2)[/itex]. Where [itex]K[/itex] is some constant that depends on the moment of inertia of the flyweel (roughly proportional to the mass of the flywheel) and the units that you want to use and [itex]\omega[/itex] is the rate at which the flywheel is spinning.

    Since power is work divided by time you can figure that the amount of power that is going into the flywheel is going to be
    where [itex]t[/itex] is the amount of time it takes you to accelerate. Clearly, the power that goes into the flywheel is larger in the lower gears because the car is accelerating more quickly so [itex]t[/itex] is smaller.

    This is of course, not practical for your car, but it occurs to me that a flywheel's primary function is usually to provide some minimum angular momentum. So, it's concievable that a flywheel with moving parts could alter it's moment of inertia to maintain some minimum angular momentum (or minimum energy) rather than having a constant moment of inertia. In that case the amount of power (and torque) that goes into the flyweel could possibly be reduced to zero. A sufficently sophistcated control system might even be able to use the flywheel as energy storage for a sort of regenerative braking.
  12. Nov 17, 2004 #11
    So really I have to work with some time variable... since eveyr car accelerates a lil differently and it is tough to get an accurate time for acceleration in 1 gear, i was hoping to avoid it but so be it... thanks again guys.
  13. Nov 17, 2004 #12


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    Since I drive a 94 Probe with the 2.5l v6, I have been following this conversation with some interest. A very fun car to drive, I do not race, but would be interested in hearing some of your times. I do not usually push it hard, but it is also hard to hold it back, I have no trouble getting into traffic, my commute includes an intersection where I am at a stop sign, waiting to get onto a busy 5 lane hiway. I don't need much of a hole!

    I would recommend that instead of converting everything to English units, go the other way convert to metric, then when all is done do a final conversion to English. It just simpler to keep track of metric units.

    I would think that the fly wheels are close enough to a uniform disk, that you can use that as the MI with only small errors.

    We have:
    [tex] \tau = I \alpha [/tex]

    where [itex] \tau [/itex] = torque and [itex] \alpha [/itex] is the angular acceleration.


    [tex] \alpha = \frac {\tau} I [/tex]

    Since You are reducing I , [itex] \alpha [/itex] will increase.

    This gets you the easy part, it is much harder to translate this into a meaningful time to RPM while pushing (or pulling) the car down the road.

    Oh, yeah, to see how to enter symbols simply click on a formula, a tex box will pop up showing the text entered, also there is a thread about LaTex at the top of the General Physics forum with more details.
  14. Nov 17, 2004 #13


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    Unfortunately, the flywheel also serves as the contact point with the friction plate of the clutch, so it needs to be mechanically tough and rigid since it transfers power from the engine to the drive train.
  15. Nov 17, 2004 #14


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    Russ, get your caclulator out and do that conversion again. 200mm = 8in so 250mm ~ 10in
  16. Nov 17, 2004 #15
    Keep the good stuff coming, I am trying to get an EXACT diameter, i can always rep lug in the numbers u know? I will have a full write up, and will deffintly link it.

    Intergal catch me on AIM some time: mugen2200
  17. Nov 18, 2004 #16
    250mm=9.84252 inches
  18. Nov 18, 2004 #17


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    As long as you're breaking out the calipers, you might also want to measure the inner diameter. (Assuming the flywheel is not solid.)
  19. Nov 18, 2004 #18
    Thats kinda the problem. On the outermost edge is the ring gear where the starter would engage, not much weight on those teeth. Inside that is a large portion of the mass about 2 inches thick (diameter that is) then inside that is the friction surface that is thinner still (this is really the rest of the mass), and inside that still is the bolt holes, no much mass there and finall a 1.5 inch diameter hole for the input shaft... hardly uniform in ANY way

    For an example look here: http://www.unorthodoxracing.com/ultral.html
  20. Nov 19, 2004 #19
    If you can take the flywheel out you can experimentally determine its moment of inertia:

    Put the flywheel on a horizontal axle (around which it should be able to rotate with the least possible amount of friction), attach a weight of 1 kilogram via a piece of rope to the flywheel as shown in the attached drawing and measure with a stopwatch how long it takes for the 1kg to fall 1 meter (do this 10 times, or more, and take the average)
    call this time [itex]t[/itex] and call the radius of the flywheel [itex]r[/itex]

    You know that the number of radians that the flywheel rotated in this time [itex]t[/itex] is

    [tex]\frac{1}{2\pi{r}} * 2\pi = \frac{1}{r}[/tex]

    and that the number of radians that the flywheel rotated in these [itex]t[/itex] seconds is also



    [tex]\frac{1}{r} = \frac{1}{2}\alpha{t^2} \Leftrightarrow \alpha=\frac{2}{rt^2}[/tex]

    since you also know that


    and that

    [tex]\tau = mgr = gr[/tex]
    (because m=1kg, if you use another weight fill in that weight)

    it follows that

    [tex]I =\frac{1}{2}gr^2t^2[/tex]

    just fill in [itex]g[/itex] (9.81), [itex]t[/itex] (the time for the 1kg to fall 1 meter) and [itex]r[/itex] (the radius of the flywheel) and you have its moment of inertia.

    Attached Files:

    Last edited: Nov 19, 2004
  21. Nov 22, 2004 #20
    I know it is old but i got more info now....

    Ok guys i still haven't gotten numbers to work out for me.

    1: How can i calculate Radians from RPM and time, say i have an acceleration of 3,000 RPM in 1.5 seconds, how do i convert?

    2. I know moment of inertia = .5Mr^2 but russ's numbers don't work out so i need to make sure i got this right.
    He has
    Now in my book .5*9lbs*3inches^2=1.1 but by my math it equals 40.5 so i divided the 9lbs by 23 to convert ot mass and got:


    As well for the 23lb flywheel he got 2.9 and I get 3.2. Some one lemme know if i am missing something please?

    My final figures ar as follows:
    Weight stock 23lbs
    Weight light 9lbs
    Diameter is 11.5 Inches, radius 5.75
    So i SHOULD get 11.847 and 4.63 respectively, at least by my math... check me out guys and hook a brotha up!
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