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Car physics

  1. Mar 6, 2008 #1
    Hi,

    I'm making a car simulation game not just a simple one, one that closely matches the kinds of Gran Turismo, therefore, it has to be extremely precise and realistic and use real life physics.

    I have been searching around for information regarding this subject and was able to get some info here and there about parts, about maths but it's hard to put it all together correctly and that's where you guys could help me out.

    The first thing i'm trying to establish and confirm is the following question:

    Speed of an object such as a car is directly related to the rotationnal speed of the tires right? If a tire is 15in radius, thus he is grossly 94.25in circumference and if i roll the wheel once, the car should then move by 94.25 inches right?

    If this is true, then i can assume that a car's engine turning at 3000 RPM or 50 Main shaft revolutions by second, will spin the wheel by 50 turns each seconds. In between that comes the transmission which may drastically lower this ratio to make the car pickup inertia more easily.

    So if i have a transmission that has a coeficient of 4.00 in the first gear, the calculation should be the following if i am not wrong:

    (These numbers are grossly calculated to remove the numerous decimals but will be kept in the final calculations.)

    Engine Revs / s = 50
    Gear 1 ratio = 4.00
    Wheel Revs / s = (50 / 4) = 12.5
    Distance covered / s = (94.25 * 12.5) = 1178 inches (or 98 feet)
    Speed = 66 mph

    If i put that in miles per hour it makes the car roll at 66 miles per hour.... This makes no sense right? I mean, a car in the fifth gear could do that speed around 3k RPM, not at gear 1. Same thing again, with gear 5 having a ratio of 0.98:

    Engine Revs / s = 50
    Gear 1 ratio = 0.98
    Wheel Revs / s = (50 / 0.98) = 51
    Distance covered / s = (94.25 * 51) = 4806 inches (or 400 feet)
    Speed = 272 mph

    What am i not grasping correctly in your opinion?
     
  2. jcsd
  3. Mar 6, 2008 #2

    rcgldr

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    Homework Helper

    Your gear ratios are tall. In a Corvette Z06, which can reach 61mph at 7000 rpm in first gear, first gear has a ratio of 2.66 to 1, while the differential gearing is 3.42 to 1, for an overall gearing of 9.10 to 1. The rear tire size is 325/30/19, the specificication include an overall diameter of 26.7 inches, and 783 revolutions per mile.

    At higher speeds, the driven tires have to opposed rolling resistance and aerodynamic drag, and the tire tread deforms before and during contact with the pavement and then recovers aftewards. This reduces the effective diameter by a small amount at higher speeds, maybe 1% or so.
     
  4. Mar 7, 2008 #3
    Hey thanks for the reply but that was not it but it did raise some new questions. The original problem was the wheel circumference calculation, i thought the R## was the radius of the wheel which is wrong. I can tell you how to calculate it but i think it is beyond the scope of this post.

    My new question that arises is the following, what is a differential gear? You are saying that at one point, what i am aware of so far in the car movement mechanics is the following:

    Engine -> Revs
    Revs -> Attenuation from clutch pressure plate / Torque converter attenuation
    Converter -> Transmission Gearing
    Transmission -> Wheels
    Wheels -> Road

    Where does the differential go in there and what is it?
     
  5. Mar 7, 2008 #4
    ps: Indeed the gear ratios where high, i was trying to match the speed of a toyota corolla at certain RPM and found that the only way was to achieve this by tweaking the transmission gear ratios... But now that the wheel circumference is good, i can put much more logical and reallistic numbers in the gear ratios and everything falls into place.
     
  6. Mar 7, 2008 #5
    Well.. yes.. but that is true considering pure rolling. In that case, it can be said that:

    [tex]
    v_cm = r\omega
    [/tex]

    As you said, you want to use real-life physics.. in that case, pure-rolling is something that never happens. However, all objects try to attain the state of pure rolling.

    When working with a car, one major force you need to account for is friction. That and other forces combined, you can get to the fact that for a wheel, the motion will be a combination of the translation motion of the wheel and rotational motion of the wheel. So, you can say that:

    [tex]
    v_cm = v + r\omega
    [/tex]

    Do note that, [itex]v_cm[/itex] will definitely be equal to the translational velocity of the car. This equation becomes even more complex when you consider the car under acceleration. The frictional constant, isn't a constant in true sense. The frictional constant is a function of the speed and shape of the acting body. This function can however be approximated to a simple mathematical function.. something that may not be true.. but for a sim.. will be adequate.
     
  7. Mar 7, 2008 #6
    Yeah that i was aware of, i knew that there are all kinds of factors that may affect the distance, what i'd like to point out though, right away, i am not a mathematician. Just a more than average programmer that knows what he's doing, so usage of VcM = RW is like pure mystery for me.

    Please use less mathematical formulas and more explanations :) I like your description of pure-roling though, i can understand that.
     
  8. Mar 7, 2008 #7
    So, if i undestand your post correctly, friction can affect the speed of the car? If so how? Is it calculatable?
     
  9. Mar 7, 2008 #8
    it definitely does.. Frictional force is calculated using:

    [tex]
    f_s = \mu_s N
    [/tex]

    here, 'N' is the normal force i.e. the force exerted by the object [in our case the car on the surface]. The variable [itex]\mu_s[/itex] is a characteristic of the pair of surfaces in contact and is different for different surfaces. As i said earlier, this is considered to be a constant but well.. in the case of physics of fast moving objects like a car, it changes with change in the speed and shape.
     
  10. Mar 7, 2008 #9
    Look up all that goes into translating engine rpm into wheel rpm, you missed the differential in your example, which drops speed by its divisor effect. If the engine is turning at a certain rpm, then the car WILL be going at the speed indicated by the net gearing.

    The question of whether the engine WILL turn at that rpm is a function of the power output versus the power required to move the car at the speed suggested by the net gearing, which is primarily moving air out of the way.
     
  11. Mar 7, 2008 #10
    Ok, so i'd need to evaluate the friction characteristic for asphalt which would give me the "Ug" sign (Not sure if it's a Ug, don't even know how you are displaying this). Can you tell me where to find such an information?

    Also, what is N, the weight of the vehicule in this case? I would also need to take into account the pressure of the tires and the shock absorber's value i guess?
     
  12. Mar 7, 2008 #11
    Thanks regor60, don't know if i updated about the speed yet so here goes what i have now.

    Name Corolla
    Weight 2850,00
    Inertia 0,00
    RPM Increase Max 0,00
    Base Torque 162
    At Base RPM 4000
    RPM 6000
    Torque 243
    Power /s 24300
    Shaft rev / s 100,00
    Torque converter applied friction 100%
    Torqueconverter conversion ratio 100%
    Flywheel conversion ratio 100%
    Transmission rev / s in 100,00
    Active gear 5
    Gear 1 Ratio 316%
    Gear 2 Ratio 190%
    Gear 3 Ratio 1310%
    Gear 4 Ratio 88%
    Gear 5 Ratio 72%
    Gear 6 Ratio 0%
    Transmission rev / s out 138,89
    Differential ratio 265%
    Revolutions after differential ratio 52,41
    Applied brake force 0%
    Revolutions after braking 52,41
    Tire Quality P
    Tire width (mm) 215,00
    Tire W/h Ratio (%) 45%
    Tire Radial size 17
    wheel size length cm 62,53
    m/s 32,77
    m/h 117981,13
    km/h 117,98
    mile/h 73,74

    As you can see, to match the speed of a corolla at 3000 RPM (i own one and i do about 120 km/h) i have to put the engine RPM at 6000 following all the formulas from the excel sheet i have. So something is definetly missing here, i have a 0.5:1 factor missing somewhere in between.

    What do you think that could be in terms of mechanics?
     
  13. Mar 7, 2008 #12
    Oh and please, lets just keep the conversion about car speed calculation for now, thanks for the help rohanprabu, but i'd like to focus only on one aspect at a time.

    When the speed calculation works, i'll focus on calculating the accelleration potential...
     
  14. Mar 7, 2008 #13

    Stingray

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    Science Advisor

    It's not clear from your list of data what the tire diameter is. If it's 63 cm, the speed at 6000 rpm in 5th gear would be (52 rev/s) (3.14*.63 m/rev) = 100 m/s. Of course, the engine could never get the car going that fast. Although most cars will not reach redline in top gear, this gearing looks too far off to be right. Searching around a bit, I think your axle ratio is wrong.

    As a general warning, the level of physics in something like Gran Turismo is at a far higher level than the calculations we're discussing. If you're still interested in this sort of thing, there's a lot of documentation over at racer.nl. Someone there has already done what you're proposing.
     
  15. Mar 7, 2008 #14
    thanks, i'll look it up, note that i want to make a racing sim but it's an online one, so i'm not looking at doing an extreme racing sim, just one that is not a online, click and flip a coin in hopes to win, i want some really mechanical aspects and allow my players to tune the cars.

    I'll go check the site you are talking about
     
  16. Mar 7, 2008 #15
    I've looked it up, it's not showing much other than presenting the racing engine, i found a part discussing car physics, but so far i know as much as is presented.

    After reading your post, i've tried to find out what you mean by 52 rev/s, i can't seem to produce this, where do you get that from?

    6000 RPM = 100 Rev/s
    After gear 5 at 82% (or 0.82:1), i get 121.95 revolutions,
    After differential at 345% (or 3.45:1) i get 35.35 revolutions this is far from the 52 you are stating...

    (PS i may have changed the numbers trying to find data on the net, so you don't have the same numbers as me anymore)

    Finaly, yes the tire size is simple to calculate, the formula is:

    (Wheel Radius (15) * 2.54 (to get CM)) + ((185mm * 65%)/10) and this gives us ... O_O

    Oh ****, i got it, this is stupid, i am not having the circumference here, i'm simply calculating the tire's height, i need to multiply it by PI.

    OMG, YES!!! That was the missing part, thats why i had crappy km/h. Now were talking... The speed for my car at 3000 RPM is now 124km/h which is much more logical.

    Thanks

    Ok, now the second part which i have NO CLUE or almost what to do, power calculations. Now that i know how fast i can go i need to calculate how a car can accelerate based on it's power output.
     
  17. Mar 7, 2008 #16

    Stingray

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    Science Advisor

    This seems to have some more detailed discussion:

    http://racer.nl/prgusers.htm

    There are (or were) descriptions of the Pacejka model for the tires somewhere too. I think some older versions of the software are actually open source, which may be useful to you.

    If the wheels aren't spinning, a reasonable first approximation to the acceleration is found from
    [tex]
    m a v = \epsilon P_{\mathrm{eng}} - P_{\mathrm{drag}}
    [/tex]
    m is mass. a is acceleration. v is velocity. [itex]\epsilon[/itex] is a driveline efficiency (near 1). [itex]P_{\mathrm{eng}}[/itex] is the engine power at the appropriate rpm. The last term represents power lost to aerodynamic drag and rolling resistance of the tires. It can be approximated by
    [tex]
    P_{\mathrm{drag}} = \alpha v+ \beta v^3
    [/tex]
    for appropriate constants [itex]\alpha[/itex] and [itex]\beta[/itex]. These terms can be estimated by searching around for tire (first term) and aerodynamic (second term) data.
     
  18. Mar 7, 2008 #17
    I know, that a car has Torque@RPM, that produces torque in a matter of foot-pounds or SI units. I know i have first to calculate how much power the engine can output at the RPM it's reving. What i am not sure is if the RPM of the engine affects the torque at the same time.

    For example, if a car can output 162ft-lbs at 4000 rpm, the total power output, as i've read about it, is 648000 SI per minute or in this case 10800 SI per second. Now do we have to do a cross calculation based on the RPM to find the torque or is the torque always the same at different RPM? My guess is that the torque changes with the RPM but then that would mean that the multiplicator has a double edge, you would get very low SI when the engine is idling.

    700 rpm * 162 / 4000 = 28.35 SI * 700 RPM / 60 seconds = 330.7 SI... This is almost 3% of the optimal possible SI capable by the engine.

    Can anyone confirm that?
     
  19. Mar 7, 2008 #18
    Thanks for the info stingray, the data presented in that page is about the engine itself, i did peak at the car physics at the bottom but this is way to mathematical for me, i need to understand first then i can maybe start playing around with formulas.

    I'm not a mathematician, just a simple programmer that is maybe more advanturous than others :)
     
  20. Mar 7, 2008 #19
    Ok if i understand your post stingray, the movement of a car is equivalent to it's mass multiplied by acceleration multiplied by velocity and this must equivalent to the SI units generated by the engine and i'd have to remove the drag from this.

    Taking into account that i don't have any information relative to drag i will simply assume it is 0 for now and try to understand what you are telling me.

    If the velocity of my car is 0, because we are not moving and the mass is 2850 pounds, and the SI units generated from my engine are 330 units, i can calculate the acceleration by doing the following:

    Mass * Accel * Velocity = SI
    Mass * Accel = SI / Velocity
    Accel = SI / (Mass * Velocity)

    Problem is, with this formula, if velocity is at 0, this creates a division by 0 error. Note that my math courses are quite far away, is this the right way to alter a formula to isolate a new result?
     
  21. Mar 7, 2008 #20
    Another thing i'm wondering, the engine will still be able to rev up but slowly at the first calculation pass creating more power on the second pass and so on. So if i have a strong enough power, i can eventually move right? That would be how a car could eventually defeat hills and wind drag?
     
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