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How to calculate max velocity of a vehicle from torque and RPM?

  1. Nov 16, 2009 #1
    Hey guys,

    I'm kinda simulate vehicle dynamics. I got gear ratios, torque and RPM values. I find how to calculate instant accerelation from

    Ok now I got acc. :wink: I can calculate velocity from acc but my problem is how to calculate losses. I cannot constantly increase vehicle's speed. :approve: At certain speed it will stabilize and won't increase or decrease if I keep car's RPM constant... So basicly I need to calculate max velocity of a vehicle at instant RPM and torque... :bugeye:

    Thank you all.. :smile:
  2. jcsd
  3. Nov 16, 2009 #2
    Losses aren't trivial to model.

    You will likely have to specify them. Drag and rolling resistance will be your main components. For that you need to know nothing about the engine, but the car/vehichle itsself. If you want to incluse drive train losses, you will have to specify them. These will most likely be rule of thumb estimates. Such as friction losses, etc.

    You can also calcualte velocity purely from

    (engine speed* gear ratios * final drive *) wheel circumfrence or
    wheel rpm * circumfrence

    This is assuming no slip (purely rolling condition). A more realistic slip ratio is 1.1 so for every 11 rotations of the wheel you go forward 10 wheel circumfrences.

    I understand what you are trying to show with your graph, but it's pointless to make a distinction that HP and Torque are two seperate things as they are so closely linked, and either can be used to show the same thing. It's good that you are trying to show that torque is the real driving 'force'.

    Eg Saying torque is important but power isnt is pointless, unless you make the caveat that you have set gear ratios.

    If you don't have set gear ratios power is the only thing that matters, as that shows you what you can do in a given time. More power = more that can be done. When you nail the gear ratios down, it makes sense to talk in terms of torque only.

    ps thats a pretty nice spreadsheet.
    Last edited: Nov 16, 2009
  4. Nov 16, 2009 #3
    In order to determine the maximum velocity of an automoble, you first need to determine the power loss due to drag and rolling forces, as pointed out above.

    For automobiles, the drag power is the largest, and varies as speed-cubed:

    Pdrag = (1/2) ρACdv3

    See http://en.wikipedia.org/wiki/Drag_(physics [Broken])

    while rolling resistance power is

    Prolling =~0.01 mgv
    where tire rolling resistance coefficient is ~ 0.01

    Then match max engine power (torque x 2 pi RPM/60) to provide the necessary power. If, in an Excel run, the vehicle reaches maximum velocity before the engine reaches max power output, you will need a higher gear ratio.

    Understanding the power loss due to drag, rolling reststance, etc., and knowing the engine brake (dynamometer) power, and using the acceleration rate vs velocity

    Pacc = m a v

    (mass acceleration velocity)

    is sufficient to calculate elapsed times, final velocities, etc.

    Bob S
    Last edited by a moderator: May 4, 2017
  5. Nov 17, 2009 #4
    Thank you for your replies guys...

    That spreadsheet I posted before does not belong to me, I found it on here. I've used torque, car mass, gear ratios and RPM to calculate vehicle's accerelation.
    I actually want is max velocity that's all. Seems like this method is useful for me:

    (engine speed* gear ratios * final drive *) wheel circumfrence or
    wheel rpm * circumfrence

    Let me ask you about this formula,

    1. engine speed is RPM value right? :redface:

    2. I have gear ratio and differantial ratio. And I multiply these two value to find a final ratio. In the formula above there's "final drive", what is final drive? In a regular car what value can I use? :frown:

    3. here's how I found acc. Please correct if anything is wrong:
    result torque = differantial ratio * chosen gear ratio * torque (at given RPM)
    accerelation = result torque / mass of vehicle

    So when calculating velocity, is it smt like this?:
    result torque = differantial ratio * chosen gear ratio * torque (at given RPM)
    velocity = result torque * 2pi * wheel radius :confused:

    Thank you guys. :smile:
  6. Nov 17, 2009 #5
    1. Yep engine speed is rpm.

    2. Diff ratio is final drive. (its just a different way of saying it)

    3. F = ma finds acceleration. Thrust at contact patch is F.
    What you have found is wheel torque, to convert a torque a force you need a distance. T = Fd.

    Thrust = wheel (resultant) torque * radius. (centre of wheel to ground.)

    Velocity = Engine RPM * gear ratio * diff ratio * wheel circumfrence.

    You don't need torque to find velocity.
  7. Nov 17, 2009 #6
    Ok I am not sure if I am doing it right.
    I have shared a spreadsheet: http://spreadsheets.google.com/ccc?key=0ArPyqArltO-KdGxqX3l6WGJfTjUyTEwyUTZidkwtRHc&hl=en
    This spread sheet is editable.
    Spreadsheet includes RPM, torque, calculated gear ratios, HP, at the end velocity from formula but I guess I messed up at some point. I am using SI units (N, m etc..)


    And secondly I did a litle research about drag power. I found CdA is approximately between 0.4-0.8 m2 for regular cars. And air density is 1.2kg/m3 so final value is:
    Fd = -0.3 * velocity2 (velocity is in m/s)

    Ok we may use these calculations also. I got torque, gear ratios, HP, RPM, wheel radius and Fd... I guess I need to subtract engine force on car (I mean the result force to make the car move forward) - Fd....

    Well I kinda lost in here :)

    Thank you...
  8. Nov 17, 2009 #7
    Ok, sorry about this I should have made it more clear.

    When referring to speed the ratio is a reduction ratio. So for example if you have a drive of 5:1 the engine is spinnign 5 times faster. However gears wrt to torque are multiplication ratios, 5:1 will increase torque 5 times. So if you've specified 5:1 as the gear ratio, you must use the inverse for speed reduction. ie 1/5.

    So although I put both as multiply, I didn't make it clear that one was a reduction. I've edited the spreadsheet but can't seem to save it.

    Edit also you've used, RPM directly which will give you mters travelled in a minute. You need to /60 to get seconds.
    Last edited: Nov 17, 2009
  9. Nov 17, 2009 #8
    Ok I guess you have edited spreadsheet. I see you change gear ratio as 1/gear ratio. And Divide RPM to 60 for converting m/min to m/sec.. So good till this point. But as you can see at table below km/h table, this car reach 105km/h at 1st gear which supposed to be a normal car, and at 5th gear I see 439km/h wow :eek:

    Seems like I need to cut these values a litle bit. Do you have any proposition for that? What can I use to calculate air resistance, tire resistance etc...

  10. Nov 17, 2009 #9
    I noticed that when I was editing it. Did you make these ratios up or use real data?

    The diff seems a bit short, and the gears seem slightly too far apart.
    I've edited them to give slightly more sensible answers.

    your power outputs are calculated incorrectly. either do

    Engine torque * engine RPM *(2pi/60) = Power
    Wheel torque * Wheel RPM * 2pi/60 = Power.

    The values should be the same.

    and the units need changing. You are calcualting power from Nm and radians/second. So your power output is in kW not horsepower
    Last edited: Nov 17, 2009
  11. Nov 17, 2009 #10
    I have a ford escort (2000) which is 90hp and I found gear ratios of ford: 3.35 1.99 1.33 1.00 0.68 (http://www.vibratesoftware.com/html_help/html/Ford/Ford_Transmissions_Main.htm#5%20Man [Broken])
    I have changed these values on spreadsheet.

    I might messed up with rpm-torque relationship cause I do not have real values. I am not sure these values belong to a standart car engine. And 1kW = 0.98630HP so it does not make too much difference when power is converted to HP. I still see e.g 1065155HP at cell M12 which does not make any sense.. :cry:

    And still there are absurd values at velocity table.. :frown:
    Last edited by a moderator: May 4, 2017
  12. Nov 17, 2009 #11
    I dont mean the relationship is messed up just that the way you have calcualted power is wrong. You are multiplying a torque figure at the wheel by an engine rpm. When you should be multiplying engine rpm by engine torque.

    I've edited the first column of the power to calcualte engine power output. That gives sensible figures. All other power outputs should be the same as this column.

    Also the wheel looks a bit big for an escort (its 23 inches across). Plus you need to put a correct diffratio in, it'll probably be between 3.5 and 4.2.
    Last edited by a moderator: May 4, 2017
  13. Nov 18, 2009 #12
    Ok yesterday thanks to Chris I finally got max velocity at certain gear and RPM...
    Now I got accerelation and max velocity value.. I don't wanna stop here, and I need more information and I think I can only get it from here somebody who knows about this topic.

    I realized I need to adjust my accerelation values with actual speed. For example if you are doing 20kmph at first gear, and if you gear up to fourth gear car engine may stop or shake the car you know. So you won't get accerelation, you may even break something in transmission. Like you cannot start a car at fifth gear right? So you need velocity to use gears effectively. My accerelation values only depended on gear ratio, torque at given RPM, vehicle mass..

    How can I involve current velocity to accerelation value? :bugeye:

    Another issue is let's say you are travelling at 3rd gear and 4000RPM. When you shift up to 4th gear RPM value decreases to between 2000-2500.. How can I calculate smt like that? :eek:
    Last edited: Nov 18, 2009
  14. Nov 18, 2009 #13


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    You can, but you'd be slipping the clutch for a bit, and acceleration would be slow. On a Top Gear episode about the Corvette Z06 (you can find a few of these on youtube), Jeremy Clarkson claims to do a 0 to 160 mph run in 5th gear, but looking at the gauges the car was in 4th gear, which redlines at 175 mph at 7000 rpm, and Jeremy mentions the 175 mph speed in that segment, so my guess it was done in 4th gear, although the early scene shows him shifting into 5th gear. The Z06 reaches it's top speed of 198 mph in 5th gear at just over 6300 rpm, the rpm of it's peak power. The engine has adequate power at 1000 rpm to do this, but you'd have to slip the clutch until the car reached 32 mph for the engine to run at 1000 rpm without slipping the clutch.

    The point here is that for maximum speed, the car needs to be geared so that maximum speed is reached at the rpm of peak power for an engine, not the engine's redline.

    Just multiply 4000 by (4th gear ratio) / (3rd gear ratio).
  15. Nov 18, 2009 #14
    My problem is I need to adjust accerelation value. If you start your car at 4th gear it won't accerelate as fast as 1st gear right? So how can I adjust my acc. values?
  16. Nov 18, 2009 #15


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    Depends on how accurate you want this to be. The force at the rear tires = (engine torque) x (overall gearing) x (efficiency factor) / (tire radius). For efficiency factor .85 would be a good approximation. You could use a spread sheet to calculate acceleration versus speed (based on torque versus rpm) versus gear, assuming that you shift up at redline or when the force versus speed curve drops below the curve for the next gear.
  17. Nov 18, 2009 #16
    You have to determine two things. 1: Your power required which is the force available at the contact between the drive wheel tyres and road and is known as 'tractive effort'. The ability of the drive wheels to transmit this effort without slipping is known as 'traction'. Hence usable tractive effort never exceeds traction. The tractive effort relate to engine power as per formula in site below.

    And 2: Your resistance or power required. Also formula for it is in site below.

    In order to accelerate a vehicle, climb a hill, or pull a trailer, or a combination of them; your power available must exceed power required.

    You can’t simply ignore resistance in the real world because drag varies as the square of the velocity and power by the cube. In other words, to double your speed you need to increase your power by a factor of eight!

    The part about final drive is for some vehicles, such as tractors, that have high torque often have a planetary gear between differential and wheels. You can ignore it for cars or leave it as 1:1.

    Chapter 31 of the following site has a lot of information:

    Another book, which explains this, is “Principles of highway engineering and traffic analysis by Fred Mannering”, which is a book used to design highways. In chapter 2 of the book he gives detailed information of acceleration requirements for vehicles. I can’t find the book on the web but the following site contains parts of the book, starting on page 9:

  18. Nov 18, 2009 #17
    You should have an adjusted wheel torque value (which you do). The 4th gear value will be much lower than the 1st gear value. You don't need to adjust it from that sense.

    However so far no losses have been included, they would reduce the accceleration.
  19. Nov 18, 2009 #18
    @Jeff Reid I'm kinda rookie on these things. I am a software engineer, and I do not know much about cars. I want to simulate a vehicle using engine HP, gear ratios, and torque. Actually I kinda don't get what you mean, sorry. :confused:

    @nucleus That crankshaft web site is wow, but there's too much info in it. I want basic simulation. As I understood you want me to calculate power and use that values. It might be a good idea to do that but I need more help. :cool:

    1. First of all I would like to ask you both should I use SI metrics (Nm, m etc) or English/American metrics (lb, feet, etc). Beacuse at the end I want to obtain instant accerelation at certain RPM and gear, and instant velocity at certain RPM and gear. It is easy to conver mph to kmph. What do you think on that? :frown:

    2. After deciding units, what I have is: gear ratios, RPM, torque. I can calculate hp (kW) from here. And I can calculate the torque at wheels. And the drag power against car at certain velocity. :redface:

    3. After calculating acc and velocity, in near future I would like to have engine brake it user is not touching anything the car slows down and it also depends on gear and velocity. And hitting brake will slow down car, I think this won't be a problem. One more thing if the car is moving uphill/downhill there's gravity factor which won't be a problem if I can calculate over all force on car right :smile:

    Can you make comments on these 3 topics? I probably need our help when I start to calculate smt. It would be best if I can open a editable spreadsheet so anybody can make comments. Thank you guys.
  20. Nov 18, 2009 #19
    I assume you have having issues with thinks like clutch slip?

    If you try to start your car in 4th from idle. It probably won't transmit enough power to turn the wheel (or possibly spin it becuase it will be rotating too quickly) and as a result no motion at the wheel means the engine must also stop, so you stall.

    What slipping the clutch does is allows the engine to remain above idle revs, but transmit less power and rpm along the transmission. It's when you bring the clutch up through the bite point slowly. It basically acts as a constantly variable gear ratio, the higher you lift it, the more powre it transmits.

    The problem with this is it put alot of wear and starin on te clutch.

    You don't been to worry about altering acceleration for differnt gears, becuase we've already dont that by calcualting wheel torque in each gear.

    1. Use whatever units you feel most comfortable with it's easy to convert. Normally I stick with 1 typre for all calcualtions then convert the end results (stops roundning and conversion errors).

    2. yep Use dragpower to work out speed, but you can also work out drag force, which will alter acceleration. (in reality yiu can do it either way but I like to use force for acceleration).

    3. I don't get this question.
    Last edited: Nov 18, 2009
  21. Nov 18, 2009 #20
    Hey Chris,

    I just posted a comment before you did. Can you make comments on that? What would you suggest? Should I start over, this time calculating forces on car and let the calculation determine max speed at certain gear and RPM?
  22. Nov 18, 2009 #21
    I've just edited the post. Don't restart, what you've got now is fine.

    You just need to add force and acceleration cells. There is no point doing it the other way, as if the model is correct you'll come up with the same answer.
  23. Nov 18, 2009 #22
    It is not easy to find SI unit examples and explanations on internet maybe I should stick with English units all the time.

    I have added accerelation at the bottom of sheet (http://spreadsheets.google.com/ccc?key=0ArPyqArltO-KdGxqX3l6WGJfTjUyTEwyUTZidkwtRHc&hl=en)
    Sheet is editable. I divided output torque to car mass value to find accerelation, is it correct? I believe this accerelation value is in m/s2, I am not sure about that.

    And I am a bit confused actually:
    F = m . a So I got a and m if I multiply car mass with acc I found F force on car. Which is torque value in my calculation.

    But tork = F . r F force, r radius of wheel.. So smt is wrong...

    About my previous 3rd question: If the car is moving uphill with slope of 30 degrees the car has a negative force since there's gravity. How should I include those later.. I guess we should go step by step to we don't get confused...
  24. Nov 18, 2009 #23

    You need to convert torque to force. T = Fd
    So you take your torque value and divide by rolling radius. This gives the force at the contact patch.

    Torque / mass gives units of = m^2/s^2

    Also car mass is in kg. 1100kg is the mass of the car.

    For the third question, we can deal with that later. Goping up a hill is a set force pushign backwards determined by mass gravity and slope angle. Thats really easy to do.

    F = mg sin (angle)
  25. Nov 18, 2009 #24
    Ok let's see

    T = F.d so F = T/d

    and F = ma so T/d = ma

    therefore a = T/(d . m)

    I guess 1100kg is weight of a car so I should divide it by 9.8 gravity, right?
  26. Nov 18, 2009 #25
    you just use 1100.
    The acutal weight of the car is 1100*9.8.

    This is one of the things where common usage of the word weight is not strictly correct in a scientific sense.
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