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Homework Help: I need help to figure out how to turn acceleration data into Horsepower/Torque

  1. Dec 19, 2008 #1
    I have a vehicle with some known info...
    weight: 3400lbs
    Speed at a specific time: see data in chart linked below
    overall gear ratio: 2.922
    Tire size: 25.68 inch Radius
    RPM at speed: see data in chart linked below

    heres the data
    http://spreadsheets.google.com/ccc?key=pxiQyAOYb5vb6jOOMdoQuOg" [Broken]

    where do I start??
    I dont mind doing math....but I need some help understanding what I need to do and where I need to start.

    I probably want to do this over a 1 second sample so that I dont have to do any x1000 or /1000 math in there...
    I can convert the speed into Kph and the mass I can do in kg if it makes it easier

    I hope to come up with a basic understanding of the math involved and do a little chart of Horsepower and torque vs RPM in the end
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 19, 2008 #2
    I tried to a car analysis at one time. Here is a link with a lot of formulas:

    http://www.rdrop.com/~/larry/download/formulas.pdf [Broken]

    See if this helps. I couldn't see your car data.
    Last edited by a moderator: May 3, 2017
  4. Dec 19, 2008 #3

    not really...I can find all of those formulas and I understand them on their own....
    its making them work together to get my formula here...

    I know the mass and I know that it accelerates a certain amount over a given time....
    how do I calculate horsepower for that given time...
    I can set it at 1 second or 10 seconds....I dont care about teh interval...al I want to know is how to get from acceleration of 12mph in one second to horsepower value for that tme
    Last edited by a moderator: May 3, 2017
  5. Dec 19, 2008 #4
    Did you try looking for any dyno chart data for your stock vehicle online? This could simplify your task tremendously.

    I'll take a better look at your question when time permits.
  6. Dec 19, 2008 #5


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    Staff: Mentor

    If you know the acceleration and mass, you know the force. If you know the force and distance, you know the work. If you know the work and time, you know the power. And 1hp is about 750W (I don't remember the exact conversion). Does any of that help?
  7. Dec 19, 2008 #6
    Acceleration of the car is caused by a force. When the force acts over a certain distance it does work. If you assume that the acceleration is constant, then the force is constant: force = mass x acceleration. When you accelerate your car and it moves over a distance d, work is done: Work = force x distance and has units of joules (energy). Now, it takes your car some amount of time to traverse this distance. Power is the rate at which work is applied: power = work / time and has units of joules per second (or watts). Horsepower conversion: 1 hp = 750 watts (approx.).

    Don't forget that the car has to overcome numerous frictional forces so this may complicate your analysis.

    Does this give you a better idea?
  8. Dec 19, 2008 #7
    I did this for a math experiment to test horsepower of cars and compare them to manufacturers data.
    I fitted a curve to the data as a function of time, then used that as position function x(t). from there there are many ways to get to power as a function of time; remember a car's power changes with time (it is not constant).
  9. Dec 19, 2008 #8

    I re-did the URL
    you should be able to see it now....I forgot to put it in share mode
    Last edited by a moderator: May 3, 2017
  10. Dec 19, 2008 #9

    I get this......
    I just have a hard time with understanding the units
    force = mass x acceleration
    what units are force?
    is mass LBS or KG
    what is the acceleration unit?

    so work(joules) = (mass x acceleration) x distance

    how can I get the distance? tire diameter math and speed correct?
    and does that need to be in inches? feet? meters?

    so...Power(watts) =((mass x acceleration) x distance) / time(in seconds?)
    and that whole thing...
    watts * 0.00134102209 = Horsepower
    or watts / 745.699872 = Horsepower

    am I correct in the math formula??
    if yes..then I just need help with the units of measure being used in my variables
    Last edited: Dec 19, 2008
  11. Dec 19, 2008 #10
    Force should be in Newtons or pound-force. You need to get the mass of your car instead of weight. I would go with metric units for now. Covert your weight to Newtons and then divide by local gravity 9.81 m/s2 should be fine.
  12. Dec 19, 2008 #11
    so my mass should be
    3400 * 4.4482216 = 15123.95344 newtons / 9.81 = 1541.6874046890927624872579001019
    rounded = 1541.6874 (what unit???)

    and what about the acceleration part...
    what formula is my acceleration? and what are my units for acceleration?
  13. Dec 19, 2008 #12
    acceleration should be in meters per second squared. But you do not have acceleration, you have the vehicles speed (or velocity) at various times.

    You need to come up with a formula or function that relates the sample velocities to time.
  14. Dec 19, 2008 #13
    so could I have just done

    1 pound = 0.45359237 kilograms (google)
    and taken 3400lbs * 0.45359237 = 1542.214058 kilograms
    even though it doesnt quite match up to the previous math 1541.6874 kg?

    as far as the acceleration...
    difference in velocity over difference in time ? correct??
    this formula just popped back in my head...
    so (dV2-dV1)/(dT2-dT2)
    velocity in meters/second and time in seconds
    which gives me the acceleration in m/s^2

    so...Power(watts) =(([mass_in_kg] x [(dV2-dV1)/(dT2-dT2)]) x [meters_traveled]) / time(in seconds)
  15. Dec 19, 2008 #14
    Looks pretty good to me. (dV2-dV1)/(dT2-dT2) will get you the average acceleration and hence the average force and hence the average power.

    You might be able to get a better approximation of acceleration by plotting a portion of your data in excel and finding a regression line and then using that to find the work done. But, that is only if you want to.
  16. Dec 19, 2008 #15
    I was planning on that :)
    I have plans on using the data to solving for horsepower in vehicles at the dragstrip...
    lets me know if the vehicle is running it tip top shape or if we need to adjust some things to get a little more power out of it..
    we know what it does at its best...and we know what to adjust to compensate for various weather/climate conditions to get back to that....we've only figured that stuff out from the dyno....if we can come up with a dyno on the dragstrip then we can run a better race

    one thing I dont get
    Power(watts) =(([mass_in_kg] x [(dV2-dV1)/(dT2-dT2)]) x [meters_traveled]) / time(in seconds)

    what is my bold type time in seconds?
    where do I get that value?
    Last edited: Dec 19, 2008
  17. Dec 19, 2008 #16
    I believe that should be the same time as used in the formula for acceleration. Think about it, you want to know how much Work (Force*displacement) this thing does PER second (Power). So if your force was calculated over a certain time interval (and then work with a corresponding distance interval) than you need to divide that work by THAT time interval to get Power over that interval.
  18. Dec 19, 2008 #17

    so a recap to make sure this is right

    ((([mass_in_kg] x [(dV2-dV1)/(dT2-dT1)]) x [meters_traveled]) / [(dT2-dT1)seconds])* 0.00134102209 = Horsepower

    Torque = (Horsepower x 5252)/RPM

    then I can plot them both in excel and do some comparisons :)
    Last edited: Dec 19, 2008
  19. Dec 19, 2008 #18
    ok...so I think I did some bad math somewhere...
    7.52 3551 22.37 2.922 0.62 0.07 0.65 0.83
    7.65 3605 22.99 2.922 0.62 0.13 1.34 1.71
    obviously my HP math is wrong somewhere


    I also re-updated the Data in the google documents page
    you are looking at data in line 10/11
  20. Dec 19, 2008 #19
    I am confused by your data chart. The "distance traveled" column... what exactly is that? It obviously is not the Total distance traveled at that time.... is it the distance traveled between each interval?
  21. Dec 20, 2008 #20
    yes distance traveled for a given dT2-dT1

    is that the correct one to use or is that my error?
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