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Car Forum / Driving, Maintenance, Tuning / Maintenance and Repair / May 2007HORSEPOWER FROM OBD2 DATA
Hello I"m writting a software who get some data in my ECU via the odb2 connector and interface. I can get rpm / speed / and elapsed time from the begining of run I see on internet some software who dyno the car with a rpm/time measure and cars specification ( wheigt / tire / gear ratio and SAE correction ) , i want to include this dyno part too, but i dont know the correct formula in order to calculate the horspower from the rpm/ time and car specs . Ps: ( i can't have the torque info like on German ECU .. ) If somebody can help me, it would be a pleasure ! Thanks, Fl.dubois > Hello > [quoted text clipped - 13 lines] > > Thanks, Fl.dubois You start by using the RPM and other specifications and calculate MPH which is your velocity. You then calculate the change in velocity which is you acceleration. You now have acceleration and time you need to calculate force. Stan > fl.dub...@gmail.com wrote: > [quoted text clipped - 23 lines] > > - Afficher le texte des messages précédents - Tanks Stan, i've search formula for what you say and result is : velocity = displacement/time (m/s ) acceleration = ( v2 - v1 ) / t (m/s²) average acceleration = (v2-v1) / (t2-t1) force = mass * acceleration ( n ) POwer = force * velocity OU work / time work(nm) = force * displacement = mass * acceleration *displacement so .. P = (mass * accel * displacement ) / time for my probleme .. work= (POwer * 5252) / RPM or HP = (work*rpm)/5252 is it correct ?? ( i guess i need to add the dyno ratio factor and Scx ratio ) > > fl.dub...@gmail.com wrote: > [quoted text clipped - 38 lines] > is it correct ?? ( i guess i need to add the dyno ratio factor and > Scx ratio ) Keep in mind that the result will be applied, or wheel, horsepower, not brake horsepower. Also you must work with small time increments and seperately compute horsepower over each short time interval. Sort of the nurmerical approximation to integration. On 17 avr, 15:24, Don Stauffer in Minnesota <stauf...@usfamily.net> wrote: > > > fl.dub...@gmail.com wrote: > [quoted text clipped - 45 lines] > > - Afficher le texte des messages précédents - time increment is about 3hertz between two data , i guess i would plane the graph with bezier algo > > fl.dub...@gmail.com wrote: > > [quoted text clipped - 38 lines] > is it correct ?? ( i guess i need to add the dyno ratio factor and > Scx ratio ) I would start with force = mass * (average acceleration) (v2-v1) / (t2-t1) > > > fl.dub...@gmail.com wrote: > [quoted text clipped - 43 lines] > > - Afficher le texte des messages précédents - I think i start with it too, and hp = force * displacement , so Hp = mass * average accel * displacement . For the displacement, i find this : distance or 1 RPM : d = Ct / (Grv * Grd) = diam * Pi/(Grv * Grd) where Ct = circunference of tire, Grv = gear ratio Grd = differential ratio accept 2 time and rpm data , t1,t2,rpm1,rpm2 so displacement = ( ( rpm2 - rpm1 ) * d ) / ( t2 - t1 ) ) ( in fact i count the number of rpm betwen the elapsed time t2-t1 , but i ask if i must extrapole the data because time result is a millisecond, not second .. this point is obscur, because my result is a displacement for a interval T, not a seconde ) exemple : gear ratio : 1,66 differential : 3,23 tire diameter : 0.62738 1.145 sec 1450 rpm 1.340 sec 1620 rpm d for 1 rpm = 0.62738 * 3.14 / ( 1.66*3.23 ) = 0.367 m => 1620-1450 = 170 rpm => 170 * d = 62.4m in 0.195s what is my velocity for the exemple ? 62,4 or the ratio of 62,4 for 1 second ?? > > > > fl.dub...@gmail.com wrote: > > [quoted text clipped - 71 lines] > what is my velocity for the exemple ? 62,4 or the ratio of 62,4 for 1 > second ?? average acceleration = (v2-v1) / (t2-t1) force = mass * average acceleration hp = ((((force * trie_radius) / total torque mult. [gear ratios]) * RPM) / 5252) > > > fl.dub...@gmail.com wrote: > [quoted text clipped - 41 lines] > I would start with > force = mass * (average acceleration) (v2-v1) / (t2-t1) Keep in mind the HP to accelerate is only true total HP at very low speeds. The higher we go, the more HP is used to overcome drag. One way to count for drag is to do coast-down tests. Measure the deceleration at various speeds, and add to HP calculated to acceleration data. > > > > fl.dub...@gmail.com wrote: > > [quoted text clipped - 48 lines] > deceleration at various speeds, and add to HP calculated to > acceleration data. I understand, but before he does coast-down tests or calculates aero drag using Cd and frontal area or tire rolling resistance he need to get the HP calc working or all this does not matter. re all I come back cause i have a little matter . Speak with real value: car definition : Weight : 1300 kg 3rd gear ration : 4.48 : 1 Differential : 1 ( i guess ) tire dimension : 175/65 R14 and these 2 obd2 value : ( time;rpm) 0,682926833629608;2029 0,852926835417747;2076 so .. ( in a metric system ) /!\ 1 pouce = 25.4 millimetre diameter wheel = (((width *0.Aspect_ratio)*2) +(jante_diameter *25.4) diameter = (175*0.65*2)+(14*25.4) = 583.1 mm = 0.5831 m so circonference = PI * d = 1.831 m after we compute the velocity ! at 1000 rpm, speed is : 1000(rpm) * ( 1 / Rv ) = Z tr/min Z * 60 = Z tr/h Z*circonference_of_wheel = V en m/h V/1000 = V en Km/h simplification : ((1000/Rv)*60 * Ct )/1000 = V soit (1/Rv) * 60 * Ct = V in hm/h @ 1000 rpm for X rpm : (X * (1/Rv) * 60 *Ct )/1000 = V en km/H @ X rpm with Z,X=rpm, Rv = ( gear ratio * diff ) this formula is good, i check her with my technical book of my car , i get the right speed at 1000 rpm ! so .. for our 2 obd2 value !! ( get in 3rd gear ) to translate km/h to m/S i use the formula : (Speed / 3600 ) * 1000 0,682926833629608;2029 --> (2029 * (1/4.48) * 60 *1.831) / 1000 = 49,75 km/h - > 13.819 m/s 0,852926835417747;2076 --> 50,90 km/h -> 14.138 m/s now i use the acceleration formula : a = (v2-v1) / (t2 - t1) = 0.31 / 0.17 = 1.88 m/s² we know that force = mass * acceleration : 1300 * 1.88 = 2444 newton and work(nm) = force * displacement = 2444 * 0.31 = 757.64 newton.meter or 757.64/1.356 = 558.7 lb-ft i know that i havent applie the wind drag factor yet, but this value looks me too high . Is my reasonement wrong ? or the wind factor very important to get correct value I know that my car cant have more than 185 nm of torque ! Thanks, and sorry for my poor english > re all > [quoted text clipped - 59 lines] > > Thanks, and sorry for my poor english I have not had time to go through your math. This is what I get. I changed the space to - to keep it lined up but it also makes it a little harder to read. Stan ---------------------------Rear------Aero-----Rolling------------Rear-W---Accele --RPM------MPH---Velocity--Wheel----dynamic---Resist.--Elapsed----Horse---ration ------------------ft/sec--Torque---Drag---HP----HP------Time------Power---in-G's -2029.0--30.944---45.384-1277.15-----1.561----3.547-----.6829----493.40---2.0655 -2076.0--31.661---46.436--131.03-----1.672----3.630-----.8529-----51.79----.1922 > > re all > [quoted text clipped - 73 lines] > > - Afficher le texte des messages précédents - Thanks a lot stan, but im unable to read your post, desciption is up side down . I "ve find a mistake in my reasonement, work = force * displacement, and i get the difference of velocity instead of real displacement . I see you find the right torque and correct horspower, how do you get the displacement ?? For displacement, i have this formula, but result seem's to be unreal ! distance for 1 RPM : d = Circonference / (Gearr * differential) = 40.8 cm , we have 47 rpm, it do 19.2m in 0.17 sec .. my car is a fly :D > > > re all > > [quoted text clipped - 84 lines] > distance for 1 RPM : d = Circonference / (Gearr * differential) = 40.8 > cm , we have 47 rpm, it do 19.2m in 0.17 sec .. my car is a fly :D I have a program that I wrote and what I posted is it's output. I did post this earlier to you. > average acceleration = (v2-v1) / (t2-t1) > force = mass * average acceleration > hp = ((((force * trie_radius) / total torque mult. [gear ratios]) * RPM) / 5252) In your case v2 = 46.436 and v1 = 45.384, t2 = .8529 and t1 = .6829 Stan Ok, thank you very much stan, your post was very helpfull i' have find my mistake in the displacement, results are now correct Hope this thread could help any people . I can now work on the wind drag factor and the wheel adherence to correct these value at any speed . Hello guys I come back cause i need your point of view for find the correct environemental factor who influence the real horsepower. the last formula give us the engine horsepower . We can now substract : - the rolling resistance ( force in newton ) - the wind resistance ( force in newton ) - transmission loss ( coeff of 0.85 on the torque i think ?? ) - moment of inertia ( wheel + transmission + engine ) convert to energy ?? the last point is my probleme, i dont know if i must use the wheel moment of inertia , just transmission + engine i guess ? but how find them ?? Do you see another factor to substract ?? thanks > Hello guys > [quoted text clipped - 18 lines] > > thanks > the last formula give us the engine horsepower . Wrong. You should have calculated the wheel thrust / torque which you translated back to engine torque / hp minus all frictional (trans. / rear) losses. To get engine HP look back at the coast down test that Don S. talked about some time ago. Stan > > Hello guys > [quoted text clipped - 27 lines] > > - Afficher le texte des messages précédents - Ok, you've right, this is wheel horsepower, engine horsepower will be same with loss . But for the coast down test, car would have an acceleration, and a deceleration, its not very userfriendly for an embedded application, im just trying to resume the coast down test factor without a deceleration process
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