Just had a thought, wanted to throw it out here for discussion. There's a fair amount of people with data logging capabilities. Having data such as time, RPM, speed, gear ratios, tire circumference, weather etc available it ought to be possible to figure horsepower/torque from there and plot it. My math on this is rusty (will do some reading though), but if we have all the information it should be doable. Would be cool to be able to do 'dyno runs' with various mods, boost levels etc and be able to tell how the performance was affected immediately. Thoughts?
very possible... but for accuracy u will need to know: 1) exact weight with u in the car 2) effects of drag: this will be difficult to determing accurately if ur car has any aero mods but will be very important since accurate dyno plots are generated in high gear then again, who says dyno's are 100% accurate... definitely doable 1) as long as u can determine velocity (v) versus time (t) with high enough resolution, u can differentiate it to determine acceleration (a): a = dv/dt 2) force (f) acting in the forward direction for a car with mass (m): f = m*a 3) torque (wT) at the wheels of total radius (r) including tires: wT = f*r 4) apply correction factor (c) accounting for gear ratios, etc to determine engine torque (eT): eT = c*wT 5) hp = T*rpm/5252 im ignoring drag... anyone think of other factors?
Thanks.. I have seen some 'laptop dyno' stuff but they tend to charge a good bit for it, and it's not doing anything you couldn't put together on your own with some effort. I figure that if you datalog rpm and speed only you ought to get some fairly high resolution on the results with a datalogger. Where I'm a bit stuck is gear ratios; what the heck do you do with them? One possible solution for drag... if you do a coastdown in neutral and log that, you should be able to get the amount of hp lost in the drivetrain and due to drag, which should actually let you use that correction to calculate flywheel hp based on your input data... drag, drivetrain loss etc would be out of the equation then, more or less.
To make myself more clear... push clutch in at 100mph or whatever and coast down to 20-30 or something, and build a correction table from the data that would represent drivetrain and drag losses... this way you could really get flywheel hp, and might be more accurate than guesstimating drag losses only (since we have no way to really measure them) to get accurate correction data for whp compared to a dyno.
Continuing on... From what I gather, hp = Force (lb) * Velocity (mph) / 374 F = ma as you said, so... hp = ma (converting units would be needed) * Velocity(mph) / 374 And you can get mass easily enough, acceleration could be calculate from the delta speed divided by delta time between two data points... right? So if the units are right, it should be possible to get hp from logging only velocity and time... and you want rpm too to graph it and calculate the torque. Think that approach might work? Then to get flywheel, you do the same calculations for a coastdown and simply add that curve to the previous one for flywheel hp.
woah buddy dont use the lb always convert to kg and for mph its in meters per second, also dont use the weight use mass weight equals= mass*gravity mass in kg gravity is 9.81m/(s^2) F is in Newonts (N) im in AP Physics right now so IM me or PM me if you need anything, cause i told my teacher on the first day that i like cars and almost everything that you could in kinematics used cars in the example, lol and for the velocity and accel thing i think you need to setup some speed traps, and use your gear ratios the thing is, accel is never the same, you could figure accel at each point but you would require a dyno chart with exact tq at each rpm, lol so thats not viable onl thing icould think of is first find the Fn(normal force, force perpendicular to the surface), to do that ya need Fg(m*g) to get Fn do this ...|\ ...|.\ ...|..\ ...|...\ <---get that, use trig, this Fg should be parallel to Ff, hence per ...|....\ pendicular to the surface(this also =Fn) ...|......\ ...|.......\ ...|_____\ because Fd(force of drag) is a resisting force you need the force that is resisting for each speed, ehhhhhhhhh oh yea using Fn you could derive Ff (force of friction) i think its Ff/Fn=u(coefficent of friction) im not sure how much the frictional force (Ff=u*Fn) would play a roe but maybe a tad bit well if im right then +F(hehehe, sum of all forces)=Ftq-Ff-Fd to find Ftq you need the velocity, the accel, the mass, the radius of your wheel AND tire to the millimeterand because the force is not equal (if it was then you would be not accelerating) we are fucked, lol give me a while and ill think of it, if i had seen this earlier, i could have done something constructive yesterday, instead of whoring up forums and playing video games, and playing bloody knuckles in the cold, lol (im an idiot i already told you that)
good point about units... make sure everything is in the same convention... SI is my preferred way to go
Well, if we can come up with some good math I'd be happy to put together either a spreadsheet or VB/VC# program to crunch the input from TARI and plot dyno charts...
also, torque is the cross product of force and radius, not multiplication, tires deform on acceleration, so it'll be F*R*sin(80 degrees) or so, you'd have to look up the shearing coefficient for rubber to determine that number accurately. so that's a small inaccuracy in addition to drag which is a big force that needs to be taken into account. the best way to measure torque and power would be to do a timed and distance measured run, 60 to 80 or so. get your kinetic energy at both points, so KEi=m(60)^2, KEf=m(80)^2 W=deltaKE=KEf-KEi this is your work which divided by the time it took would be your average power over the 60-80 run in J/seconds or watts, the closer you get the 2 speeds together, the more accurate. you can then convert to horsepower if you wish. W=-deltaU(potential energy) Force = differentiate potential energy U with respect to x. or use W=f dot product displacement, so measure the distance it took for your 60-80 run and multiply f*s*cos(angle of incline) to get force Drag force is usually modelled with Fd=1/2Cd*A*rho*v^2 Cd=drag coefficient for air A= cross sectional area(measure it on the subie front, not going to be accurate at all because it aint a cube, any AE's help plz.) rho=density of air v=velocity of object Subtract this from what you got from the run and you have net force. cross product this with the radius of your tires, we'll negate the afformentioned shearing forces accounting for drivetrain losses and you have your torque at the wheels. whoopdy doo...just dyno your car people, phuc physics.
^^^Im with him^^^^ How much does it cost to get dynoed. 40-120 depending on were you go. Although if you get that formula working let me know It would be cool to check your # agianst a dtno.
Well, several other individuals and companies have done it before (Mike Chaney from the f-body mailing list did a dyno program back in the late 90s) so it is certainly doable. No, it's not going to be 100% accurate, and neither is a stationary dyno. But I remember even Mike's pretty simple program was within 5hp of measured dyno #s for the cars they tried it on. Besides, sometimes it's fun to do things just to do them, y'know? Sure, you can go fork out money to somebody else to do everything, but then you don't get to dink around with things yourself, figuring out the best route through trial and error... and that's some serious fun to me. RamblinWRX: Thanks for your input. I was planning to use the delta speed between two input data points and apply some smoothing to the results to get a higher resolution hp curve. Sound feasible? (We're talking maybe 5-10 data points per second hopefully...)
Seems to me this might be able to be simplified. If all moose wants is to be able to calculate a number that represents torque (and from it get HP) and be able to re-calculate that number to see the difference when he makes a change I don't think you need to worry too much about drag as long as you measure without a big change in wind velocity and do the measurement at a lower speed. Can't tire deflection and the shearing coefficient be assumed to be constant as long as you have not changed tires or pressure? I know i'd like to have a spreadsheet like this that I could use to play with. --Subscribed--
Well, ideally I'd like for it to be pretty close to accurate #s.. I know what you are saying about seeing the difference and I agree that you wouldn't need to worry about that.. heck, you could just graph mph/time and get an idea of whether things got better or worse! My thinking was to ditch the aero/friction stuff by doing the coastdown measurement and add that curve to the one measured during acceleration, which should take drag/friction/drivetrain losses out of the equation more or less...
we could go into intense details to get it very accurate... but what margin of error are we willing to accept here... most dynos vary by a decent amount anyway... who is to say which one is accurate
Well, I think the real decision is whether to get into drag and driveline efficiency and friction and everything to get a close estimate on whp, or whether to do the coastdown thing and just shoot for flywheel... thoughts?
alright i did all the calculations using jason's 1/8 mile time from scoobymile.org, here goes data- 1/8 mile=201 m tf=7.968 s vf=88.09mph=39.4 m/s unladen mass of sti = 1495 kg 1 watt = .001341 hp 17 inch wheels = .43 m wheels coefficient of static friction car tire to asphault=.9 work-enegy derivation for power: total work=deltaKE+work from friction delta KE = 1/2 m v^2 = 1160389 Joules work friction = (.9)(1495 kg)(9.8)(201m)= 2652898 J average power (not maximum) = total work/time = 370472 W = 496 hp instantaneous power is something i can't do without a full time plot, sry, i cant do calculus from 1 data value. for torque: take only the work the motor does, so negate frictional forces(you dont want to have to factor the torque of the ground back on the wheel in later). W = F dot S, assume drag strip to be flat= F*s net force = W/s= 5773N *optional* (this can be broken down Fengine-Fs-Fd= 5773N Fengine-muS-1/2(Cd)(A)(rho)(v^2) Fengine=5773+(.9)(14665)+1/2(.3)(2.38 m^2)(1)(9.19)^2 = 21986 N which from there you can figure out wheel torque by the mechanical advantage(gear ratios), however, there are probably 3 gear changes...sooo, we shouldn't do this way. *optional* just take the net force 5773N, divide the force evenly across 4 wheels=1443N/wheel -to get torque= force cross product radius (no deformation)=F*r=1443*(.43m)=620n-m or 457 lb-ft of torque. is this around what jason's numbers are? 1/8 mile - 7.968 @ 88.09 - 496hp, 457 lb-ft torque discounting drive train losses sound right? try my method out on your drag strip times and see what you get, i'm probably leaving something out cos those seem a bit high.