All,<p>I've been working on a PC based DYNO. One that will collect data as well as provide valuable information. I need some help. For those that say, how do you do that... Well, you've heard of the G-Tech right, how about the $3K Vericom. They all use what's called an accelerometer. You can get those from Analog Devices for about $70 or less. The other stuff you need you can get from Radio Shack for < $5. How about that! And best of all you can gather data on a portable and then use it on your PC for performance gathering.<p>Ok, so here's what I need. I have all the math except one formula. The Accelerometer gives you G's. So from that and an accurate time, I have<br>Velocity<br>Distance<br>Speed<br>Time<p>Samples over .02/s should be way good enough and are what the Big bucks versions do. From the above, I can do 0-60, 60-0, 50-70, anything you want. I can do 1/4 miles runs etc.<p>BUT, I need to know how to give HP from the above. I know that it needs the vehicle weight or you could use 1000 lbs and give HP per 1000 lbs.<p>So, you math nuts, help me out, I've got HP calcs for 1/4mile runs but they don't work.<p>If you need a head start, download the "profile" program from the http://www.vericom.org site, it has a bunch of formula and sample data, but I can't get to the HP as it's listed in the sample data using the formula they show. There must be something obvious. I'm going to go post on sci.physics as well.<p>If you're looking to use some of that gray matter that you learned in college or high school... chime in.<p>I promise when I'm finished, I'll provide source code, schematics, documentation and sources for the accelerometer for you all to build one as well, they are really very easy!<p>yea, yea, yea, I know, get a life... Well, I didn't see anyone else doing this, so I figured I would.<p>Alan

I have a Masters degree from MIT in Mechanical Engineering so you can trust me on this one.<p> Power is defined as the time derivative of the object's kinetic energy which in this case is (1/2)*(mass)*(velocity)**2. Performing the minimal math required, power = (mass)*(velocity)*(acceleration).<br> I think that you can realize that the power so-derived is not a constant, but varies during the duration of your test run. So if you want to output the power as a single number, this should be a time-averaged value of the above calculation.<br> So the question is really about the units:<br>If the vehicle weight (W) is in pounds, the speed (V) is in ft/sec, the acceleration (A) is in ft/sec**2, then the instantaneous power is given by: (P)=(W)*(V)*(A)/32.2 . The power units will be in ft-lb/sec. I think the conversion to HP is 550 ft-lb/sec=1 HP, so you would divide the above answer by 550. Please look up this conversion, It's been a while since I've worked in these units.<br> I'd be happy to provide any further info.<br> <br> <br>

Well, heres what i know.<p>Work is equal to force * displacement<br>W=f*x<br>By this the units are slug*ft./s² (slug is English unit for mass)<p>Power is equal to the change in work per the change in time<br>p= delta W / delta t (no delta symbol in ascii)<p>1 HorsePower is equal to 550 ft.*lb./s (if you divide works units by seconds, then remember that a lb is equal to a slug*ft/s², you get ft.*lb./s)<p>since you have distance and time and acceleration you should be able to calculate horsepower. you need to take small sections of your 1/4 mile time to get the plot (like breaking up the area under functions w/ rectangles in calc class to get a riemann sum) <p><br>I'll put a sample post attached to this (and i hope I can do the math right)

This can get a bit complicated to try to get exact answers, but let's start with the basics. Let's first assume that there's no aerodynamic drag on the car and also that there's no driveline friction in the car. You have velocity and acceleration for each sample period from your measurements which, combined with info about your car, is all you'll need. Using the time-honored Force=MassxAcceleration (F=mA) formula you can calculate how much force must be applied at the tire's contact patch each sample period (given your car's weight as mass).<p>For simplicity's sake let's assume your car is only being driven by one wheel instead of 2 (or four in the case of quattro). In the end the forces at each wheel add up to the total force (and hence power) anyway. So that one driven wheel is exerting the force necessary to accelerate the car at the measured acceleration rate. That force (at the tire/ground contact point) is acting at a distance from the center of the wheel, hence there is a torque around the wheel's axle. That torque is simply the force (let's assume in lbs) acting at a distance of the tire's radius (assume in feet). The tire's radius can be estimated (ignoring tire deformation from the car's weight) using the tire's size. The tires on my A4 are 205/55R-16.<br>So the sidewall height is 205mm x .55 = 112.75mm ~= 4.439in.<br>The radius of the wheel is 16/2 = 8in for a total tire radius of 8+4.439 = 12.439in = 1.037ft.<p>So you now have the torque about the drive wheel's axle. The final formula you need is:<br>HP=Torque x RPM / 5252<br>(assuming torque in ft-lbs, otherwise the 5252 constant will be different).<p>The only number we haven't figured out yet is the RPM of the tire, but that's easy.<br>The circumference of the tire is PI x diameter.<br>Circumference = PI x 2 x radius = 3.14159265359 x 2 x 1.037ft = 6.516ft.<br>So at 60MPH the tire would be rotating:<br>(60mi/hr) x (5280ft/1mi) x (1rev/6.516ft) x (1hr/60min) = 810.3 RPM (revs/min)<p>Voila! Now you know how much horsepower the wheel is putting to the ground! If there's no driveline friction then the engine must be putting out the exact same amount of horsepower. (Power does not get altered by gearing, but torque does!)<p>To get more accurate answers you'd have to know your car's drag coefficient and frontal area (combined with the measured velocity) to find out how much extra force the tire must put to the ground to overcome the aerodynamic drag. Plus to get true engine horsepower you'd have to know how much friction was in the driveline. Anyway, hopefully this provided you with the type of info you were looking for. Good luck!<p>Jim Meyer<br>'98.5 2.8QMS

A car weighs 3500 lbs. It starts with acceleratoins at 0. its jerk is 1f/s³(this won't be constant either for your idea). Example for first 4 seconds. 3500lbs/32.15 = 108.9 slugs. <br>F=ma<br>a=jerk * time<br>jerk = 1<br>so..<br>a = t<br>f = m*t<br>delta Work = m*t*x<p>x = at²<br>a = t<br>x = t³<br>delta work = m *t * t³<p>power = delta work / delta time<br>power = m * t * t³ / t<br>power = m * t³ <p>so HP = m*t³/550<br>HP in this instance at any time t = 108.9*t³/550.<br>Obviously you don't an equaton like this, but you do from you earlier post you should have enough info to get good data points if you have small enough intervels for time. Hope any of this random info helps (if this indeed doesn't work, well i'm sorry, this is all this young 12grader knows, ask me 4 years from now when i have my physics major)

i have worked for 2 different companies using the AD devices for 2 different applications:<br>dynamic weight determination, and impact and crash reconstruction.<p>Maybe tomorrow i will look at the calculations, but i am feelling lazy today.<p>however:<p>i know that some of the AD parts are being obsoleted (we are doing replacements now) so choose carefully.<br>The other problem i ran into in the past, is that the AD devices have a fairly high noise floor. In my apps i could average or integrate out the noise, but something to be aware of.<p>good luck<p>1996 a4 2.8 qm<p>RAndall (EE)<br>