


#1




Is there any formula or way to estimate increases in horsepower or torque after engine modifications based on improved acceleration times? Suppose a vehicle had an engine transplant, exchanging an engine with a known horsepower and torque for another engine with unknown horsepower and torque (which nevertheless peaked at about the same rpm). Could the improvement in 060 times be used to estimate the output of the replacement engine?
The reason for my curiosity is that I have a ~5,650 lb. Gelaendewagen that had a lowcompression engine (M110, twin cam) designed for thirdworld countries' low octane gasoline. I just replaced it with a custom long block that was built following the engines found in the high performance Euro sedans  it includes custom, highcompression pistons and high performance camshafts. The engine builder promised me a lot of performance improvement, but frankly, I didn't expect (or dare hope) as much as I'm seeing. When the truck was stilll fairly new, it had a 060 time of about 17andchange seconds. With this new engine, with just a couple of thousand miles on the clock, I'm doing consistent 13 second 060's on the same road, and using the same timing and shifting techniques as I did before. The engine originally had about 154 hp and 167 lbft torque, and I expected that the new engine would merely be similar to the regular 185 hp of the Euro SEL sedans (and both engines peak their hp and torque within a couple of hundred rpms of each other, so this shouldn't be a factor). However, I find it hard to believe that such a reduction in 060 times, from 17.5 (average) to around 13.1 seconds, can be explained by a mere 20% boost in power. The engine builder promised a hotter engine  I think he underestimated his work, and I think that you can appreciate my curiosity in trying to estimate what my new engine is really putting out. Is there any formula or relationship that can give such an estimate of the new power output? I hope that you can help me  even a rough estimate is welcome. Thank you very much. 
#2




Hi Ken,
I think the easiest way to accomplish what you want is to borrow or buy a GTech Pro. It's a portable accelerometer that will extrapolate HP from acceleration runs to 60 and the quarter mile. It should be able to give ballpark numbers. Good luck. 
#3




GTech Pro ?
re: "...a GTech Pro. It's a portable accelerometer that will extrapolate HP from acceleration runs to 60 and the quarter mile. It should be able to give ballpark numbers...."
I've seen ads for (what I think is) this gadget  it contains accelerometers to record performance, but to estimate horsepower?? Really? There must be so many variables (the car's weight, its Cdx (can't use subscripts  I mean its aerodynamics), etc. The Gelaendewagen is so heavy ( > Chevy Suburban) and so unaerodynamic (sort of like a brick, facing broadestside into the wind), these must be factors when estimating horsepower  wouldn't you think? The GTech's "ballpark" figures might register 25 horsepower :) , if even that much :) . However, I can see its value  it's too bad that I didn't have a GTech BEFORE I did the engine upgrade, to get baseline figures. Even if the GTech can't consider the GWagen's weight and aerodynamics, at least, comparing the before and after biased power estimates would have let me estimate the real power increases by applying the same relationship. It wouldn't be purely accurate, but at least it would be a better estimate. But getting back to be original request, by starting with baseline horsepower/torque values and an acceleration time, and comparing this to new acceleration figures, I thought that there might be some formula or mathematical relationship (raising the horsepower to some power or exponent (or something like that) that might provide some estimate. Ever hear of such a relationship? In any case, thanks very much  I enjoyed hearing from you. 
#4




From the GTech (www.gtechpro.com) website:
(at the bottom is the formula for calculating RW hp) The heart of the GTECH/Pro is a precision silicon accelerometer. An accelerometer is a sensor that measures acceleration, also known as GForce. GForce is what's keeping you in your seat as you are reading these words and if it wasn't for the seat you would be accelerating to the ground at 1G (32 feet per second per second). GTECH/Pro measures your speed and distance traveled by integrating acceleration over time. Basically, if you know how fast you are accelerating for a certain time period you'll know how much your speed has changed after that time period. So, if you start of from zero speed then you'll know what your speed is after every time period. These time periods are very small (2.5 milliseconds) and that's how GTECH/Pro maintains its accuracy. Distance is measured in the same fashion, if you know how fast you are going for a certain time period you will know what is the distance traveled during that time period. GTECH/Pro also measures delivered Horsepower, sometimes called Rear Wheel Horsepower. This horsepower includes the loss of power through the drivetrain which is usually 1015%. The formula for Horsepower is speed times acceleration times the weight of the vehicle. 
#5




F = M x A
Hi. Thanks for coming back, Glen. I'm glad that we can "talk", so to speak. Let's pursue this further....
The only trouble I have with that estimated "rear wheel horsepower" that you mentioned is that it must be a generalization that assumes "average" car weight (or mass) and "typical" Cdx (aerodynamics). My reasoning is that any realized acceleration is a function of not only the force applied (e.g., torque at the driven wheels) but also the mass (or what laymen commonly think of as its "inertia" to be overcome) of the object. In other words, given a certain amount of torque available, the heavier vehicle is not going to accelerate as fast as the lighter one; whereas, given two vehicles with identical weight, and engines with different torque output, the one with the lighter engine will not accelerate as quickly. In both cases, two vehicles accelerate at different rates, so how would that meter differentiate between a case of different weight vehicles, or different size engines? The converse of this question is, how does it know how much power is generated if it doesn't know such things as the vehicle's weight? Another consideration is that horsepower has to overcome resistances which, in the GWagen, must include a large amount of air resistance, especially increasing (squared) as speeds rise. And this resistance must be FAR greater than that of a car. Is there a means of inputting, to the GTech meter, the vehicle's weight and Cdx? Looking forward to hearing from you again. And thanks. 
#6




The GTech DOES require you to input vehicle weight to calculate HP. I would imagine that Cd is insignificant when accelerating 60 MPH. You would still need to subtract the approx. 1520% parasitic drivetrain loss to estimate crankshaft hp. On the G, it may even be more like 25%, but you would definitely be 'in the ballpark.'
ps. It's good 'talking' to you too...it's been a long time. Still have both W124s? 
#7




re: "...It's good 'talking' to you too...it's been a long time. Still have both W124s?..."
Yep, and the 201 that my daughter drives when she's home from school. But I'm glad to hear that the device DOES ask to input vehicle weight  it makes sense, and it makes the device's estimate more credible to me. I'll think I WILL ask around, locally, to see if anyone has one. Best regards, Ken 
#8




#9




It IS a great site
Thanks Kuan. I'm going to fool around with these functions and see what I can come up with.
Thanks again. Ken 
#10




For people with an aptitude for audio electonics there is an amazingly accurate method of measuring changes in horsepower. It does this by measuring the exhaust beat frequency over a constant acceleration course. You can achieve precision of <1%. Do a search on Home Dyno. You need a digital recorder and their program to analyze. This seems to be the most accurate (cheap method) for same car comparisons to determine whether performance mods have an effect. It does not do car to car comparisons very accurately.
[Edited by Lou Nielsen on 04052001 at 01:45 PM] 
#11




Attempts at calculations
Kuan,
Both of the formulae at the site that you recommended ( http://vette.ohioracing.com/hp.html ) treat the estimation of horsepower as a CUBED (i.e., raised to the power of 3) function. It would then stand to reason that my decrease in time (to 60 mph), from 17.4 to 13.0 seconds), comparable to the time to the quarter mile in the second graphed equation, also relate to the new horsepower as a cubed function. So, my Factor (for the increase in horsepower) would likely be (5650/(time2*5.825)^3)/(5650/(time1*5,825)^3)), where time1=17.4 sec., and time2=13 seconds. This is reduced to (time1/time2)^3 Does this sound about right? Well, this gives me a factor of 2.39 ! Meaning that to improve my 060 times, I increased my horsepower 2.4X. Or, that I've got 2.4*150 = 360 hp in my new GWagen. A nice dream, but these formulae don't hold water. I'd sooner believe a simple 33% increase in horsepower (i.e., 17.4/13 = 1.33), or 206 hp, but there's no basis to this simple  at least, not that I know of. Back to the drawing boards! :) But thanks, everyone. 
#12




What's Exhaust Beat Frequency
Hi, Lou.
Very interesting. But what exactly is exhaust "beat frequency"? Is that the audible pulsations in the exhaust note? Wouldn't that simply be a function of rpm (frequency that exhaust valves oven and close)? Or is there more to this? Please elaborate. It sound interesting. It does seeem, though, that it won't work for me because I no longer can get BEFORE data to compare to the AFTER vehicle that I now have  I'm assuming that it can make comparisons, for degree of improvement. Looking forward to hearing from you. 
#13




Another attempt at calculations
You know, what about this simple function?
1/((17.4/13)^3) This gives me a factor of 0.417. Could this be the proportion of horsepower increase? About 64 hp more? This would mean that I've got about 218 hp, which sounds about right. I'm going to have to think about this a little longer. Regards to all. 
#14




Ken, precisely. The beat frequency method counts the discreet explosions over a known time interval. You have to use the same section of road to normalize the elevation changes, etc. The program then counts the beat frequency and compares it to the next test. The differences in the rate of change is then calculated to a difference in HP. The weight of the vehicle, CD, the gear ratio, tire diameter, are all entered into the program. You need a computer, a microphone, and a digital recording device.
There is another post that mentions a similar method that measures spark impulses. 
#15




s / * / / /
Ken,
The * should have been a /. So 5650/(t2/5.825)^3)/(5650/(t1/5.825)^3 BUT, I actually took time to read the whole page I posted and the initial post. It makes more sense to me now but doesn't help you one bit. The formulae are intended for use with 1/4 mile times, not 060 times. Seems like the reason 1/4 mile times are used is because the distance travelled is known. I went to the library last night to try and figure out this problem. No luck but I did learn something: One horsepower is defined as the ability to move 33,000 pounds one foot in one minute. This might be why I can't find a formula which gives us hp for 060 times. I also figured out a coupla things that might be interesting to note. Using: Initial velocity = 0 Final velocity = (6*5280ft)/3600s=88ft/s then divide by 17.4s yields 5.05ft/s Second set of figures gave an acceleration of 6.77ft/s. This means average velocity for the original engine was (17.4/2)*5.05 = 43.9ft/s compared to (13/2)*6.77 = 44ft/s! The difference in average velocity is not much, so yeah, I don't think you have 350hp! So can we assume that it takes 150hp to move 5650lbs 763.86 feet in 17.4 seconds? If so, how many hp does it take to move 5650lbs 572 feet in 13 seconds? I'm having some difficulty here, someone jump in please? Hmmm... Kuan (shoulda stayed in skule) 
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