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Old 04-13-2010, 04:03 PM   #11 (permalink)
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...the basic constant-power approximation (CPA) equations for ET and MPH from vehicle WT and HP are:

ET = 5.82 * (WT / HP )^(1/3)

MPH = 232 * ( HP / WT )^(1/3)

...or, approximately: ET = 1350 / MPH

...let me know if you want to know *where* the constants 5.82 and 232 came from.

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Old 04-13-2010, 05:02 PM   #12 (permalink)
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1 horsepower is about equal to 746 watts.

HowStuffWorks "How Horsepower Works"
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Old 04-13-2010, 06:50 PM   #13 (permalink)
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According to Watt, one HP is 33,000 foot pounds of work per minute

HP and torque curves cross at 5252 rpm, not open for debate.

http://vettenet.org/torquehp.html
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Old 04-13-2010, 07:38 PM   #14 (permalink)
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Horsepower is a measure of work done over time. Torque is a measure of turning force. An engine will maintain max vehicle speed at its horsepower peak, where it's doing the most work per unit of time. It will accelerate the hardest at its torque peak, where it's turning the driveshafts the hardest.
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Old 04-13-2010, 07:42 PM   #15 (permalink)
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My transmission has more than one gear too, makes a huge difference. Coupled to a smaller engine and body makes for more smiles (or less frowns) at the pump.
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Old 04-13-2010, 08:05 PM   #16 (permalink)
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Quote:
Originally Posted by comptiger5000 View Post
It will accelerate the hardest at its torque peak, where it's turning the driveshafts the hardest.
Are you sure? Since horsepower is a measure of work done per unit time, I think it's analogous to acceleration. You want the most work done per unit time to accelerate the vehicle as quickly as possible. If you had 500 ft-lbs of torque, but only 1 rpm at your disposal, the vehicle wouldn't accelerate very fast. 500 ft-lbs with 5000 rpm will get you moving. You need the combination of torque and speed (rpm), which equals horsepower, to get the acceleration. So the vehicle should accelerate the hardest at its horsepower peak (assuming you can keep the engine there while the vehicle accelerates). A CVT would be great. Or am I missing something?
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Old 04-13-2010, 08:24 PM   #17 (permalink)
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you must be correct, that peak POWER (assuming you get the gearing right) is where best acceleration happens. Though I'm happier with most efficient, not fastest, acceleration.
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Old 04-13-2010, 08:45 PM   #18 (permalink)
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As much as I love my Jeeps own I-6 for it's torque, but I definately pay at the pump for it as everytime I take off from a stop, it eats a ton of gas whether I'm super easy or not on the throttle, but at least it gets good mpg cruising.
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Old 04-13-2010, 08:52 PM   #19 (permalink)
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Yeah, that'll happen when gearing doesn't match the torque curve... I keep telling people that my truck only really ever needed 3 gears... get going, keep going, and really go.

None of those gears would ever need to see any higher than 2400-2500 RPM.. Hell, I can run my I6 down to 700ish before it even feels like it's having any sort of difficulty moving the truck. I'd imagine that ~1,000 RPM would be the perfect cruise location for Gerald, which is why I still want to find a gear splitter for the rear axle. As it sits, I cruise at 2k ish @ 60 MPH... consequently, that's my torque peak... With a splitter, I could be in 5th(2) at 60-65 running ~1,000 RPM, and if I'm towing or need the extra torque, I could simply switch back to 5th(1) and be right back at my peak torque number, and it wouldn't affect my 4wd parts, either. I could run 4x4 with the splitter on (1), which is a simple 1:1 gear interface.
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Old 04-13-2010, 09:03 PM   #20 (permalink)
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I think the rule of thumb is that gearing changes are marginalized without a corresponding aerodynamic change, it moves you left (rpm) on the bsfc chart, but since the power requirements are basically the same it moves you up the load axis. So if you happened to be southeast of the bsfc sweet spot then you might get lucky, but generally it doesn't work like that.

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