Author

Topic: HP/Torque Vs. MPH

JAAZZY Gearhead Posts: 777 From: Bay Area, CA Registered: Apr 2001

posted 08092001 07:08 PM
I figured somebody here would probably know the answer to this question.Let's say you had a 300rwhp & 320rwtq. The max hp was obtained at around 6000rpm but the motor reved freely up to 6800rpm which would translate to about 170mph. Okay now let's say you built another larger motor with around 370rwhp & 380rwtq with the max hp coming at about 5400rpm but the car only reved easily up to 6200rpm. Would it pull a lower top speed a higher top speed or about the same? Everything except the motor is the same. This is not a test or anything I am really just curious.
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steve'66 Gearhead Posts: 8826 From: Sonoma,CA,USA Registered: Mar 2000

posted 08092001 07:45 PM
Jas,If your car ran 170 @ 6800 rpm, it will run 155mph @ 6200 rpm. Increased HP and torque will just make it accelerate quicker. You're gonna need a taller gear to take advantage of the increased power at lower rpms. SteveW
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JAAZZY Gearhead Posts: 777 From: Bay Area, CA Registered: Apr 2001

posted 08092001 08:09 PM
I am guessing then that this is true in all scenarios. I was thinking about this in a couple of different ways but if that is true I will need to change gears eventually. Who know I might even go to a 6spd if things work out well in the next 18 months.BTW, the hp/torque numbers are hypothetical since the car has never been on a dyno. quote: Originally posted by steve'66: Jas,If your car ran 170 @ 6800 rpm, it will run 155mph @ 6200 rpm. Increased HP and torque will just make it accelerate quicker. You're gonna need a taller gear to take advantage of the increased power at lower rpms. SteveW
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Mike_R Gearhead Posts: 176 From: Indianapolis, IN 46237 Registered: Feb 2001

posted 08102001 09:15 AM
I think the engine with more horsepower and more torque will always pull a better MPH regardless of what RPM it occurs at. This is assuming you're gearing is adjusted accordingly like Steve mentioned.This is an issue I've really been thinking about alot lately too. Do I build a 513 cubic incher that I shift at say 6500 or a 460 cubic incher that I shift at say 7000 RPM. It seems that either can make good peak horsepower, but is there a clear advantage in the extra cubic inches? I'm talking striktly in the horsepower and torque department, not durability. If I could sort of open up a related question, if two engines basically have similar horsepower across the board, but one has generally more torque across the board will it run stronger in the 1/4 mile?
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SG236 Gearhead Posts: 416 From: Jasper, TN, USA Registered: Dec 2000

posted 08102001 10:44 AM
Mike, I'm currently building a 513" stroker. My primary concern is for the car to run around 6.00s in the 1/8 and not have to turn the car to hard so it will last. I plan on shifting the car a 6400 according to my figures. I initially thought I would just build the 460. My issue was cost. I believe it is just as cheap to build a 513. I am going with the 6.8 rod, 4.440 bore and 4.14 stroke. If you look at the cost of maintenance in the future I believe the 513 is the better way to, at least in my situation. Just my .02
 Russ Hood 70s Maverick S/Pro https://mustangsandmore.com/ubb/SG236.html
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bluestreek Gearhead Posts: 1724 From: Athens,GA Registered: Jul 2001

posted 08102001 11:15 AM
Horsepower = Torque * RPM /5252Torque = Horsepower * 5252/RPM HP, Torque, and RPMs are directly related to each other.. If you compare 2 engines with the same amount of HP and one has more Torque, then the higher Torque engine would have to be making more Torque at a lower RPM.. Theoretically, if you gear the car with the higher Torque number, to keep the RPMs in the "Peak Power" level of the engine, then the higher Torque engine will always accelerate faster given the same set of circumstances. Also consider this: If you compare 2 engines and both have the same amount of Torque and HP, and one has "Peak Power" in a higher RPM range, then you will be able to generate more Power to the rear wheels because of Torque multiplication coming from the ability to use a better gear ratio.. [This message has been edited by bluestreek (edited 08102001).]
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Mike_R Gearhead Posts: 176 From: Indianapolis, IN 46237 Registered: Feb 2001

posted 08102001 12:02 PM
I think we could probably all agree that more cubic inches when possibly or legal is a good thing, but do you guys feel that there is a direct correlation between raising cubic inches and raising horsepower (ie the old 1 hp per cubic inch added rule)? I always heard if you take two engines prepared exactly the same but one is 20 cubic inches larger than the other that it would have about 20 more horsepower as a general rule. This crazy Desktop Dyno has me questioning that theory. It seems to show that you raise torque with added cubic inches but horsepower stays the same or even goes down. However when I look at the articles on larger cubic inch engines they always seem to make a bunch more horsepower than smaller engines with similar combos so that would tend to make me stick with my belief that big cubic inches will lead to big horsepower if you feed them enough. So how do you guys feel? Is adding cubic inches alone directly add horsepower or does it just give you more torque and more potential for horsepower. Sorry for the long message, it's one I've been trying to clear up in my mind.
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bluestreek Gearhead Posts: 1724 From: Athens,GA Registered: Jul 2001

posted 08102001 01:04 PM
I like Blacktop dynos better!!
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Daniel Jones Gearhead Posts: 813 From: St. Louis, MO Registered: Aug 99

posted 08102001 03:21 PM
> HP, Torque, and RPMs are directly related to each other.. Yes. See the tech tips section for a detailed explanation: https://mustangsandmore.com/ubb/DanJonesAerodynamics.html >If you compare 2 engines with the same amount of HP and one has more Torque, >then the higher Torque engine would have to be making more Torque at a lower >RPM.. Correct. >Also consider this: If you compare 2 engines and both have the same amount >of Torque and HP, and one has "Peak Power" in a higher RPM range, then you >will be able to generate more Power to the rear wheels because of Torque >multiplication coming from the ability to use a better gear ratio.. No. Rear wheel torque is what accelerates a vehicle. Horsepower, being the rate at which torque is produced, is the indicator of how much potential torque multiplication is available. In other words, horsepower describes how much engine rpm can be traded for tire torque. It doesn't matter where in the RPM band the horsepower is made. The word "potential" is important here. If a car is not geared properly, it will be unable to take full advantage of the engine's horsepower. It's fairly basic physics to derive the equations for top speed. A vehicle will continue to accelerate until the total external resistive forces (aerodynamic drag and rolling resistance) cancel out the propulsive force As a first approximation, lets ignore the effects of rolling resistance and concentrate on aerodynamic drag. The power required to overcome the drag force can be expressed as: Preq = D * V where: Preq = power required D = aerodynamic drag V = velocity Drag is usually expressed in terms of nondimensional coefficients: D = Cd * A * q where: Cd = coefficient of draq A = frontal area q = dynamic pressure and: q = (rho * V**2)/2 where: rho = air density (a function of temperature and altitude) Substituting into our original equation yields: Preq = Cd * A * (rho * V**2)/2 * V = (Cd * A * rho * V**3)/2 Using typical units: A in ft**2 V in MPH Preq in HP Noting: 1 HP = 500 ftlb/sec rho = 2.3769E03 lb*sec**2/ft**4 (sea level standard day) 5280 feet = 1 mile 1 hour = 3600 seconds Assuming I've done the math correctly, the equation reduces to: HPreq = 6.817E06 * (Vmph**3) * Cd * A Rearranging yields: / / HPreq Vmax = 52.739 * \ 3 /  \ / Cd * A \/ Plugging in some numbers for a cars frontal area (say 30 square feet) and drag coefficient (0.32) and solving for 175 MPH: HPreq = 6.817E06 * (175**3) * 0.34 * 30 = 373 HP (rounded to the nearest horsepower) We've ignored the effects of rolling resistance but it typically scales with velocity raised to something a fraction over one, so the effect should be small compared to the aerodynamic drag. As a crude rationality check, Mad Dog Antenucci competed in the 165 MPH class at the Silver State run in his Pantera. He was running a 377C that made something on the order of 400 HP at the rear wheels and was able to average right at 165 MPH, hitting something like 175 MPH in spots, IIRC. Given the equation above, that seems to be in the right ballpark. Panteras are fairly slick and Ford supposedly had one in the wind tunnel along with GT40. I think the published number for a skinny tired 1971 model was 0.29. Wider tires and wings increase that a bunch. Dan Jones
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bluestreek Gearhead Posts: 1724 From: Athens,GA Registered: Jul 2001

posted 08112001 12:17 AM
Dan, I think you misunderstood my intentions..My statement was targeted at a simple, short distance, "drag race" comparison, where two cars are exactly the same, but one has an engine that produces the same HP and Torque at the CRANKSHAFT, but in a higher RPM range.. The engine that can produce it's peak HP at the highest RPM range can take advantage of using a more powerful gear ratio to complete the 1/81/4 mile..It will also use more air and fuel to feed the higher rpms and, once it is coupled with an optimum gear ratio, will produce more "rear wheel power".. It won't be the most efficient, but it'll the quickest!!That's my story and I'm stickin to it!!
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steve'66 Gearhead Posts: 8826 From: Sonoma,CA,USA Registered: Mar 2000

posted 08112001 12:46 AM
There are two different topics to this thread. Jas's original question was about maximum attainable top speed from his Stang. If it ran 170 @ 6800rpm before what would it run with more hp and torque at 6200rpm. That was easy, 155mph. Jas has run the Silver State deal, and wants to know the effect of a new higher hp/torque combo that is limited in rpm. The result is a slower, but quicker car. Unless he changes gear ratios to take advantage of the lower rpm hp and torque. I loved your analysis Dan, and I'm sure Jas did too! It's great info Now how fast it'll be in the 1/4 mile is a different ? SteveW
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bluestreek Gearhead Posts: 1724 From: Athens,GA Registered: Jul 2001

posted 08112001 01:11 AM
Sorry, I got off track.. This is one of those topics that make you go Hmmmmmmmmm???
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JAAZZY Gearhead Posts: 777 From: Bay Area, CA Registered: Apr 2001

posted 08112001 05:12 AM
Actually I read the replys with interest but didn't respond because I didn't have much to add. I've actually read quite a bit about the relationship of HP/Torque. I remember it for about a minute each time.
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bluestreek Gearhead Posts: 1724 From: Athens,GA Registered: Jul 2001

posted 08112001 09:40 AM
Dan, I did some research and found out that you were right and I was wrong. I stand corrected.. However... it is possible to have two comparative engines with the same amount of Torque, and one can be more powerful than the other.. Here's how: HP = Torque x RPM / 5,252 kW = Nm x speed / 9549) Because the maximum torque for a given displacement is constant, the power per cubic inch is directly proportional to engine speed. Since engine maximum speed is inversely proportional to engine stroke, the shorter the stroke, the more power per cubic inch the engine can develop.
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Daniel Jones Gearhead Posts: 813 From: St. Louis, MO Registered: Aug 99

posted 08112001 12:05 PM
>Now how fast it'll be in the 1/4 mile is a different ? Newton's Second Law of Motion covers that as well and states that the sum of the external forces acting on a body is equal to the rate of change of momentum of the body. This can be written in equation form as: F = d/dt(M*V) where: F = sum of all the external forces acting on a body M = mass of the body V = velocity of the body d/dt = time derivative For a constant mass system, this reduces to the more familiar equation: F = M*A where: F = sum of all the external forces acting on a body M = the mass of the body A = the resultant acceleration of the body due to the sum of the forces For top speed, the same basic equation applies but we only care about one point, that being th point at which the car stops accelerating. For drag strip acceleration, you have to worry about the summation of all the points on the RPM curve between each shift. In mathematics, this summation is known as the integral of the curve. If you read the article I noted earlier (https://mustangsandmore.com/ubb/DanJonesTorqueVsHP.html), it derives the above then goes on to relate crankshaft horsepower to rear wheel torque. Plagarizing from my own article: "Knowing the basic physics outlined above (and realizing that acceleration can be integrated over time to yield velocity, which can then be integrated to yield position), it would be relatively easy to write a simulation program which would output time, speed, and acceleration over a given distance. The inputs required would include a curve of engine torque (or horsepower) versus rpm, vehicle weight, transmission gear ratios, final drive ratio, tire diameter and estimates of rolling resistance and aerodynamic drag. The last two inputs could be estimated from coast down measurements or taken from published tests. Optimization loops could be added to minimize elapsed time, providing optimal shift points, final drive ratio, and/or gear spacing. Optimal gearing for top speed could be determined. Appropriate delays for shifts and loss of traction could be added. Parametrics of the effects of changes in power, drag, weight, gearing ratios, tire diameter, etc. could then be calculated. If you wanted to get fancy, you could even take into account the effects of the rotating and reciprocating inertia (pistons, flywheels, driveshafts, tires, etc.). Relativistic effects (mass and length variation as you approach the speed of light) would be easy to account for, as well, though I don't drive quite that fast." >I did some research and found out that you were right and I was wrong. I >stand corrected.. However... it is possible to have two comparative engines >with the same amount of Torque, and one can be more powerful than the other..
Correct. In the HP versus Torque article, I given an example of how a lower torque engine can have more power and therefore better acceleration than a higher torque but lower power engine. >Because the maximum torque for a given displacement is constant, I think you mean "if" the torque is constant. Displacement does not necessarily determine torque. >the power per cubic inch is directly proportional to engine speed. The cubic inch part is not important. The important part is that the same torque, developed at two different speeds, will result in two different horsepower numbers. The one developed at higher RPM will make more power. >Since engine maximum speed is inversely proportional to engine stroke, >the shorter the stroke, the more power per cubic inch the engine can >develop. Two main issues here. The first is a structural issue. A shorter stroke implies a slower piston speed and therefore less load on a crankshaft for a given RPM, all other things equal. Secondly, for a given displacement, a shorter stroke has a larger bore which means more room for larger valves and the potential for better breathing. There are lots of other issues involved but those are the two basic ones. Dan Jones
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bluestreek Gearhead Posts: 1724 From: Athens,GA Registered: Jul 2001

posted 08122001 01:26 PM
Thanks Dan, I knew there was a logical explanation.. Sometimes I tend to generalize on certain things, not really understanding the specifics of "how and why".. It's a good thing I had clean feet.. Later, Dan
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