rabbit_diesel Posted October 2, 2010 Share Posted October 2, 2010 (edited) http://craig.backfire.ca/pages/autos/horsepower Wow! An even better article about horsepower and torque! Whoever said "High average power wins races.", is right. In my opinion, that is what we should be classing upon, instead of torque; and this is what the fellows were aiming for when they put torque in the rules. (Edit: though they may have been aiming for it, they missed the mark.) Will Edited October 2, 2010 by Guest Quote Link to comment Share on other sites More sharing options...
001GTSM3 Posted October 2, 2010 Share Posted October 2, 2010 (Not having been around at the time, I do not know personally; and the threads from that period are no longer on-line.) Does anyone remember what year the GTS Challenge was begun? I did a bit of looking, and the earliest mention that I found on the web, was early in 2003. Will Wow, guess I'm now that old guy that raced back in the day. The first GTS race was run in April 2003 at Mid-Oh. The wt/hp rule set was not in place, rather cars were classed based on where the fit in other series such as BMWCCA, PCA, SCCA, etc. -Sean Tillinghast Quote Link to comment Share on other sites More sharing options...
cstreit911 Posted October 2, 2010 Share Posted October 2, 2010 Although that article also states " It does not matter if the engine makes power by revving high or making a lot of torque, because drivetrain gearing can be used to adjust the torque and revs proportionally. " which seems to contradict your point? I think I understand the point though. If my engine makes 200# of torque at 1000 RPM, to generate speed from it, it must be geared down, this reducing the available or delivered torque to the wheels. (because the trans is a force multiplier of divider) If my engine makes 200# of torque at 9000RPM the same torque is delivered to the trans, but because it is at a higher RPM I don't need to gear down as much so more is delivered to the rear wheels. Is that the essence of it? Quote Link to comment Share on other sites More sharing options...
rabbit_diesel Posted October 2, 2010 Share Posted October 2, 2010 (edited) Although that article also states " It does not matter if the engine makes power by revving high or making a lot of torque, because drivetrain gearing can be used to adjust the torque and revs proportionally. " which seems to contradict your point? Actually, my point agrees with his. When he writes: "It does not matter if the engine makes power by revving high or making a lot of torque [...]", he is meaning for us to think of two engines of the same horsepower, such as one engine making 200 HP at 8000 RPM, and the other making 200 HP at 4000 RPM. He goes on to say that in both cases the gearing will adjust things such that the car will accelerate at the same rate with either engine. This is precisely what I have been trying to say, and is the reason why I think that we should NOT be averaging TQ with HP (to include TQ, upsets the relationship). If my engine makes 200# of torque at 1000 RPM, to generate speed from it, it must be geared down, this reducing the available or delivered torque to the wheels. (because the trans is a force multiplier of divider) If my engine makes 200# of torque at 9000RPM the same torque is delivered to the trans, but because it is at a higher RPM I don't need to gear down as much so more is delivered to the rear wheels. You are getting closer to understanding. Please let me describe a particular example. Most cars have wheels of such a diameter that 1000 RPM of the wheels will make it go in the vicinity of 60 MPH (more or less depending on the actual diameter). For the sake of this discussion, let's assume it is so for this mythical car. 1) If an engine makes a peak of 200 HP at 1000 RPM, it will have 1050 ftlb of torque at 1000 RPM [using the formula TQ = (HP * 5252) / RPM]. At 60 MPH, this mythical car will need no ratio change in the gears (in other words the ratio is 1:1) and all 1050 ftlb are delivered directly to the wheels. The 1050 ftlb at the wheels will produce some particular acceleration of the car at 60 MPH, depending on the vehicle weight. 2) If an engine makes a peak of 200 HP at 9000 RPM, it will have 116.7 ftlb of torque at 9000 RPM. At 60 MPH, the mythical car will need a ratio change of 9:1 in the gears in order to make the wheels turn at 1000 RPM. The 9:1 gearing while decreasing the shaft speed on the output, will be increasing the torque on the output; the torque on the output will be (9 * 116.7 ftlb) = 1050 ftlb at 1000 RPM. The same amount of torque and speed at the wheels as above. Because torque at the wheels is the same, and the weight is the same, the acceleration at 60 MPH will be the same. For the sake of continued discussion, let's assume that both engines have a peak torque which is the same as their torque at their peak HP (a situation known as 'no torque rise'). If we apply the GTS rules to the engine in case #2, the (TQ+HP)/2 does not apply, and this engine can be used as is. If we apply the GTS rules to case #1, the (TQ+HP)/2 will apply, and the engine will be assigned a 'power value' of 625. In order for this fellow to stay in the same class, he will have to restrict the engine to 64 HP. Since this engine makes its peak power at 1000 RPM and it has no 'torque rise', it will now have a peak torque of 336 ftlb ---> (336 + 64) / 2 = 200 assigned 'power value'. The problem with this is that since his real power is about one third of case #2, his acceleration will also be one third. In addition, his top speed will be greatly reduced. With a third the acceleration, and greatly reduced top speed, how is he expected to compete well with the other cars in his class? [Edit: If instead of restricting in order to stay in class, the car is moved up several classes, it would be competing against other cars with about three times the acceleration capability and much higher top speed.] ------------------- If the GTS rules were comparing the torque available at the wheels at some particular speed of the wheels, there would be no problem. Notice that: 'torque at speed' is another way of saying 'power' (real power), according to the formula HP = (TQ * RPM) / 5252. Will Edited October 2, 2010 by Guest Quote Link to comment Share on other sites More sharing options...
KTL Posted October 2, 2010 Share Posted October 2, 2010 Chris, You've got the right idea, but don't give the high rpm torque too much credit. There's an advantage in that you don't have to gear down as much to reach that torque. However the high torque peak gives up accelerative force in the lower range. Lets consider another engine type for a second. Think about why those little Baby Grands can be quick despite their severe lack of torque. Yes they're very light but they also have to rev like mad to make their power. Kind of opposite of the diesel, yet they still can make their power work for them- through gearing. Lastly, i think this is where the diesels have some difficulty. Think of an extreme case like a semi. They have 18 gears because they have a narrow rev span, make ginormous torque and need to be able to travel at reasonably high wheel speeds on the highway. However I don't see that same handicap with the racing diesels because their rev range is better and they don' need to maintain massive torque mutiplication to move its load. Quote Link to comment Share on other sites More sharing options...
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