70 replies to this topic

[Cnj]

I think Jim gets to grin and bear it for a while.  I generally like the guy, but I think he has earned a substantial amount of public shaming, even on his own site.

Rob,

I understand your point but I'm not here to defend Jim or anyone else. My view is that I'll always take rational forum debate and practical proposals for better/strong compliance enforcement over a move to public humiliation - which at best is temporal, emotional, unevenly applied - and which I doubt leads to the intended outcome.

CNJ

[RWP80000]

"Except that we are not drag racers racing in a vacuum so adding 100 pounds would be ridiculous given the disparity it would create for all the time we spend with the steering wheel turned and/or foot on the brake.  Also that math doesn't take any aero into account.  Just the acceleration calculation alone that should be used for setting weight (with no regard for handling) is a very complicated calculation. There are software calculators out there for this if someone wants to do it right.  This is why a restrictor plate is a more reliable equalizer so that all race at the same weight."

 

Tom,

From your quote I see you disagree with the methodology.  I also see I am behind the times as they are now drag racing in a "vacuum".  There is no denying I am "old school" but I think you are trying to make it much more complicated than it needs to be.  As for the math not taking in any "aero", why should it?  For the factor we are trying to compensate/correct (difference in engine power output) aero doesn't matter, unless you are adding the weight to the car using a sprint car wing. (Maybe your thinking inertia.)

 

Some of my "old school" basic equations of motion are as follows;

 

Force (F) = Mass (M) x Acceleration (A).  (F=MA) This can also be rearranged to define Acceleration (A) = Force (F) / Mass (M)  (A=F/M)

Another basic equation is that Work (W) = Force (F) x Distance (D) or from the above Work = Mass (M) x Acceleration (A) x Distance (D) ie (W=MAD)

The next equation involved here is Power (P) = Work (W) divided by Time (T)  (P=W/T) replacing "W" with "MAD" then Power (P) = MAD/T

 From this last equation (P=MAD/T) you can now see that time (T) will always reduce your Power (P) no matter how (MAD) you are. 

 

If you don't like working with weight to power relationship how do you feel about weight per displacement.  With the situation of the .010/.25 mm overbore, does the current 15 pounds penalty seem appropriate?  In the case of overbore we are dealing with an increase in engine displacement.  If we look at a the weight to displacement ratio for the base 1.8L in a 99 car, it is 2400 pounds divided by 1839.60 cc for a ratio of 1.3046 pounds per cc.  When you calculate the new displacement for the .010 overbore you get 1850.70 cc for a net gain of 11.1 cc. Multiply the 11.1 time the 1.3046 ratio and you get 14.48 pounds of added weight.  This seems reasonable because the type of dimensional change made would not be expected to radically change the engine horsepower output characteristics thereby supporting a weight change directly proportional to the change in displacement.

 

If you now go back and analyze the engine displacement change on the basis of power output assuming an arbitrary but realistic 125 hp for the base motor with a 1839.6 cc displacement you get a 0.06795 HP per cc ratio. Multiplying this ratio times the additional 11.1 cc's shown above yields a 0.7542 expected HP increase, or a total 125.7542 HP.  Now using weight to power ratio here with the baseline engine of 125.0 HP at 1839.60 cc's, and a weight of 2400 pounds, the weight to power ratio is 19.2.  Multiplying the 19.2 ratio by the 125.7542 total HP results in 2414.48 pounds.  So we have arrived at the exact same weight correction for the overbore based on either the proportional change in displacement or the proportional change in HP using one common assumption the the overbore does not radically change the engine horsepower output characteristics but rather assumes they are proportional in both cases.

 

 Since we add 14.48 pounds for what could be a 0.7542 projected HP increase, is it that unrealistic to anticipate a potential (14.48 lbs. times 5.0 hp/0.7542 hp)=96 lbs. penalty?

(Even though I did originally raise concern for using a peak horsepower value in making this type of estimate.)

[38bfast]

All your equations do not take in acount HP vs areo drag. At about 60 mph the hp to weight curve intersects hp to areo curve. So above 60 you are not fighting mass you are fighting areo to accelerate the car. So adding weight to the car only works for down low. Weight also affects Cornering Gs and braking .

[RWP80000]

All your equations do not take in acount HP vs areo drag. At about 60 mph the hp to weight curve intersects hp to areo curve. So above 60 you are not fighting mass you are fighting areo to accelerate the car. So adding weight to the car only works for down low. Weight also affects Cornering Gs and braking .

 

The point I am trying to make is that at the speeds where the aero is in play it is a constant HP requirement and all you are doing is using more of what power you have to overcome the drag.  Lets just say that the HP drag at a typical average lap speed of 75 mph is 25 HP and your Plunge cut Pro head makes 130 hp while your 5 HP reduced stock head engine is 125 HP.  You still have a 5 HP difference for accelerating the vehicle mass, it is just now that you have 105 hp available for the Plunge cut Pro head and 100 HP available for the Stock head with the stock head when adding weight equalize the power to weight ratio.

 

The weight to power ratio for the stock head using the "average lap aero speed" becomes  2400 ls/100 HP=24 lbs/HP. Using 24 lbs/HP times 105 HP available for the Pro head, the weight now becomes 2520.  Now you have a case for a 120 lb weight penalty for the 5 HP delta. As you can see I think including aero only magnifies the disparity and while the main issue here is we are calculating on a Peak HP at the rear wheels.  Using flywheel HP as on an engine dyno and  eliminating all drivetrain losses is necessary when using weight to power ratio for determining added weight. It should also be based on the difference in HP under the curve for the typica RPM operating range and better yet based on a RPM histogram in order to properly weight the time typically spent along the power curve.  But where does that information exist?

[Tom Sager]

The point I am trying to make is that at the speeds where the aero is in play it is a constant HP requirement and all you are doing is using more of what power you have to overcome the drag.  Lets just say that the HP drag at a typical average lap speed of 75 mph is 25 HP and your Plunge cut Pro head makes 130 hp while your 5 HP reduced stock head engine is 125 HP.  You still have a 5 HP difference for accelerating the vehicle mass, it is just now that you have 105 hp available for the Plunge cut Pro head and 100 HP available for the Stock head with the stock head when adding weight equalize the power to weight ratio.

 

The weight to power ratio for the stock head using the "average lap aero speed" becomes  2400 ls/100 HP=24 lbs/HP. Using 24 lbs/HP times 105 HP available for the Pro head, the weight now becomes 2520.  Now you have a case for a 120 lb weight penalty for the 5 HP delta. As you can see I think including aero only magnifies the disparity and while the main issue here is we are calculating on a Peak HP at the rear wheels.  Using flywheel HP as on an engine dyno and  eliminating all drivetrain losses is necessary when using weight to power ratio for determining added weight. It should also be based on the difference in HP under the curve for the typica RPM operating range and better yet based on a RPM histogram in order to properly weight the time typically spent along the power curve.  But where does that information exist?

Ralph's comments above are correct.

 

We're certainly getting way off the Sebring topic here, apologies for that.  If we are talking about using weight to handicap and equalize acceleration between 2 similar (or not similar), we need a bunch of data about the cars.  Best I can remember the software I've seen asks for the following inputs: Dyno torque and HP curves, car weight, frontal area, drag coefficient, gear and final drive ratios, tire diameter, redline or upshift RPM and time for driver to shift.  With all this acceleration curves can be produced and at the speeds at which we race. Even though the evaluation in our case is to compare 2 cars with the same attributes (let's say a '99 vs another '99), you still need the inputs to calculate acceleration.   

 

Problem with this is that if you get acceleration equalized, you are then faced with calculating or evaluating what the weight difference will do to handling and lap times. A very difficult task to say the least and if you are searching for what weight will make cars equal in terms of lap times then you'll end up with a slightly heavier more powerful car and a slightly lighter less powerful car.  

 

Restrictor plate just makes equalization a lot easier than weight.  If power can be made the same then the cars can weigh the same and drive the same. 

[Steve Scheifler]

What's often forgotten about these small HP differences is the degree to which aero/drag nullifies them at higher speeds. Because drag increases at the square of velocity, the difference in acceleration that 4hp makes at 50 mph is reduced to the equivalent of 1hp at 100 mph. Not sure how that fits into your debate, but worth keeping in mind.

Generally speaking, I side with Tom in that it is best to do as much as possible to bring actual power close, and use weight as a last resort. When weight is used, the impact on traction (cornering speed and braking) is significant and needs to be considered.

[Rob Burgoon]

That doesn't sound right to me. If at 100 mph it takes 100hp to knock the air out of the way, one guy has 20 net HP, the other has 24 net HP. I think that would be quite obvious without a draft.

[Steve Scheifler]

Your math looks OK, but I think the difference is that the power to accelerate from 1 mph to 2 mph is less than required to accelerate from 101 to 102. Right? So although the car with more power is still accelerating faster, the difference in the rate of change diminishes as drag increases with the square of velocity. Or have I left something out?

[RWP80000]

Steve,

I agree in principle with what you stated but I would suggest the word "Work" in place of "Power" as "Work" encompasses all the elements involved to change the state of the vehicle 1 mph  (from either 1 vs 2 mph  or 101 vs - 102 mph).

 

Power is defined as the Rate (time) of doing Work. Work can be calculated from "Power X Time" , "Force X Distance" or "Mass X Acceleration X Distance".  So the "Work" term accounts for the reduced power available, increased time to accelerate as well as the increased distance over which all of this is occurring.  Mass would be the only thing remaining constant here.

[Steve Scheifler]

Agreed. I am a stickler on the point that choice of words matters. "Work" then.

[Andrew Charbonneau]

lol what a joke im going with stock head and restrictor plate out should help me

[Rob Burgoon]

So assuming a 4:1 rear ratio, in 4th 1:1, car 1 has 480 ft/lbs of force, car 2 has 500 ft/lbs of force.  At 100mph 400 ft/lbs are used to knock the air out of the way.

 

Car 2 is accelerating using 100ft/lbs of static torque, Car 1 80 ft/lbs.  Car 2 appears to be accelerating 25% harder than car 1.

 

At 20mph in 4th (heh), car 1 accelerates with car 1 has 480 ft/lbs of force, car 2 with 500 ft/lbs of force.  Car 2 appears to be accelerating 4% harder than car 1.  Same percentage if they are in 2nd gear.

 

F=ma, no matter what speed you're doin.  But boy will you notice a horsepower gap at high speed.

[Steve Scheifler]

More good math, but percentages can give an inflated impression. Take it to the extreme where car 1 hits terminal velocity and car 2 still has those 20 lb-ft (if you want to be correct) of torque left with which to do work. 20 vs zero is REALLY impressive as a percentage, but now calculate how much difference that makes when trying to push your cute little brick through the air at over 100 mph. Because the NEXT mph requires SO much more work than those first few mph, that infinitely higher percentage of available torque makes very little difference, and before you know it it's time to brake.

And the gearing IS important, because at 20 mph you are of course in a lower gear where the difference multiplier is higher. The percent difference is the same as in higher gears, but the absolute difference is greater. Meanwhile drag is lower so the extra torque accomplishes more useful work at low speed rather than wasted effort pushing against the air.

[Steve Scheifler]

To take the numbers out all together and paraphrase what Rich was saying about work, as the speed increases the rate of change decreases (due to gearing/up-shifts and drag) so in absolute terms the difference in the rate of change also decreases, from both gearing and drag.

At least that sounds reasonable to me. ¯\_(ツ)_/

[Andrew Charbonneau]

So assuming a 4:1 rear ratio, in 4th 1:1, car 1 has 480 ft/lbs of force, car 2 has 500 ft/lbs of force.  At 100mph 400 ft/lbs are used to knock the air out of the way.

 

Car 2 is accelerating using 100ft/lbs of static torque, Car 1 80 ft/lbs.  Car 2 appears to be accelerating 25% harder than car 1.

 

At 20mph in 4th (heh), car 1 accelerates with car 1 has 480 ft/lbs of force, car 2 with 500 ft/lbs of force.  Car 2 appears to be accelerating 4% harder than car 1.  Same percentage if they are in 2nd gear.

 

F=ma, no matter what speed you're doin.  But boy will you notice a horsepower gap at high speed.

I ask rob do you really git it you maroon are you coming to Fla for a ass beating 

[Andrew Charbonneau]

did you all forget it is racing  im driving a noble power plant miata

[Ron Alan]

And the gearing IS important

 

Hmmm...would a 4.44 to 4.7  diff help a 1.6? Higher rpm for same speed out of corners...torque?

[Steve Scheifler]

Diffs giveth, and they taketh away. It depends on the track overall and corner by corner. While in the same gear as your competition you definitely have the edge, but with a lower gear you also hit redline at a lower speed and shift sooner. Now you are at a gearing disadvantage until he up-shifts. So depending on the combination of exit speeds and distance to the next corner you may end up with a net gain or a net loss. Add to that the possibility that you need to shift more total times per lap or find yourself at the wrong RPM at exit more or less often, and there are a lot of variables making the cars less the same rather than more, even if lap times are closer in some cases.

[Ron Alan]

Diffs giveth, and they taketh away. It depends on the track overall and corner by corner. While in the same gear as your competition you definitely have the edge, but with a lower gear you also hit redline at a lower speed and shift sooner. Now you are at a gearing disadvantage until he up-shifts. So depending on the combination of exit speeds and distance to the next corner you may end up with a net gain or a net loss. Add to that the possibility that you need to shift more total times per lap or find yourself at the wrong RPM at exit more or less often, and there are a lot of variables making the cars less the same rather than more, even if lap times are closer in some cases.

Welcome to the NA 1.8 world

 

Centrifical clutch and belt(snowmobile!)???  Sorry...just looking for ways to get David(and others)their torque! 10-1 helps!

[Rob Burgoon]

More good math, but percentages can give an inflated impression. Take it to the extreme where car 1 hits terminal velocity and car 2 still has those 20 lb-ft (if you want to be correct) of torque left with which to do work. 20 vs zero is REALLY impressive as a percentage, but now calculate how much difference that makes when trying to push your cute little brick through the air at over 100 mph. Because the NEXT mph requires SO much more work than those first few mph, that infinitely higher percentage of available torque makes very little difference, and before you know it it's time to brake.

And the gearing IS important, because at 20 mph you are of course in a lower gear where the difference multiplier is higher. The percent difference is the same as in higher gears, but the absolute difference is greater. Meanwhile drag is lower so the extra torque accomplishes more useful work at low speed rather than wasted effort pushing against the air.

 

Ok, I think you're right.  I just played with this hp calc, and you save more time doing a drag from 40mph than from 90mph with 2 extra hp.

 

I haven't fully wrapped my head around "why" yet, but I'll chew on that.

 

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