Questions on Rob Raymer's Turbo kit???
10-31-2002, 02:08 PM
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Re: Questions on Rob Raymer's Turbo kit??? </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by SS00Blue:
<strong>All true, but it's just not that complicated when you look at the torque being produced at the crankshaft (since we all know HP is just a theoretical, RPM-based number). It does not require doubling cylinder pressure, and therefore compressive force on the piston/rod/bearing combination, to double torque produced because of the advantage of the torque arm when the torque arm is extended. SC-</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">It doesnt matter how horespower is produced, thats what you are not understanding. Here is a very simply relationship:
hp = tq X rpm X 5250
Since 5250 is a constant and we keep rpm the same, the ONLY way to double horsepower is to double torque. Anyone with a basic SE background will know that the torsional stress applied to the crankshaft by the powerstroke is equal to the torque times the radial distance from the crank centerline (conveniently the stroke in this case) divided by the rotational modulus of elasticity (Tr/J). Since the ONLY thing changing in this equation is that the torque applied to the crankshaft is doubled, the torsional stresses on the crankshaft will double. Consequently, the only way to double torque applied with the same stroke is to double the force which the piston applies, and this also doubles the compressive stresses in the rod etc. Since you are speaking from 80 years of engineering common sense, where am I wrong here?
<strong>All true, but it's just not that complicated when you look at the torque being produced at the crankshaft (since we all know HP is just a theoretical, RPM-based number). It does not require doubling cylinder pressure, and therefore compressive force on the piston/rod/bearing combination, to double torque produced because of the advantage of the torque arm when the torque arm is extended. SC-</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">It doesnt matter how horespower is produced, thats what you are not understanding. Here is a very simply relationship:
hp = tq X rpm X 5250
Since 5250 is a constant and we keep rpm the same, the ONLY way to double horsepower is to double torque. Anyone with a basic SE background will know that the torsional stress applied to the crankshaft by the powerstroke is equal to the torque times the radial distance from the crank centerline (conveniently the stroke in this case) divided by the rotational modulus of elasticity (Tr/J). Since the ONLY thing changing in this equation is that the torque applied to the crankshaft is doubled, the torsional stresses on the crankshaft will double. Consequently, the only way to double torque applied with the same stroke is to double the force which the piston applies, and this also doubles the compressive stresses in the rod etc. Since you are speaking from 80 years of engineering common sense, where am I wrong here?
10-31-2002, 02:15 PM
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Re: Questions on Rob Raymer's Turbo kit??? </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Fenris Ulf:
<strong>It doesnt matter how horespower is produced, thats what you are not understanding. Here is a very simply relationship:
hp = tq X rpm X 5250
Since 5250 is a constant and we keep rpm the same, the ONLY way to double horsepower is to double torque. Anyone with a basic SE background will know that the torsional stress applied to the crankshaft by the powerstroke is equal to the torque times the radial distance from the crank centerline (conveniently the stroke in this case) divided by the rotational modulus of elasticity (Tr/J). Since the ONLY thing changing in this equation is that the torque applied to the crankshaft is doubled, the torsional stresses on the crankshaft will double. Consequently, the only way to double torque applied with the same stroke is to double the force which the piston applies, and this also doubles the compressive stresses in the rod etc. Since you are speaking from 80 years of engineering common sense, where am I wrong here?</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Well, only in one area:
HP = Tq*RPM"/"5250.
And you've got me. I spoke of stress on an engine, and you've defined the entire engine as the crankshaft, but in spite of this, you're still only correct as the additive stress of an increase in torque only pertains to the area between the #8 rod journal and the crankshaft flange <img border="0" title="" alt="[Roll Eyes]" src="images/icons/rolleyes.gif" />
SC-
I would love to discuss this more, but I am put off by your snotty nose.
And BTW, I did show that an increase in stroke could be part of the equation, where you invented your own constraint on stroke size, which makes your assertion on rod forces falacious.
<small>[ October 31, 2002, 02:20 PM: Message edited by: SS00Blue ]</small>
<strong>It doesnt matter how horespower is produced, thats what you are not understanding. Here is a very simply relationship:
hp = tq X rpm X 5250
Since 5250 is a constant and we keep rpm the same, the ONLY way to double horsepower is to double torque. Anyone with a basic SE background will know that the torsional stress applied to the crankshaft by the powerstroke is equal to the torque times the radial distance from the crank centerline (conveniently the stroke in this case) divided by the rotational modulus of elasticity (Tr/J). Since the ONLY thing changing in this equation is that the torque applied to the crankshaft is doubled, the torsional stresses on the crankshaft will double. Consequently, the only way to double torque applied with the same stroke is to double the force which the piston applies, and this also doubles the compressive stresses in the rod etc. Since you are speaking from 80 years of engineering common sense, where am I wrong here?</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Well, only in one area:
HP = Tq*RPM"/"5250.
And you've got me. I spoke of stress on an engine, and you've defined the entire engine as the crankshaft, but in spite of this, you're still only correct as the additive stress of an increase in torque only pertains to the area between the #8 rod journal and the crankshaft flange <img border="0" title="" alt="[Roll Eyes]" src="images/icons/rolleyes.gif" />
SC-
I would love to discuss this more, but I am put off by your snotty nose.
And BTW, I did show that an increase in stroke could be part of the equation, where you invented your own constraint on stroke size, which makes your assertion on rod forces falacious.
<small>[ October 31, 2002, 02:20 PM: Message edited by: SS00Blue ]</small>
11-01-2002, 09:27 AM
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Re: Questions on Rob Raymer's Turbo kit??? </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by SS00Blue:
<strong> </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Fenris Ulf:
<strong>It doesnt matter how horespower is produced, thats what you are not understanding. Here is a very simply relationship:
hp = tq X rpm X 5250
Since 5250 is a constant and we keep rpm the same, the ONLY way to double horsepower is to double torque. Anyone with a basic SE background will know that the torsional stress applied to the crankshaft by the powerstroke is equal to the torque times the radial distance from the crank centerline (conveniently the stroke in this case) divided by the rotational modulus of elasticity (Tr/J). Since the ONLY thing changing in this equation is that the torque applied to the crankshaft is doubled, the torsional stresses on the crankshaft will double. Consequently, the only way to double torque applied with the same stroke is to double the force which the piston applies, and this also doubles the compressive stresses in the rod etc. Since you are speaking from 80 years of engineering common sense, where am I wrong here?</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Well, only in one area:
HP = Tq*RPM"/"5250.
And you've got me. I spoke of stress on an engine, and you've defined the entire engine as the crankshaft, but in spite of this, you're still only correct as the additive stress of an increase in torque only pertains to the area between the #8 rod journal and the crankshaft flange <img border="0" title="" alt="[Roll Eyes]" src="images/icons/rolleyes.gif" />
SC-
I would love to discuss this more, but I am put off by your snotty nose.
And BTW, I did show that an increase in stroke could be part of the equation, where you invented your own constraint on stroke size, which makes your assertion on rod forces falacious.</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">First off the total stress on a motor means NOTHING. When a stress exceeds the yield strength of a material, thats when you have a problem (i.e. something breaks!). Who cares about an increase in stroke? You said that horsepower can be increased without the torque being increased by lengthening the stroke (and also stated that better flowing heads etc don't increase the stresses in a motor, which is again bullshit). This is a completely false statement, since torque is directly related to horsepower as stated above (and corrected by JAS <img border="0" title="" alt="[Wink]" src="gr_images/icons/wink.gif" /> ). I said lets assume that the stroke doesnt change because we are talking about forced induction here, not strokers (not that it matters since you are wrong either way). So no, my assertion on stresses on the rod and the rest of the rotating assembly are not falacious. Sorry you are put off by my "snotty nose" LOL, but I hate to see my fellow LS1techers fed false info like your "20% stress increase rule of thumb"...
<strong> </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Fenris Ulf:
<strong>It doesnt matter how horespower is produced, thats what you are not understanding. Here is a very simply relationship:
hp = tq X rpm X 5250
Since 5250 is a constant and we keep rpm the same, the ONLY way to double horsepower is to double torque. Anyone with a basic SE background will know that the torsional stress applied to the crankshaft by the powerstroke is equal to the torque times the radial distance from the crank centerline (conveniently the stroke in this case) divided by the rotational modulus of elasticity (Tr/J). Since the ONLY thing changing in this equation is that the torque applied to the crankshaft is doubled, the torsional stresses on the crankshaft will double. Consequently, the only way to double torque applied with the same stroke is to double the force which the piston applies, and this also doubles the compressive stresses in the rod etc. Since you are speaking from 80 years of engineering common sense, where am I wrong here?</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Well, only in one area:
HP = Tq*RPM"/"5250.
And you've got me. I spoke of stress on an engine, and you've defined the entire engine as the crankshaft, but in spite of this, you're still only correct as the additive stress of an increase in torque only pertains to the area between the #8 rod journal and the crankshaft flange <img border="0" title="" alt="[Roll Eyes]" src="images/icons/rolleyes.gif" />
SC-
I would love to discuss this more, but I am put off by your snotty nose.
And BTW, I did show that an increase in stroke could be part of the equation, where you invented your own constraint on stroke size, which makes your assertion on rod forces falacious.</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">First off the total stress on a motor means NOTHING. When a stress exceeds the yield strength of a material, thats when you have a problem (i.e. something breaks!). Who cares about an increase in stroke? You said that horsepower can be increased without the torque being increased by lengthening the stroke (and also stated that better flowing heads etc don't increase the stresses in a motor, which is again bullshit). This is a completely false statement, since torque is directly related to horsepower as stated above (and corrected by JAS <img border="0" title="" alt="[Wink]" src="gr_images/icons/wink.gif" /> ). I said lets assume that the stroke doesnt change because we are talking about forced induction here, not strokers (not that it matters since you are wrong either way). So no, my assertion on stresses on the rod and the rest of the rotating assembly are not falacious. Sorry you are put off by my "snotty nose" LOL, but I hate to see my fellow LS1techers fed false info like your "20% stress increase rule of thumb"...
11-01-2002, 09:29 AM
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Re: Questions on Rob Raymer's Turbo kit??? </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Cool28:
<strong>Wo well this is totally off topic anymore but I dont really see any flaming going on. So it stays open. I think its a pretty good thread(even though off-topic).</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Sorry about that Todd, just trying to set things straight for everyone else or learn something in the process <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" />
<strong>Wo well this is totally off topic anymore but I dont really see any flaming going on. So it stays open. I think its a pretty good thread(even though off-topic).</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Sorry about that Todd, just trying to set things straight for everyone else or learn something in the process <img border="0" title="" alt="[Smile]" src="gr_stretch.gif" />
11-01-2002, 02:07 PM
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Re: Questions on Rob Raymer's Turbo kit??? Here's a link to an excerpt from Maximum Boost by Corky Bell. I think pages 5-9 address this topic to a certain degree:
http://www.amazon.com/exec/obidos/tg...=5#reader-link
http://www.amazon.com/exec/obidos/tg...=5#reader-link
11-02-2002, 01:35 AM
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Re: Questions on Rob Raymer's Turbo kit??? <img border="0" title="" alt="[Confused]" src="images/icons/confused.gif" /> <img border="0" title="" alt="[Eek!]" src="gr_eek2.gif" /> My brain hurts. <img border="0" alt="[kaboom]" title="" src="graemlins/gr_bomb.gif" />
This will probably feel better: <img border="0" alt="[Banging Head]" title="" src="graemlins/gr_banghead.gif" />
'Los
This will probably feel better: <img border="0" alt="[Banging Head]" title="" src="graemlins/gr_banghead.gif" />
'Los
11-02-2002, 01:26 PM
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Re: Questions on Rob Raymer's Turbo kit??? Ferris nice refute, even a peasant like myself could see you are correct!
<small>[ November 02, 2002, 01:26 PM: Message edited by: cantdrv65 ]</small>
<small>[ November 02, 2002, 01:26 PM: Message edited by: cantdrv65 ]</small>
11-03-2002, 08:31 PM
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Re: Questions on Rob Raymer's Turbo kit??? </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by cantdrv65:
<strong>Ferris nice refute, even a peasant like myself could see you are correct!</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Thanks for the kind words <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />
<strong>Ferris nice refute, even a peasant like myself could see you are correct!</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Thanks for the kind words <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />
11-04-2002, 05:17 PM
#71
Re: Questions on Rob Raymer's Turbo kit??? ummm... hey ya'll check out these pics of my new single turbo kit.
http://www.ls1motorsports.com/Single_Turbo.htm
well pretty much <img border="0" title="" alt="[Wink]" src="gr_images/icons/wink.gif" /> wonder how long before i'll be getting the call for the rest of the money..
http://www.ls1motorsports.com/Single_Turbo.htm
well pretty much <img border="0" title="" alt="[Wink]" src="gr_images/icons/wink.gif" /> wonder how long before i'll be getting the call for the rest of the money..
11-07-2002, 03:49 PM
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Re: Questions on Rob Raymer's Turbo kit??? With regard to Nitrous vs Turbo, which engine would last longer at WOT on an Engine dyno? A hypothetical 700rwhp (15psi) Turbo car or a hypothetical 450rwhp (150 shot) Nitrous car? For the nitrous car, you'd have to assume that you don't run out of nitrous.
My question isn't a jab at anyone. I'd actually like to know because that would tell me which system is more robust.
BTW: I've seen 1000°C (1800°F) EGT on an engine dyno before during an engine test. The glowing exhaust manifold was pretty cool. <img border="0" title="" alt="[Cool]" src="gr_images/icons/cool.gif" />
<small>[ November 07, 2002, 03:53 PM: Message edited by: Steve98TransAm ]</small>
My question isn't a jab at anyone. I'd actually like to know because that would tell me which system is more robust.
BTW: I've seen 1000°C (1800°F) EGT on an engine dyno before during an engine test. The glowing exhaust manifold was pretty cool. <img border="0" title="" alt="[Cool]" src="gr_images/icons/cool.gif" />
<small>[ November 07, 2002, 03:53 PM: Message edited by: Steve98TransAm ]</small>
11-09-2002, 06:31 PM
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Re: Questions on Rob Raymer's Turbo kit??? I got tired of reading the first page so I skipped to the last but has anybody mentioned the GREAT cooling effect nitrous has on the combustion chamber before ignition. Nitrous vaporizes at -127F. Then you get supercharger and turbocharger that cavitate in the air and produce intake air temp higher then 300 degrees. Now analyze that. Cylinder pressure is Cylinder pressure regardless of what makes it. Also have you ever noticed how racecars run 400 shots on 14:1 compression motors. You dont see racers running boost on that kinda compression especially enough boost to make up for a 400hp shot. Personal I like turbos, most of my cars have had FI or I made them FI, bootles are for babies, dont take it personal but I would hate to have to fill up all the time. Supercharges waste to much energy parasiticly so I think NOS is the easiest Turbo second and Superchargers last.
11-09-2002, 08:45 PM
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Re: Questions on Rob Raymer's Turbo kit??? </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Also have you ever noticed how racecars run 400 shots on 14:1 compression motors</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Your comparing apples to oranges.
Your comparing injecting a oxygen rich gas vs compressing normal air.
Huge difference in engine setup.
well a boosted motor against a n20 motor at the extremes in racing, the boosted motor usually will out power n20 motors.Forget your little 400 shot.We're talking 1000+ hp shots.
Look at IHRA PRO MOD, low 6 sec cars. That class is blower vs. n20 monsters.
N20 cars get a few setup advantages to stay competitive against the blown cars or the blower cars walk away from them.
N20 offers 0 cooling in the combustion chamber.
It cools off the intake charge only!
At 500+ degrees during compression/combustion the n20 gets split to release the oxygen and nitrogen.
Intercoolers and water/alky system will take all the heat out of the charge in a boosted motor.
Ahh but since this thread is so far off topic now, remember the true top dawg of poweradders.
Nitro methane will smoke either method if running gasoline.
Generally N.A. injected nitro cars run with or better than Blown alcohol cars.
That is an NHRA class I forget the name of.
Now that stuff will be HARD on parts.
Steve
<small>[ November 09, 2002, 08:46 PM: Message edited by: SJH ]</small>
Your comparing injecting a oxygen rich gas vs compressing normal air.
Huge difference in engine setup.
well a boosted motor against a n20 motor at the extremes in racing, the boosted motor usually will out power n20 motors.Forget your little 400 shot.We're talking 1000+ hp shots.
Look at IHRA PRO MOD, low 6 sec cars. That class is blower vs. n20 monsters.
N20 cars get a few setup advantages to stay competitive against the blown cars or the blower cars walk away from them.
N20 offers 0 cooling in the combustion chamber.
It cools off the intake charge only!
At 500+ degrees during compression/combustion the n20 gets split to release the oxygen and nitrogen.
Intercoolers and water/alky system will take all the heat out of the charge in a boosted motor.
Ahh but since this thread is so far off topic now, remember the true top dawg of poweradders.
Nitro methane will smoke either method if running gasoline.
Generally N.A. injected nitro cars run with or better than Blown alcohol cars.
That is an NHRA class I forget the name of.
Now that stuff will be HARD on parts.
Steve
<small>[ November 09, 2002, 08:46 PM: Message edited by: SJH ]</small>
11-10-2002, 11:24 AM
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Re: Questions on Rob Raymer's Turbo kit??? First off how do you figure injecting a gas that is way below zero doesnt cool the combustion chamber, also if you acctully read my post I also said before ignition not after.
Second, the class you were thinking of is Federal mogal dragsters or as they used to be called Top Alcohol. If you look in the NHRA rule book you would see that the NA nitromethane injected cars get a HUGE weight break to make up for the 40-50psi of boost that the superchaged alchohol cars run. Also nitromethane runs at a A/F ratio of about 4-5:1 on top of, if you have ever got nitromethane on your hands you would notice it has a real low heat of vaporization point meaning, if you got it on your hands it would have the same cooling effect as rubbing alchohol.
Third, FI have to have way less advance and CR because since there is another atmosphere more air comes into the CYL and this can cause almost a deiseling effect because dynamic compression becomes so high, unlike NOS were your compression is your static compression or what is built into the engine.
Second, the class you were thinking of is Federal mogal dragsters or as they used to be called Top Alcohol. If you look in the NHRA rule book you would see that the NA nitromethane injected cars get a HUGE weight break to make up for the 40-50psi of boost that the superchaged alchohol cars run. Also nitromethane runs at a A/F ratio of about 4-5:1 on top of, if you have ever got nitromethane on your hands you would notice it has a real low heat of vaporization point meaning, if you got it on your hands it would have the same cooling effect as rubbing alchohol.
Third, FI have to have way less advance and CR because since there is another atmosphere more air comes into the CYL and this can cause almost a deiseling effect because dynamic compression becomes so high, unlike NOS were your compression is your static compression or what is built into the engine.
11-10-2002, 11:51 AM
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Re: Questions on Rob Raymer's Turbo kit??? nitro is is around 2-1 a-f.
It can be used even at 1-1.
Methanol is 5-1.
I used n20 for almost 20 years now.I like it a lot. But bottle filling gets old.
Combustion means spark happened.
The charge cooling effect of n20 is great, no argument there.It's tremendous. I used to see sub zero IAT's by the 1/8th with 200+ dry shots.
It allows a ton more fuel to be burned and this is where the heat comes from.
But during the combustion stoke forget about any cooling effect.Maybe you mean it has a bit of cooling during intake stroke. Probably a bit.
Why do you think piston makers recomend the largest ring gaps for n20 motors?
Because of the intense heat and burn in the chamber from a n20 enviroment.Whatever little cooling that takes place from the n20 during the intake stroke is quickly lost during combustion.
Sure if you compare a boosted motor with a 300 degree intake temp the n20 motor will be easier to tune.
N20 is an oxidizer and not a fuel.
It allows more fuel to be burned to make more power.
More power means more heat is generated.
It can be used even at 1-1.
Methanol is 5-1.
I used n20 for almost 20 years now.I like it a lot. But bottle filling gets old.
Combustion means spark happened.
The charge cooling effect of n20 is great, no argument there.It's tremendous. I used to see sub zero IAT's by the 1/8th with 200+ dry shots.
It allows a ton more fuel to be burned and this is where the heat comes from.
But during the combustion stoke forget about any cooling effect.Maybe you mean it has a bit of cooling during intake stroke. Probably a bit.
Why do you think piston makers recomend the largest ring gaps for n20 motors?
Because of the intense heat and burn in the chamber from a n20 enviroment.Whatever little cooling that takes place from the n20 during the intake stroke is quickly lost during combustion.
Sure if you compare a boosted motor with a 300 degree intake temp the n20 motor will be easier to tune.
N20 is an oxidizer and not a fuel.
It allows more fuel to be burned to make more power.
More power means more heat is generated.
11-12-2002, 07:01 AM
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Re: Questions on Rob Raymer's Turbo kit??? </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by SJH:
<strong> </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Also have you ever noticed how racecars run 400 shots on 14:1 compression motors</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Your comparing apples to oranges.
Your comparing injecting a oxygen rich gas vs compressing normal air.
Huge difference in engine setup.
well a boosted motor against a n20 motor at the extremes in racing, the boosted motor usually will out power n20 motors.Forget your little 400 shot.We're talking 1000+ hp shots.
Look at IHRA PRO MOD, low 6 sec cars. That class is blower vs. n20 monsters.
N20 cars get a few setup advantages to stay competitive against the blown cars or the blower cars walk away from them.
N20 offers 0 cooling in the combustion chamber.
It cools off the intake charge only!
At 500+ degrees during compression/combustion the n20 gets split to release the oxygen and nitrogen.
Intercoolers and water/alky system will take all the heat out of the charge in a boosted motor.
Ahh but since this thread is so far off topic now, remember the true top dawg of poweradders.
Nitro methane will smoke either method if running gasoline.
Generally N.A. injected nitro cars run with or better than Blown alcohol cars.
That is an NHRA class I forget the name of.
Now that stuff will be HARD on parts.
Steve</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Speaking on the NA nitro injected cars they run flacky at best and RARELY win a race against a blown alcohol car. Now a BLOWN nitro engine is a different animal indeed and tend to make TWICE the power of a blown alcohol engine, hence the different class.
<strong> </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Also have you ever noticed how racecars run 400 shots on 14:1 compression motors</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Your comparing apples to oranges.
Your comparing injecting a oxygen rich gas vs compressing normal air.
Huge difference in engine setup.
well a boosted motor against a n20 motor at the extremes in racing, the boosted motor usually will out power n20 motors.Forget your little 400 shot.We're talking 1000+ hp shots.
Look at IHRA PRO MOD, low 6 sec cars. That class is blower vs. n20 monsters.
N20 cars get a few setup advantages to stay competitive against the blown cars or the blower cars walk away from them.
N20 offers 0 cooling in the combustion chamber.
It cools off the intake charge only!
At 500+ degrees during compression/combustion the n20 gets split to release the oxygen and nitrogen.
Intercoolers and water/alky system will take all the heat out of the charge in a boosted motor.
Ahh but since this thread is so far off topic now, remember the true top dawg of poweradders.
Nitro methane will smoke either method if running gasoline.
Generally N.A. injected nitro cars run with or better than Blown alcohol cars.
That is an NHRA class I forget the name of.
Now that stuff will be HARD on parts.
Steve</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Speaking on the NA nitro injected cars they run flacky at best and RARELY win a race against a blown alcohol car. Now a BLOWN nitro engine is a different animal indeed and tend to make TWICE the power of a blown alcohol engine, hence the different class.
11-13-2002, 10:12 PM
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Re: Questions on Rob Raymer's Turbo kit??? Thats weird, a injected nitro car won the NHRA lucas oil championship this year. His name was art gallant and he drives the pep boys car.
