HOW DO YOU ADD TORQUE?
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From: assville, MI, USA
Re: (zad5)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by zad5 »</TD></TR><TR><TD CLASS="quote">the reason why they are torqueless is because one the bore size it sonly 1.6liter second.
.....
basically all honda motors are designed to rev higher in trade off of torque number but we can get away with it due to our cars beign light and the gearing. the gearing of the b16 is considered pretty short. i personally wouldnt stroke the motor u can if u want but i wouldnt. bore it out to 2.0 84.5mm if you want torque.</TD></TR></TABLE>
Surprised no one jumped on this yet
You do know that the B16's bore is the same as the B18B/B18C1/B18C5 right?
The difference is the other factor that changes displacement, stroke. So if you stroke a B16 you pretty much end up with a B18
Bore it out to 84.5 wouldn't make it a two liter.
.....
basically all honda motors are designed to rev higher in trade off of torque number but we can get away with it due to our cars beign light and the gearing. the gearing of the b16 is considered pretty short. i personally wouldnt stroke the motor u can if u want but i wouldnt. bore it out to 2.0 84.5mm if you want torque.</TD></TR></TABLE>
Surprised no one jumped on this yet
You do know that the B16's bore is the same as the B18B/B18C1/B18C5 right?
The difference is the other factor that changes displacement, stroke. So if you stroke a B16 you pretty much end up with a B18
Bore it out to 84.5 wouldn't make it a two liter.
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Re: (StyleTEG)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by StyleTEG »</TD></TR><TR><TD CLASS="quote">
Surprised no one jumped on this yet
You do know that the B16's bore is the same as the B18B/B18C1/B18C5 right?
The difference is the other factor that changes displacement, stroke. So if you stroke a B16 you pretty much end up with a B18
Bore it out to 84.5 wouldn't make it a two liter.</TD></TR></TABLE>
This is not true. The B16 and B17 share one block casting, while the B18a/b/c share a completely different casting; the primary difference between these blocks is the deck height. If you stroke a motor, you have a worse R/S ratio... the way to improve the R/S ratio is by increasing the deck height. This yeilds longer rods for any given stroke, thus decreasing the angle of force against the cylinder walls (it does not, however, do anything for piston speeds and crankshaft/rod loads at TDC and BDC.)
Surprised no one jumped on this yet
You do know that the B16's bore is the same as the B18B/B18C1/B18C5 right?
The difference is the other factor that changes displacement, stroke. So if you stroke a B16 you pretty much end up with a B18
Bore it out to 84.5 wouldn't make it a two liter.</TD></TR></TABLE>
This is not true. The B16 and B17 share one block casting, while the B18a/b/c share a completely different casting; the primary difference between these blocks is the deck height. If you stroke a motor, you have a worse R/S ratio... the way to improve the R/S ratio is by increasing the deck height. This yeilds longer rods for any given stroke, thus decreasing the angle of force against the cylinder walls (it does not, however, do anything for piston speeds and crankshaft/rod loads at TDC and BDC.)
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Re: (db2integra)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by db2integra »</TD></TR><TR><TD CLASS="quote">This is not true. The B16 and B17 share one block casting, while the B18a/b/c share a completely different casting; the primary difference between these blocks is the deck height. If you stroke a motor, you have a worse R/S ratio... the way to improve the R/S ratio is by increasing the deck height. This yeilds longer rods for any given stroke, thus decreasing the angle of force against the cylinder walls (it does not, however, do anything for piston speeds and crankshaft/rod loads at TDC and BDC.)</TD></TR></TABLE>
I said pretty much, I didn't say it was the same block with different/crank rods.
Increasing the deck hight doesn't yeild longer rods, it allows you the opportunity to use longer rods. And you don't have to raise deck height to use longer rods, custom pistons have given us the ability to move the wrist pin location as well
The B16 has a 77.4mm stroke, and an 81mm bore
The B18C1/5 has a 87.2mm stroke, and a *gasp* 81mm bore.
So if you stroke a B16 to a 87.2mm stroke you pretty much end up with a B18C block. Yes a better rod stroke ratio means less side loading blah blah blah, sure it makes a difference in theory but I have yet to see any solid proof that its anything to worry about beyond the theory. There are all motor b20/vtec motors that rev beyond 10k with out frequent rebuilds.
The B16 has a 1.75 R/S ratio
The B18C1/5 has a 1.58 R/S ratio
Funny the B16 has a much better R/S yet honda had no problem with the ITR reving in the 8,000+ range as well
Modified by StyleTEG at 10:01 PM 7/26/2004
I said pretty much, I didn't say it was the same block with different/crank rods.
Increasing the deck hight doesn't yeild longer rods, it allows you the opportunity to use longer rods. And you don't have to raise deck height to use longer rods, custom pistons have given us the ability to move the wrist pin location as well
The B16 has a 77.4mm stroke, and an 81mm bore
The B18C1/5 has a 87.2mm stroke, and a *gasp* 81mm bore.
So if you stroke a B16 to a 87.2mm stroke you pretty much end up with a B18C block. Yes a better rod stroke ratio means less side loading blah blah blah, sure it makes a difference in theory but I have yet to see any solid proof that its anything to worry about beyond the theory. There are all motor b20/vtec motors that rev beyond 10k with out frequent rebuilds.
The B16 has a 1.75 R/S ratio
The B18C1/5 has a 1.58 R/S ratio
Funny the B16 has a much better R/S yet honda had no problem with the ITR reving in the 8,000+ range as well
Modified by StyleTEG at 10:01 PM 7/26/2004
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From: Makati, Metro Manila, Philippines
Re: (Lsos)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos »</TD></TR><TR><TD CLASS="quote">The max torque given is at the flywheel. It is what the engine directly puts out. It tells you more about the character of the car then if they gave you the max torque at the wheels....
As for final drive and shortness of the gears...they both have equally to do with torque multiplication. You multiply the flywheel torque by the gear ratio and then by the final drive to arrive at the wheel torque. The only difference is that the final drive is fixed, and the gear ratio isn't (unless you change the final drive...)</TD></TR></TABLE>
Uh-huh. Sort-of-related question: what do dyno torque numbers represent then (the numbers are always lower than the flywheel's)? After all, this is the torque number that i'd like to think really matters, right?
As for final drive and shortness of the gears...they both have equally to do with torque multiplication. You multiply the flywheel torque by the gear ratio and then by the final drive to arrive at the wheel torque. The only difference is that the final drive is fixed, and the gear ratio isn't (unless you change the final drive...)</TD></TR></TABLE>
Uh-huh. Sort-of-related question: what do dyno torque numbers represent then (the numbers are always lower than the flywheel's)? After all, this is the torque number that i'd like to think really matters, right?
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Re: (SiRP)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by SiRP »</TD></TR><TR><TD CLASS="quote">Uh-huh. Sort-of-related question: what do dyno torque numbers represent then (the numbers are always lower than the flywheel's)? After all, this is the torque number that i'd like to think really matters, right? </TD></TR></TABLE>
Dyno torque numbers represent the torque at the flywheel, minus whatever losses occur throughout the drivetrain (I guess around 15% in our cars). If you include the 15% loss it is then called wheel torque or wheel horsepower, even though dynos actually see much higher torque than they show (over 1000lb-ft if the test was done in 1st gear). Using known gear ratios, etc. they figure out what the torque is at the flywheel.
Dyno torque numbers represent the torque at the flywheel, minus whatever losses occur throughout the drivetrain (I guess around 15% in our cars). If you include the 15% loss it is then called wheel torque or wheel horsepower, even though dynos actually see much higher torque than they show (over 1000lb-ft if the test was done in 1st gear). Using known gear ratios, etc. they figure out what the torque is at the flywheel.
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From: Makati, Metro Manila, Philippines
Re: (Lsos)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos »</TD></TR><TR><TD CLASS="quote">Dyno torque numbers represent the torque at the flywheel, minus whatever losses occur throughout the drivetrain (I guess around 15% in our cars). If you include the 15% loss it is then called wheel torque or wheel horsepower, even though dynos actually see much higher torque than they show (over 1000lb-ft if the test was done in 1st gear). Using known gear ratios, etc. they figure out what the torque is at the flywheel. </TD></TR></TABLE>
Ah, simple mind is now clear as well. Many thanks.
I wonder how many people actually know this, or am i the only ignorant one?
Ah, simple mind is now clear as well. Many thanks.
I wonder how many people actually know this, or am i the only ignorant one?
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From: Charlotte, NC
i would have to agree with what a few peeps already said- if you're looking for a quick fix in low-end acceleration without having to get involved in internal engine mods, I would track down a lower final drive, such as the JDM ITR 4.785 (I believe). That with a slightly smaller overall tire diameter would get you off the line quicker, be careful of going too much smaller in tire diameter unless you're going wider tho- going too much smaller will result in less tire contact patch on the road, and then all your additional tq multiplication will be good for is nice tire smokeshows when you try to jump a light or anything.
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Re: (stmotorsports)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by stmotorsports »</TD></TR><TR><TD CLASS="quote"> if you're looking for a quick fix in low-end acceleration without having to get involved in internal engine mods, I would track down a lower final drive </TD></TR></TABLE>
That's an encouraging thought. Thanks.
No, i don't plan on traffic light dashes in the near or far future. But I would like some more low-end grunt for my daily driver. Having some mild headworks done along with an entirely new exhaust (5-Zigen 4-2-1, 2.5" piping, custom reso and muffler, no cat). That should help some...
That's an encouraging thought. Thanks.
No, i don't plan on traffic light dashes in the near or far future. But I would like some more low-end grunt for my daily driver. Having some mild headworks done along with an entirely new exhaust (5-Zigen 4-2-1, 2.5" piping, custom reso and muffler, no cat). That should help some...
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Re: (stmotorsports)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by stmotorsports »</TD></TR><TR><TD CLASS="quote">be careful of going too much smaller in tire diameter unless you're going wider tho- going too much smaller will result in less tire contact patch on the road, and then all your additional tq multiplication will be good for is nice tire smokeshows when you try to jump a light or anything.
</TD></TR></TABLE>
Not true, tyre contact has no bearing on the grip of the tyres.
A 15 inch wide tyre will have as much grip as a 1 inch wide tyre if the tyre compound is the same.
</TD></TR></TABLE>Not true, tyre contact has no bearing on the grip of the tyres.
A 15 inch wide tyre will have as much grip as a 1 inch wide tyre if the tyre compound is the same.
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Re: (ryanstev)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ryanstev »</TD></TR><TR><TD CLASS="quote">Not true, tyre contact has no bearing on the grip of the tyres.
A 15 inch wide tyre will have as much grip as a 1 inch wide tyre if the tyre compound is the same.</TD></TR></TABLE>
I don't want to say anything, but dude, I think you're been reading too many physics books...
Are you joking or can you explain this further?
A 15 inch wide tyre will have as much grip as a 1 inch wide tyre if the tyre compound is the same.</TD></TR></TABLE>
I don't want to say anything, but dude, I think you're been reading too many physics books...
Are you joking or can you explain this further?
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Re: (Lsos)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ryanstev »</TD></TR><TR><TD CLASS="quote">
Not true, tyre contact has no bearing on the grip of the tyres.
A 15 inch wide tyre will have as much grip as a 1 inch wide tyre if the tyre compound is the same.</TD></TR></TABLE>
WRONG
for every time the contact surface is multiplied the grip/friction is multiplied by the same amount
I GUARANTEE that if you take two tires with the same compound and have one that has a surface contact of a 15"x3" rectangle compared to a 1"x3" rectangle...the larger will be superior when it comes to traction/friction
you have mistaken the total friction of something with the static friction of a substance....there is a correlation between the surface contact and the type of substance.....but if the same exact substance is found on eace tire the larger will be superior....bigger is better, sometimes
try to prove me wrong
Not true, tyre contact has no bearing on the grip of the tyres.
A 15 inch wide tyre will have as much grip as a 1 inch wide tyre if the tyre compound is the same.</TD></TR></TABLE>
WRONG
for every time the contact surface is multiplied the grip/friction is multiplied by the same amount
I GUARANTEE that if you take two tires with the same compound and have one that has a surface contact of a 15"x3" rectangle compared to a 1"x3" rectangle...the larger will be superior when it comes to traction/friction
you have mistaken the total friction of something with the static friction of a substance....there is a correlation between the surface contact and the type of substance.....but if the same exact substance is found on eace tire the larger will be superior....bigger is better, sometimes
try to prove me wrong
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Re: (ryanstev)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ryanstev »</TD></TR><TR><TD CLASS="quote">Not true, tyre contact has no bearing on the grip of the tyres.</TD></TR></TABLE>In junior-high science class the friction coefficient (mu) is constant. In the real world it's not. Especially with elastomers.
For Lsos & dirtyd... if mu were constant, the area would drop out of the equation & ryan would be correct.
For Lsos & dirtyd... if mu were constant, the area would drop out of the equation & ryan would be correct.
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Re: (JimBlake)
Overall size of the contact patch is not determined by the width of a tire. Its determined by the weight pressing down on the tire, and the tire pressure.
What does change with the width of the tire, width of the rim, etc, is the shape of the contact patch. A wider tire will have the same size patch, but it will be wider but shorter length-wise. Better for cornering, worse for braking/acceleration.
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote »</TD></TR><TR><TD CLASS="quote">
if mu were constant, the area would drop out of the equation & ryan would be correct.
</TD></TR></TABLE>
Only if you were not looking at tire temperature as well.
What does change with the width of the tire, width of the rim, etc, is the shape of the contact patch. A wider tire will have the same size patch, but it will be wider but shorter length-wise. Better for cornering, worse for braking/acceleration.
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote »</TD></TR><TR><TD CLASS="quote">
if mu were constant, the area would drop out of the equation & ryan would be correct.
</TD></TR></TABLE>
Only if you were not looking at tire temperature as well.
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From: Makati, Metro Manila, Philippines
Re: (ryanstev)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ryanstev »</TD></TR><TR><TD CLASS="quote">Not true, tyre contact has no bearing on the grip of the tyres.
A 15 inch wide tyre will have as much grip as a 1 inch wide tyre if the tyre compound is the same.</TD></TR></TABLE>
So, if my feeble gray matter can grasp this, you're saying that, all things being equal, a car with a 1-inch tire has an equal chance of getting to the finish line just as quickly as one with foot-wide rubber? Simple mind now has slightly complex headache...
A 15 inch wide tyre will have as much grip as a 1 inch wide tyre if the tyre compound is the same.</TD></TR></TABLE>
So, if my feeble gray matter can grasp this, you're saying that, all things being equal, a car with a 1-inch tire has an equal chance of getting to the finish line just as quickly as one with foot-wide rubber? Simple mind now has slightly complex headache...
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Re: (dirtyd463)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by dirtyd463 »</TD></TR><TR><TD CLASS="quote">WRONG
for every time the contact surface is multiplied the grip/friction is multiplied by the same amount
I GUARANTEE that if you take two tires with the same compound and have one that has a surface contact of a 15"x3" rectangle compared to a 1"x3" rectangle...the larger will be superior when it comes to traction/friction
you have mistaken the total friction of something with the static friction of a substance....there is a correlation between the surface contact and the type of substance.....but if the same exact substance is found on eace tire the larger will be superior....bigger is better, sometimes
try to prove me wrong</TD></TR></TABLE>
Do you have the calculation for that?
As for what I said, traction = co-efficient of friction x weight
The tyre stickyness multiplied by the weight pushing down is what determines traction. Contact area isn't in that equation, because it has no effect.
Here's a website which had some tests on this.
http://www.philohome.com/traction/traction.htm
And here's a good web page which answers this question too, the first question on the page:
http://www.physlink.com/Educat...0.cfm
for every time the contact surface is multiplied the grip/friction is multiplied by the same amount
I GUARANTEE that if you take two tires with the same compound and have one that has a surface contact of a 15"x3" rectangle compared to a 1"x3" rectangle...the larger will be superior when it comes to traction/friction
you have mistaken the total friction of something with the static friction of a substance....there is a correlation between the surface contact and the type of substance.....but if the same exact substance is found on eace tire the larger will be superior....bigger is better, sometimes
try to prove me wrong</TD></TR></TABLE>
Do you have the calculation for that?
As for what I said, traction = co-efficient of friction x weight
The tyre stickyness multiplied by the weight pushing down is what determines traction. Contact area isn't in that equation, because it has no effect.
Here's a website which had some tests on this.
http://www.philohome.com/traction/traction.htm
And here's a good web page which answers this question too, the first question on the page:
http://www.physlink.com/Educat...0.cfm
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Re: (ryanstev)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ryanstev »</TD></TR><TR><TD CLASS="quote">As for what I said, traction = co-efficient of friction x weight
The tyre stickyness multiplied by the weight pushing down is what determines traction. Contact area isn't in that equation, because it has no effect.</TD></TR></TABLE>
Like I said, too many physics books. It doesn't really work this way in the real world where materials flex and irregularities exist. The formula might remain the same, but the coefficient of friction doesn't...
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote »</TD></TR><TR><TD CLASS="quote">
Here's a website which had some tests on this.
http://www.philohome.com/traction/traction.htm
</TD></TR></TABLE>
There's a slight difference between 465g and 3000+lbs....
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote »</TD></TR><TR><TD CLASS="quote">
And here's a good web page which answers this question too, the first question on the page:
http://www.physlink.com/Educat...0.cfm</TD></TR></TABLE>
That's the first time I've ever heard that kind of explanation regarding tires. This is something I'm more accustomed to (section on coefficient of traction)
http://www.rc10.com/racerhub/t...27148
The fact that the coefficient of friction is not equal under all loads is the whole basis for suspension design. If it wasn't for that simple fact weight distribution, lowering, stiffer springs, sway bars, center of gravity, camber, double wishbones etc. etc. would make little if any difference to how fast a car corners. As we all know, they very much make a difference.
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by dirtyd463 »</TD></TR><TR><TD CLASS="quote">for every time the contact surface is multiplied the grip/friction is multiplied by the same amount</TD></TR></TABLE>
This, however, is not true either. Traction will not double with a tire twice as wide. The gains are much smaller than that, and I'm sure if the width increases past a certain point there will be little/ no benefits.
The tyre stickyness multiplied by the weight pushing down is what determines traction. Contact area isn't in that equation, because it has no effect.</TD></TR></TABLE>
Like I said, too many physics books. It doesn't really work this way in the real world where materials flex and irregularities exist. The formula might remain the same, but the coefficient of friction doesn't...
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote »</TD></TR><TR><TD CLASS="quote">
Here's a website which had some tests on this.
http://www.philohome.com/traction/traction.htm
</TD></TR></TABLE>
There's a slight difference between 465g and 3000+lbs....
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote »</TD></TR><TR><TD CLASS="quote">
And here's a good web page which answers this question too, the first question on the page:
http://www.physlink.com/Educat...0.cfm</TD></TR></TABLE>
That's the first time I've ever heard that kind of explanation regarding tires. This is something I'm more accustomed to (section on coefficient of traction)
http://www.rc10.com/racerhub/t...27148
The fact that the coefficient of friction is not equal under all loads is the whole basis for suspension design. If it wasn't for that simple fact weight distribution, lowering, stiffer springs, sway bars, center of gravity, camber, double wishbones etc. etc. would make little if any difference to how fast a car corners. As we all know, they very much make a difference.
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by dirtyd463 »</TD></TR><TR><TD CLASS="quote">for every time the contact surface is multiplied the grip/friction is multiplied by the same amount</TD></TR></TABLE>
This, however, is not true either. Traction will not double with a tire twice as wide. The gains are much smaller than that, and I'm sure if the width increases past a certain point there will be little/ no benefits.
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Re: (ryanstev)
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ryanstev »</TD></TR><TR><TD CLASS="quote">... traction = co-efficient of friction x weight
The tyre stickyness multiplied by the weight pushing down is what determines traction. Contact area isn't in that equation, because it has no effect.</TD></TR></TABLE>The coefficient of friction depends on many things, like materials, surface texture, etc. Whenever the material itself deforms under the contact pressure, mu is a fairly strong function of that contact pressure. That's true for car tires, but not really true for small, hard tires like in that first website.
Air pressure supports the weight, not the sidewall. That alone undermines the credibility of that second website. Besides, wider tires don't necessarily have a larger contact patch, it's just a different shape. Friction coefficient isn't even uniform for different shear-force directions. Shape of the contact patch, configuration of the tread blocks, and contact pressure distribution (no, it's not uniform) all contribute to that.
The tyre stickyness multiplied by the weight pushing down is what determines traction. Contact area isn't in that equation, because it has no effect.</TD></TR></TABLE>The coefficient of friction depends on many things, like materials, surface texture, etc. Whenever the material itself deforms under the contact pressure, mu is a fairly strong function of that contact pressure. That's true for car tires, but not really true for small, hard tires like in that first website.
Air pressure supports the weight, not the sidewall. That alone undermines the credibility of that second website. Besides, wider tires don't necessarily have a larger contact patch, it's just a different shape. Friction coefficient isn't even uniform for different shear-force directions. Shape of the contact patch, configuration of the tread blocks, and contact pressure distribution (no, it's not uniform) all contribute to that.
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From: Bay Area, CA
styleteg good call. i honestly didnt know they had the same bore but at the end i would still go 84.5 witht he b16 block. youll won't exactly get 2.0 liters due to the stroke as styleteg said but i would still do resleeve then stroking th emotor.
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From: Makati, Metro Manila, Philippines
Re: (JimBlake)
Wow, i never expected the thread to last this long. Cool.
I'm learning a lot, although I'm also a bit confused due to the opposing views on the traction issue. I don't think the debate is over; dirtyd hasn't responded to ryan yet...
I'm learning a lot, although I'm also a bit confused due to the opposing views on the traction issue. I don't think the debate is over; dirtyd hasn't responded to ryan yet...
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From: wolverhampton, england
Re: HOW DO YOU ADD TORQUE? (SiRP)
you should add a 4-2-1 manifold (heat wrap it) but a good one mugen, dc or spoon, also add a performance exhaust which has all the backboxes (not straight through n1 system), this creates good backpressure, make sure where pipes are bent it doesnt narrow in. Then you should a add a cold air feed, maybe heat wrap the original box and midify it a little, by adding a feed to it from the bumber (if you do this make sure the pipe is same diameter as the pipe feeding into throttle body. Dont buy a cone intake buy a itg or k&n drop in filter, you cant beat the honda design. i have also added some iridium plugs to mine.
i hav not done the manifold and have a poor n1 exhaust and i have rolling road tested 120flb and 167hp (not sir2) but it is all at the top end have gained a small amount bottom end, but pulls hard at top. with the other mods i should be 170hp mark and 124-126flb. not bad for bolt ons. make sure you reset the ecu after the mods abd drive steady for 20 mins.
i hav not done the manifold and have a poor n1 exhaust and i have rolling road tested 120flb and 167hp (not sir2) but it is all at the top end have gained a small amount bottom end, but pulls hard at top. with the other mods i should be 170hp mark and 124-126flb. not bad for bolt ons. make sure you reset the ecu after the mods abd drive steady for 20 mins.
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gnarlycs
All Motor / Naturally Aspirated
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Jul 4, 2012 06:26 AM