Low-end Torque
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Joined: Nov 2001
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From: socal
the day you sign the paperwork @ the honda dealer and drive away with your S2K, you should have already known that we won't have as much low end tq as the bigger displacement (more cyl) engine ever.
F20C is a rev-lovely engine. you will have to rev it up to get some good/fun tq to the wheel.
it's a fact, you will have to live with it or get something else. no matter how you modify it, you won't be able to get the as much low end tq as those GT does.
having said that, i still think the F20C has more than enough low end tq for traffic even w/o going into VTEC. F20C always put out at least 100lb/ft, even down low.
to me, that's what good about the S2K ... it's not for everyone~
F20C is a rev-lovely engine. you will have to rev it up to get some good/fun tq to the wheel.
it's a fact, you will have to live with it or get something else. no matter how you modify it, you won't be able to get the as much low end tq as those GT does.
having said that, i still think the F20C has more than enough low end tq for traffic even w/o going into VTEC. F20C always put out at least 100lb/ft, even down low.
to me, that's what good about the S2K ... it's not for everyone~
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Joined: Aug 2002
Posts: 611
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From: Sydney
How about a lysholm twin screw super charger?
Unlike Vortech (centrifugal) and turbos which have a latency, the lysholm screw is designed to boost from the word "Engine Start".
It' won't make it like a 5.7 Litre V8 Commodore, but since it boosts at every RPM, it should give you *something*
Unlike Vortech (centrifugal) and turbos which have a latency, the lysholm screw is designed to boost from the word "Engine Start".
It' won't make it like a 5.7 Litre V8 Commodore, but since it boosts at every RPM, it should give you *something*
Joined: Nov 2001
Posts: 432
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From: Louisville
The greatest accelerationwith a CVT would be at the torque peak, not at the horsepower peak. In other words:
F x M = A
For rotational forces, that becomes:
Torque x Mass = Acceleration
You may be doing more work (per unit time) at that higher RPM (more HP expended), but you're not going to accelerate as quickly as you would at the torque peak, where the rotational force is greatest.
Guys, this is NOT that complicated. It's HS physics.
Tim
F x M = A
For rotational forces, that becomes:
Torque x Mass = Acceleration
You may be doing more work (per unit time) at that higher RPM (more HP expended), but you're not going to accelerate as quickly as you would at the torque peak, where the rotational force is greatest.
Guys, this is NOT that complicated. It's HS physics.
Tim
Joined: Nov 2001
Posts: 432
Likes: 0
From: Louisville
OK... back to the subject at hand.
a = F/m
This being the case, the force (torque is force at a specific distance from the axis of rotation) divided by the mass, determines the rate of change in rotational velocity. In other words, you may be producing more power at a higher RPM, and therefore doing more useful work, but your acceleration MUST be less in terms of distance (revolutions) per unit time, per unit time (as in, RPM, per second).
The force is in Newtons or pounds, the mass is in grams, and the acceleration is either in meters per second, per second (for Newtons) or feet per second, per second (for pounds). It's no different with torque than it is in simple vector forces. All the same rules apply in terms of rotational acceleration.
Now, given what I've said above, can someone explain to me how
a = F/m
does NOT translate the way I've described it? If the force is less (at the crankshaft), and the mass is the same, how will the acceleration be greater?
Tim
a = F/m
This being the case, the force (torque is force at a specific distance from the axis of rotation) divided by the mass, determines the rate of change in rotational velocity. In other words, you may be producing more power at a higher RPM, and therefore doing more useful work, but your acceleration MUST be less in terms of distance (revolutions) per unit time, per unit time (as in, RPM, per second).
The force is in Newtons or pounds, the mass is in grams, and the acceleration is either in meters per second, per second (for Newtons) or feet per second, per second (for pounds). It's no different with torque than it is in simple vector forces. All the same rules apply in terms of rotational acceleration.
Now, given what I've said above, can someone explain to me how
a = F/m
does NOT translate the way I've described it? If the force is less (at the crankshaft), and the mass is the same, how will the acceleration be greater?
Tim
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Joined: Jan 2001
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From: Glendale/Burbank/LA
I'm still not sure I see your point:
Yes that's correct -- that's exactly what a = F/m is saying.
I see your *claim* that acceleration is "less" at the power peak than at the torque peak (right?), but I'm not sure what you're saying to support that. I'm disputing that and saying that maximum acceleration does NOT necessarily correspond to the torque peak. In the case of the S2000, it's the power peak.
Here's my support to back this up:
1. a = F/m --> Since Mass is fixed, Acceleration increases as Force increases. At max Force, you get max Acceleration. So for our purposes, Acceleration is directly proportional to Force.
2. F = P x Gear Ratio x Tire Radius / RPM --> The Force known as Motive Force (which moves the vehicle) is directly proportional to power (as it is to torque), and inversely proportional to the engine speed.
Example using real world numbers from my own dyno chart:
Torque chart says max torque is 146.0 ft/lb at 6588 RPM.
Power chart says max power is 220.5 HP at 8319 RPM.
At max torque (6588 RPM), power is 184 HP.
At max power (8319 RPM), torque is 136.5 ft/lb.
According to you, the car's greatest acceleration is at 6588 RPM. Let's compute the acceleration at max torque and at max power using the formula from #2 above -- but for the sake of simplifying the math, let's drop gear ratio and tire radius since they are constant for our purposes:
F = P / RPM.
At max torque:
F = 184 / 6588 = .0279
At max power:
F = 220.5 / 8319 RPM = .0265
See? And since F is proportional to Acceleration, the acceleration at max torque in this case is the greatest!! See?
.............
Okay.... where's my frickin crow?
You know, I think UL's dyno chart is wrong and that explains it perfectly since he is biased toward max torque anyway! 
Ok Tim you're cool in my book.
Originally posted by TimTheFoolMan
a = F/m
This being the case, the force (torque is force at a specific distance from the axis of rotation) divided by the mass, determines the rate of change in rotational velocity.
a = F/m
This being the case, the force (torque is force at a specific distance from the axis of rotation) divided by the mass, determines the rate of change in rotational velocity.
Originally posted by TimTheFoolMan
In other words, you may be producing more power at a higher RPM, and therefore doing more useful work, but your acceleration MUST be less in terms of distance (revolutions) per unit time, per unit time (as in, RPM, per second).
In other words, you may be producing more power at a higher RPM, and therefore doing more useful work, but your acceleration MUST be less in terms of distance (revolutions) per unit time, per unit time (as in, RPM, per second).
Here's my support to back this up:
1. a = F/m --> Since Mass is fixed, Acceleration increases as Force increases. At max Force, you get max Acceleration. So for our purposes, Acceleration is directly proportional to Force.
2. F = P x Gear Ratio x Tire Radius / RPM --> The Force known as Motive Force (which moves the vehicle) is directly proportional to power (as it is to torque), and inversely proportional to the engine speed.
Example using real world numbers from my own dyno chart:
Torque chart says max torque is 146.0 ft/lb at 6588 RPM.
Power chart says max power is 220.5 HP at 8319 RPM.
At max torque (6588 RPM), power is 184 HP.
At max power (8319 RPM), torque is 136.5 ft/lb.
According to you, the car's greatest acceleration is at 6588 RPM. Let's compute the acceleration at max torque and at max power using the formula from #2 above -- but for the sake of simplifying the math, let's drop gear ratio and tire radius since they are constant for our purposes:
F = P / RPM.
At max torque:
F = 184 / 6588 = .0279
At max power:
F = 220.5 / 8319 RPM = .0265
See? And since F is proportional to Acceleration, the acceleration at max torque in this case is the greatest!! See?
.............
Okay.... where's my frickin crow?
You know, I think UL's dyno chart is wrong and that explains it perfectly since he is biased toward max torque anyway! Ok Tim you're cool in my book.