What has more Torque???

[bikemiser]

I do know that if you have a 4 cylinder and a 12 cylinder with the same displacement the 12 cylinder will have the most HP, the 4 cylinder will have the most torque and the 4 cylinder will get much better gas mileage. My 2 cents. /forums/images/graemlins/smile.gif

 

[bikemiser]

I do know that if you have a 4 cylinder and a 12 cylinder with the same displacement the 12 cylinder will have the most HP, the 4 cylinder will have the most torque and the 4 cylinder will get much better gas mileage. My 2 cents. /forums/images/graemlins/smile.gif

 

[awlchu]

</font><font color="blueclass=small">( Wouldn't the weight of all that spinning mass have to do with torque? I know a rod and piston out of a Cummins is one heavy chunk of metal. )</font>

Actually the spinning mass has to do with inertia and the energy it can potentially have. Angular momentum increases linearly with mass and inertia (double either one will double the angular momentum). However, Doubling the RPM will increase angular momentum by 4 times! (w**2).

HP or work/time is a rate of measure on how much work can be performed.

Here's an example: I have a crankshaft spinning that has a 100 joules of energy. IF I attached a rope to the spinning mass with a 1 kg mass attached to the end of it, what is the highest distance the crank can raise the mass?

kw = potential energy
100j = mgh
100 nm = (1kg)*9.8m/s**2 * h

h = 100nm/(19.6n) = 5.1 meters maximum height the spinning crank can raise the 1 kg mass. It does not matter about the mass of the crank or the diameter of the crank.

A crank 3 meters in diameter cannot left a 1kg weight any higher than a crank 5 meters in diameter. All depends on the energy of the spinning crank (large inertia or large angular velocity).

Energy is conserved!

The size of the crank and rod are really due to strength to prevent failure under worse op conditions.

 

[awlchu]

</font><font color="blueclass=small">( Wouldn't the weight of all that spinning mass have to do with torque? I know a rod and piston out of a Cummins is one heavy chunk of metal. )</font>

Actually the spinning mass has to do with inertia and the energy it can potentially have. Angular momentum increases linearly with mass and inertia (double either one will double the angular momentum). However, Doubling the RPM will increase angular momentum by 4 times! (w**2).

HP or work/time is a rate of measure on how much work can be performed.

Here's an example: I have a crankshaft spinning that has a 100 joules of energy. IF I attached a rope to the spinning mass with a 1 kg mass attached to the end of it, what is the highest distance the crank can raise the mass?

kw = potential energy
100j = mgh
100 nm = (1kg)*9.8m/s**2 * h

h = 100nm/(19.6n) = 5.1 meters maximum height the spinning crank can raise the 1 kg mass. It does not matter about the mass of the crank or the diameter of the crank.

A crank 3 meters in diameter cannot left a 1kg weight any higher than a crank 5 meters in diameter. All depends on the energy of the spinning crank (large inertia or large angular velocity).

Energy is conserved!

The size of the crank and rod are really due to strength to prevent failure under worse op conditions.

 

[Jay4200]

</font><font color="blue" class="small">( The only thing that matters is what rpm you're measuring the torque at. If they both make 10HP@10RPM, they both have 5,250 ft/lbs of torque.)</font>

This is on the right track, but presented a little backwards, since the inital equality was HP, so what matters is at what RPM the HP was measured (semantics).

Torque is a real measured performance parameter, HP is simply what falls out of an equation, as HP = torque*RPM. Torque can be viewed as proportional to how much oomph a single firing of a single piston makes. Ideally, torque should remain more-or-less constant over a usable RPM range (pretty much true for diesels at least) so HP should ramp proportionally with RPM.

So, not enough information on the initial question. What are the displacements? More than likely, with a 3 cylinder engine vs. a 2-cylinder of the same HP rating, the 3-cylinder will probably be able to spin faster (since the pistons are likely smaller) and therefore make the same HP with less torque, but at a higher RPM. OTOH, the 3-cylinder could be a slow-revving torque monster like a diesel tractor motor, while the 2-cylinder could be a fast-revving teeny-torque motor with all of the HP made with RPMs, like a crotch-rocket motorcycle engine.

So, it could go either way - for example, my 4-cylinder Kubota L4200 makes 42HP (depending on who is doing the quoting) at ~2000RPM or so. A single-cylinder Honda CR250R two-stroke dirt bike motor also makes 42HP, but has to scream to do so.

Jay

 

[Jay4200]

</font><font color="blue" class="small">( The only thing that matters is what rpm you're measuring the torque at. If they both make 10HP@10RPM, they both have 5,250 ft/lbs of torque.)</font>

This is on the right track, but presented a little backwards, since the inital equality was HP, so what matters is at what RPM the HP was measured (semantics).

Torque is a real measured performance parameter, HP is simply what falls out of an equation, as HP = torque*RPM. Torque can be viewed as proportional to how much oomph a single firing of a single piston makes. Ideally, torque should remain more-or-less constant over a usable RPM range (pretty much true for diesels at least) so HP should ramp proportionally with RPM.

So, not enough information on the initial question. What are the displacements? More than likely, with a 3 cylinder engine vs. a 2-cylinder of the same HP rating, the 3-cylinder will probably be able to spin faster (since the pistons are likely smaller) and therefore make the same HP with less torque, but at a higher RPM. OTOH, the 3-cylinder could be a slow-revving torque monster like a diesel tractor motor, while the 2-cylinder could be a fast-revving teeny-torque motor with all of the HP made with RPMs, like a crotch-rocket motorcycle engine.

So, it could go either way - for example, my 4-cylinder Kubota L4200 makes 42HP (depending on who is doing the quoting) at ~2000RPM or so. A single-cylinder Honda CR250R two-stroke dirt bike motor also makes 42HP, but has to scream to do so.

Jay

 

[whodat90]

</font><font color="blue" class="small">( </font><font color="blueclass=small">( The only thing that matters is what rpm you're measuring the torque at. If they both make 10HP@10RPM, they both have 5,250 ft/lbs of torque.)</font>

This is on the right track, but presented a little backwards, since the inital equality was HP, so what matters is at what RPM the HP was measured (semantics).

Torque is a real measured performance parameter, HP is simply what falls out of an equation, as HP = torque*RPM. Torque can be viewed as proportional to how much oomph a single firing of a single piston makes. Ideally, torque should remain more-or-less constant over a usable RPM range (pretty much true for diesels at least) so HP should ramp proportionally with RPM.

So, not enough information on the initial question. What are the displacements? More than likely, with a 3 cylinder engine vs. a 2-cylinder of the same HP rating, the 3-cylinder will probably be able to spin faster (since the pistons are likely smaller) and therefore make the same HP with less torque, but at a higher RPM. OTOH, the 3-cylinder could be a slow-revving torque monster like a diesel tractor motor, while the 2-cylinder could be a fast-revving teeny-torque motor with all of the HP made with RPMs, like a crotch-rocket motorcycle engine.

So, it could go either way - for example, my 4-cylinder Kubota L4200 makes 42HP (depending on who is doing the quoting) at ~2000RPM or so. A single-cylinder Honda CR250R two-stroke dirt bike motor also makes 42HP, but has to scream to do so.

Jay )</font>

Torque is a force, measured as ft/lbs in this case. Horsepower is a measure of how much work that force can do over a period of time. The actual formula is hp=tq*rpm/5252 . The 5252 is an engineering constant.

 

[whodat90]

</font><font color="blue" class="small">( </font><font color="blueclass=small">( The only thing that matters is what rpm you're measuring the torque at. If they both make 10HP@10RPM, they both have 5,250 ft/lbs of torque.)</font>

This is on the right track, but presented a little backwards, since the inital equality was HP, so what matters is at what RPM the HP was measured (semantics).

Torque is a real measured performance parameter, HP is simply what falls out of an equation, as HP = torque*RPM. Torque can be viewed as proportional to how much oomph a single firing of a single piston makes. Ideally, torque should remain more-or-less constant over a usable RPM range (pretty much true for diesels at least) so HP should ramp proportionally with RPM.

So, not enough information on the initial question. What are the displacements? More than likely, with a 3 cylinder engine vs. a 2-cylinder of the same HP rating, the 3-cylinder will probably be able to spin faster (since the pistons are likely smaller) and therefore make the same HP with less torque, but at a higher RPM. OTOH, the 3-cylinder could be a slow-revving torque monster like a diesel tractor motor, while the 2-cylinder could be a fast-revving teeny-torque motor with all of the HP made with RPMs, like a crotch-rocket motorcycle engine.

So, it could go either way - for example, my 4-cylinder Kubota L4200 makes 42HP (depending on who is doing the quoting) at ~2000RPM or so. A single-cylinder Honda CR250R two-stroke dirt bike motor also makes 42HP, but has to scream to do so.

Jay )</font>

Torque is a force, measured as ft/lbs in this case. Horsepower is a measure of how much work that force can do over a period of time. The actual formula is hp=tq*rpm/5252 . The 5252 is an engineering constant.