Torque Question

[jamie g]

Max_Torque said:

Simply because it is only pressing on the crank for half as many times per cycle.

in very layman terms:

say the force on the piston is 2, and on a 2 cyl eng it happens twice per cycle (2revs) the "total" is 2 x 2 = 4


for the 4cyl, force per cyl is only 1 (half) but it happens 4 x per cycle, so 1 x 4 also = 4

hence same torque output

no that can't be right. A measured force of 'x' and then another measured force of 'x' one sceond later still only equals a maximum force of 'x', not 'x' + 'x'

[GroundEffect]

jamie g said:

But as I said before, there is no time reference in the units of measure.

Yes but go back to your previous analogy of the bike: If you have two big (let's say heavy shoes!) feet that press down on the pedal it will produce a larger torque than two smaller...but if you have 4 instead that still equal the same weight (i.e. the same capacity engine) they will average out as the same.

Also, think about this for a second: You might have a bigger force from the power stroke, but you also need more power for your compression stroke (greater mass of air) from your V2 engine.

[busta]

jamie g said:

But torque has no time or revolution reference in its value. Torque is an instantaneous value measured at one point in time.
Imagine pushing down on your bicycle pedal, the force applied is transferred into torque. In this case the piston is replacing your leg, and surely one big piston can push harder than one half the size?

Your right that torque doesn't have a time reference and if we could measure an instantaneous torque output mid combustion stroke then you would get a figure much higher than the average. However this is a pretty arbitrary figure as the engine has to work all the time, not just for 1/4 of the combustion cycle, so we actually measure the average torque output over a complete combustion cycle which is not only useful but it's much easier to measure.

[anonymous-user]

jamie g said:

no that can't be right. A measured force of 'x' and then another measured force of 'x' one sceond later still only equals a maximum force of 'x', not 'x' + 'x'

Well, in order to not confuse i didn't include the bit about "inertia"......


(basically, each firing event effectively transfers energy into the engines rotating inertia, which then "meters out" this energy (if you had no inertia (not that that would actually be possible) then you would "see" all the individual firing strokes, and each torque spike would be massive!


(take say a typical BMEP of 14bar at peak torque, work out that pressure on say a piston of 90mm diameter, then multiply that by half the crank stroke, and the out come is literally "thousands" of Nm. However, due to inertia, this pulsed output is smoothed into an output torque that is the "time averaged" firing loads.)

[jamie g]

busta said:

jamie g said:

But torque has no time or revolution reference in its value. Torque is an instantaneous value measured at one point in time.
Imagine pushing down on your bicycle pedal, the force applied is transferred into torque. In this case the piston is replacing your leg, and surely one big piston can push harder than one half the size?

Your right that torque doesn't have a time reference and if we could measure an instantaneous torque output mid combustion stroke then you would get a figure much higher than the average. However this is a pretty arbitrary figure as the engine has to work all the time, not just for 1/4 of the combustion cycle, so we actually measure the average torque output over a complete combustion cycle which is not only useful but it's much easier to measure.

If that is the answer, I'm not totally happy with it. 10lbs/ft should be 10lbs on a foot lever shouldn't it?

[icepop]

So is it nothing to do with reciprocating mass, a sort of reverse of why a 16v engine can generally rev higher than an 8v engine because of less mass/unit for the camshaft/valve train to move. The reverse then applies for 2 huge pistons vs 4 smaller ones, the force required to stop said mass from revolving is less for the 4 cylinder than the 2 cylinder, or in other words the torque is less.

Surely there must be some second year students on here that can answer this question, the bar must be closed by now

[anonymous-user]

ok, think of this scenario.

Take a heavy car wheel and tyre, mount it on a bearing, so it can spin freely.


Now, start throwing cricket balls at the outside of the tyre, so that they hit the tyre at a slight angle to the circuference. After you have thrown a few, the tyre will start to rotate, and if you throw them more often, it will accelerate and spin faster. Now, if you assume somehow that there were no frictional losses (impossible in reality) the tyre would just keep on accelerating (until it's rotational velocity actually matched that of the incoming balls) But if you applied a "brake" to slow it down until it's speed remained constant, then the torque you would measure on this brake would not be the instanteous value of that imparted by each ball coliding with the wheel, but the "average" value of all the balls hitting the tyre. Effectively, the mechanical rotational inertia of the system damps the torque fluctuations. it's the same in an engine.

[intrepid44]

icepop said:

So is it nothing to do with reciprocating mass, a sort of reverse of why a 16v engine can generally rev higher than an 8v engine because of less mass/unit for the camshaft/valve train to move. The reverse then applies for 2 huge pistons vs 4 smaller ones, the force required to stop said mass from revolving is less for the 4 cylinder than the 2 cylinder, or in other words the torque is less.

Surely there must be some second year students on here that can answer this question, the bar must be closed by now

Bah humbug... I wish, I'm meant to be doing an electronics lab report on digital circuits, despite being a Mech Eng student. I'm too tired to answer/ not entirely sure what the question is?

[GroundEffect]

icepop said:

So is it nothing to do with reciprocating mass, a sort of reverse of why a 16v engine can generally rev higher than an 8v engine because of less mass/unit for the camshaft/valve train to move. The reverse then applies for 2 huge pistons vs 4 smaller ones, the force required to stop said mass from revolving is less for the 4 cylinder than the 2 cylinder, or in other words the torque is less.

Surely there must be some second year students on here that can answer this question, the bar must be closed by now

Generally 16V engines rev higher because of volumetric efficiencies (more valve area on the cylinder head), not mechanical efficiency.

[anonymous-user]

icepop said:

why a 16v engine can generally rev higher than an 8v engine because of less mass/unit for the camshaft/valve train to move.

in a 16v engine, total valvetrain mass (and inertia) is actually more than for the equivalent 8V.............

[cptsideways]

BMEP is basically the amount of average force the piston is pushed down by, on a downstroke. Measured as pressure either PSI or Bar inside the cyclinder I think, or can be calculated from torque at the crank - see next

Torque (turning force) is measured at the crank, the stoke of the of the crank effects the torque figure (leverage of the pistons PSI on the crank)

If the stoke of a crank is 2ft (Hence 12" or 1ft of leverage from the piston) & the piston that is 1" in area pushes down with 1lb of PSI the torque at the crank will give 1ftlb of torque. Halve the crank diamter (leverage) you also halve the torque the engine makes. However it might be able to spin twice as fast so could make more bhp which is a calculation of torque x RPM

Hence why F1 engines have a tiny tiny stroke of 36 ish mm, which is why they can rev to 20k

Edited by cptsideways on Thursday 5th May 00:22

[busta]

jamie g said:

If that is the answer, I'm not totally happy with it. 10lbs/ft should be 10lbs on a foot lever shouldn't it?

No point quoting the peak if the engine can't actually produce it for a full revolution. Say an engine needs to provde a steady 100lbs/ft at the flywheel to move a car. If it can only produce this for a 1/2 a rotation of the crank, the engine wouldn't be able to run as it can't complete a revolution. To produce a consistent 100lb/ft minimum, the engine must produce significantly more on he combustion stroke.