Applying Torque vs Horsepower

[Dominant1]

A win for the Chevy cuz you wouldn't be fitting that coyote in anything.

Most of the cubes come from the stroke less from the bore... so it would fit!

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[TexasRebel]

It's also necessary to mention that the torque v. angular displacement plot is sinusoidal. With increased RPM the frequency of the sine wave increases until the fluctuation becomes negligible. This is very apparent in the old long stroke hit-miss engines, bicycles and steam locomotives.

torque is generated by the component forces of the cylinder pressure during combustion acting on the face of the piston, through the connecting rod, perpendicular to the crankshaft throw. All of this culminates into storing energy in the rotating mass of the engine, and eventually the kinetic energy of the car itself.

Torque without angular displacement is nothing. Remember from physics, force without distance means no work was done. "Revolutions per Minute" brings both time and angular displacement. Again from that physics class that might've seemed like a good nap time after lunch... work over time is power.

 

[Coolmanfoo]

It's also necessary to mention that the torque v. angular displacement plot is sinusoidal. With increased RPM the frequency of the sine wave increases until the fluctuation becomes negligible. This is very apparent in the old long stroke hit-miss engines, bicycles and steam locomotives.

torque is generated by the component forces of the cylinder pressure during combustion acting on the face of the piston, through the connecting rod, perpendicular to the crankshaft throw. All of this culminates into storing energy in the rotating mass of the engine, and eventually the kinetic energy of the car itself.

Torque without angular displacement is nothing. Remember from physics, force without distance means no work was done. "Revolutions per Minute" brings both time and angular displacement. Again from that physics class that might've seemed like a good nap time after lunch... work over time is power.

Wtf do half of those words even mean?

 

[MattW]

It's also necessary to mention that the torque v. angular displacement plot is sinusoidal. With increased RPM the frequency of the sine wave increases until the fluctuation becomes negligible. This is very apparent in the old long stroke hit-miss engines, bicycles and steam locomotives.

You are getting a bit off topic, IMO, but if you want to go this direction you need to include discussion of the flywheel. The flywheel attenuates that effect, but eats part of the torque provided by the engine, but this varies as you change the flywheel moment of inertia.

 

[Rebellion]

It's also necessary to mention that the torque v. angular displacement plot is sinusoidal. With increased RPM the frequency of the sine wave increases until the fluctuation becomes negligible. This is very apparent in the old long stroke hit-miss engines, bicycles and steam locomotives.

torque is generated by the component forces of the cylinder pressure during combustion acting on the face of the piston, through the connecting rod, perpendicular to the crankshaft throw. All of this culminates into storing energy in the rotating mass of the engine, and eventually the kinetic energy of the car itself.

Torque without angular displacement is nothing. Remember from physics, force without distance means no work was done. "Revolutions per Minute" brings both time and angular displacement. Again from that physics class that might've seemed like a good nap time after lunch... work over time is power.

'cuz #engineering

In a cross plane crank engine...wouldn't the sinusoidal effect be basically "negated" because there are 4 sinusoidal inputs of equal phase shifts? I mean, not fully negated, but the resulting tangential force would vary much less than a pure sinusoidal. The flywheel and crank (plus other rotational components) should have some equalizing effect on the peaks and valleys, leading to a more constant tangential force at the wheels.

Correct me if I'm wrong, my engine knowledge is rather limited.

 

[Dominant1]

you guys are getting to technical for me time to unsubscribe...lol

 

[vnzbd]

Very technical but very good... Following the formula posted: Torque x RPM / 5252 = horse power, How does that work with the EB when it loses so much horse power but maintains it's torque on 87 octane? If the torque and 5252 are constants does it lose RPM's on a lower octane?

 

[Rebellion]

Very technical but very good... Following the formula posted: Torque x RPM / 5252 = horse power, How does that work with the EB when it loses so much horse power but maintains it's torque on 87 octane? If the torque and 5252 are constants does it lose RPM's on a lower octane?

Don't have a EB, but I suspect what it means is that top torque (at a lower RPM) is maintained with 87, but at higher RPM it falls off quickly leading to lower top HP number. Most people post top numbers.

 

[TexasRebel]

'cuz #engineering

In a cross plane crank engine...wouldn't the sinusoidal effect be basically "negated" because there are 4 sinusoidal inputs of equal phase shifts? I mean, not fully negated, but the resulting tangential force would vary much less than a pure sinusoidal. The flywheel and crank (plus other rotational components) should have some equalizing effect on the peaks and valleys, leading to a more constant tangential force at the wheels.

Correct me if I'm wrong, my engine knowledge is rather limited.

No, the firing order and bank have nothing to do with the waveform. Each cylinder will have 440 degrees of energy consumption (exhaust, intake, compression strokes) and then 180 degrees of energy production (power stroke). In a 90 degree V8 another power stroke starts every 90 degrees so there will be some overlap. Similarly, on a 180 degree I4, only one power stroke is happening at any time, there is no overlap.

The waveform is generated by the force acting on the crank throw.

Think of a single cylinder engine (lawnmower) and the first few "pops" when it starts. As combustion begins, very little torque is generated as the connecting rod and crank throw are 180 degrees apart. All of the force (cylinder pressure * Bore^2 * pi/4) at this point is pushing directly against the main caps (sine of 180 degrees is 0, put a bicycle's crankshaft in the up/down position and press straight down on the upper pedal... no movement). As the stored energy in the flywheel turns the crank, the angle between the connecting rod and the crank throw begin to decrease, which increases the amount of force that generates torque.

Maximum torque occurs a little sooner than 90 degrees (crank rotation) after spark, when the connecting rod and the crank throw are perpendicular to each other (sine of 90 degrees is 1). There is additional variation when you consider the angle between the connecting rod and the cylinder bore which is why rod to stoke ratio matters.

The flywheel is the key. It is mechanical energy storage. It keeps the engine turning through the exhaust, intake, and compression strokes until the cylinder fires again. With no flywheel, the engine would stall under just about any load. With a very large (diameter) or thick flywheel the pulses from the torque wave are completely attenuated, but at a cost of a lower angular acceleration. The pulses still exist, but are miniscule compared to the energy stored in the flywheel.

If you're still with me, I know this seems like a tangent, but it's not. Knowing where torque comes from is necessary to understand the torque-horsepower relationship.:headbonk:

 

[crs2879]

A win for the Chevy cuz you wouldn't be fitting that coyote in anything.

Yep...seen a couple shows where shops were trying to fit a Coyote into and older mustang.....it doesn't fit without considerable fabrication.

BTW, I was not criticizing either motor. What GM has done with the pushrod V8 is pretty amazing and revving the Coyote to 7K RPM is actually enjoyable.... You can buy a GM 1500 series pickup truck with 420HP and Ford has released a TTV6 truck making 450HP.....:hail: Salad days were are living in terms of the evolution of the IC engine....may it live forever.....:cheers:

 

[Coolmanfoo]

Very technical but very good... Following the formula posted: Torque x RPM / 5252 = horse power, How does that work with the EB when it loses so much horse power but maintains it's torque on 87 octane? If the torque and 5252 are constants does it lose RPM's on a lower octane?

Excuse my ignorance but is there any reason that formula uses 5252 and would that number be the same regardless of engine specs?

 

[Rebellion]

No, the firing order and bank have nothing to do with the waveform. Each cylinder will have 440 degrees of energy consumption (exhaust, intake, compression strokes) and then 180 degrees of energy production (power stroke). In a 90 degree V8 another power stroke starts every 90 degrees so there will be some overlap. Similarly, on a 180 degree I4, only one power stroke is happening at any time, there is no overlap.

The waveform is generated by the force acting on the crank throw.

Think of a single cylinder engine (lawnmower) and the first few "pops" when it starts. As combustion begins, very little torque is generated as the connecting rod and crank throw are 180 degrees apart. All of the force (cylinder pressure * Bore^2 * pi/4) at this point is pushing directly against the main caps (sine of 180 degrees is 0, put a bicycle's crankshaft in the up/down position and press straight down on the upper pedal... no movement). As the stored energy in the flywheel turns the crank, the angle between the connecting rod and the crank throw begin to decrease, which increases the amount of force that generates torque.

Maximum torque occurs a little sooner than 90 degrees (crank rotation) after spark, when the connecting rod and the crank throw are perpendicular to each other (sine of 90 degrees is 1). There is additional variation when you consider the angle between the connecting rod and the cylinder bore which is why rod to stoke ratio matters.

The flywheel is the key. It is mechanical energy storage. It keeps the engine turning through the exhaust, intake, and compression strokes until the cylinder fires again. With no flywheel, the engine would stall under just about any load. With a very large (diameter) or thick flywheel the pulses from the torque wave are completely attenuated, but at a cost of a lower angular acceleration. The pulses still exist, but are miniscule compared to the energy stored in the flywheel.

If you're still with me, I know this seems like a tangent, but it's not. Knowing where torque comes from is necessary to understand the torque-horsepower relationship.:headbonk:

I think I understand most of it. The part I don't quite get is that, while the torque output of each cylinder is sinusoidal, when you put in a V8 configuration, wouldn't the overlap (due to the 90 degree phase shift) cause the summation of forces to approximate constant.

I'm thinking it's like the summation of 4 sinusoidal waves each with 90 degree phase shift from the last one. Perhaps my math is wrong or my understanding of the 4 cycles ain't correct.

 

[MattW]

Excuse my ignorance but is there any reason that formula uses 5252 and would that number be the same regardless of engine specs?

It is because of the units.

[U]Property[/U]       [U]SI unit[/U]             [U]English/auto unit[/U]
rev-rate       radian/sec          rpm
torque         Newton-meter        ft-lbf  (lbf = pound-force)
power          watt                hp

If you use SI units, there is no conversion factor: power (watts) = torque (Nm) X rev-rate (radian/s)

In English/auto units: power (hp) = torque (ft-lbf) X rev-rate (rpm) / 5252

 

[GT Pony]

Excuse my ignorance but is there any reason that formula uses 5252 and would that number be the same regardless of engine specs?

As mentioned in Post #58, it's because of the units used.

The formula HP = T x RPM/5252 applies to every engine in the world. It also means that HP and T are always equal to each other at 5252 RPM (you can see that on HP & T dyno graphs).

T in ft-lbs of course.

 

[TexasRebel]

Very technical but very good... Following the formula posted: Torque x RPM / 5252 = horse power, How does that work with the EB when it loses so much horse power but maintains it's torque on 87 octane? If the torque and 5252 are constants does it lose RPM's on a lower octane?

Rated torque and horsepower peaks are always accompanied by an "@ xxx RPM". Peak horsepower and peak torque never occur at the same RPM.

Torque is not necessarily a constant. 87 Octane and 91 Octane fuel burn differently. Higher Octane fuels have a slower and more controlled burn. Different burns generate different cylinder pressures with respect to time. Cylinder pressure and crank position determine torque.

I think I understand most of it. The part I don't quite get is that, while the torque output of each cylinder is sinusoidal, when you put in a V8 configuration, wouldn't the overlap (due to the 90 degree phase shift) cause the summation of forces to approximate constant.

I'm thinking it's like the summation of 4 sinusoidal waves each with 90 degree phase shift from the last one. Perhaps my math is wrong or my understanding of the 4 cycles ain't correct.

The overlap definitely smooths the torque v. crank position curve out. In fact that is why a 45 degree V16 (yep, Cadillac had one) runs extremely smooth. The reason the sine wave always exists though is because the application of force to a piston is not binary. The cylinder pressure increases as the gas expands during the burn, and decreases as the chamber volume increases after the burn. The transfer of force through the connecting rod is not constant. At top-dead-center the angle between the connecting rod and the bore is 0. In this case we use the cosine of the angle to determine axial force, and the cosine of 0 is 1. As the engine rotates the angle between the connecting rod and the bore increases. This reduces the amount of force that is transferred through the rod to the crankshaft (very slightly due to the rod length to stroke length ratio) and increases the component of the reaction force driving the piston into the bore wall (perpendicular to the bore axis). This is why engines with heavy loads will have significantly more ring ridge and cylinder wear than one that idles.

Finally, the force that is transferred through the connecting rod to the crankshaft does not uniformly produce torque. As the crankshaft rotates the angle between the crank throw and the connecting rod reduce from 180 to 0 degrees. At both TDC and BDC none of the pressure applied to the piston go toward torque generation (sine of 180 and 0 are both 0).
In an ideal world, the torque v. crank position plot would look identical for each cylinder. In a multi-cylinder engine each plot is simply phase shifted by the firing angle. If you were to have a 1 degree 720-cylinder engine, a power stroke would begin every degree the crankshaft turns. It would run incredibly smooth, but the torque v. crank position plot would still be sinusoidal. ;)

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