My experience with Hellion Top Mount..

[blakeman8192]

In a word no. The dynomometer measures the torque being applied to the roller at that RPM. So the number of combustion events required to produce that torque value at that RPM is already baked into the result.

Moreover, the power is "inferred" based off the change in torque. Or more simply put, the power is a component development of what the motor must do in order to produce the torque at that engine speed.

Which is why you see the power increasing when torque is either flat or in some cases declining. In order to maintain the torque the motor must continue to increase power output up the RPM range.

At a 1:1 drive ratio with a fixed rear ratio, the motor must increase power output all the way up the rpm range in order to maintain a desired torque output.

This isn't a difficult concept. It's why they make transmissions in the first place, to increase the torque multiplication as the engine runs out of rpm/power to sustain it as speed increases.

Your example is laughably flawed. When a motor vehicle tops out it's generally either RPM limited or torque limited (meaning it runs out of rpm to redline and can no longer accelerate or apply more power to accelerate the car any further, OR it runs out of torque required to push against the exponentially increasing wind resistance).

The Dynojet is an inertial dyno and measures the time taken for the roller to speed up, combines that with the known mass of the roller, and infers torque from there. You can get accurate torque readings without measuring RPM at all on these things. RPM is not "baked in" to the result whatsoever.

At the end of the day you need to burn fuel as quickly as possible to accelerate. You can burn that fuel with a bigger boom (torque), or by creating more booms for a given timeframe (RPM/horsepower). There's a reason why imports with low torque can hold their own against American v8s in some cases. You can also just feel car pull harder at higher RPM even if it isn't making more torque than it did at a lower RPM.

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[Angrey]

The Dynojet is an inertial dyno and measures the time taken for the roller to speed up, combines that with the known mass of the roller, and infers torque from there. You can get accurate torque readings without measuring RPM at all on these things. RPM is not "baked in" to the result whatsoever.

At the end of the day you need to burn fuel as quickly as possible to accelerate. You can burn that fuel with a bigger boom (torque), or by creating more booms for a given timeframe (RPM/horsepower). There's a reason why imports with low torque can hold their own against American v8s in some cases. You can also just feel car pull harder at higher RPM even if it isn't making more torque than it did at a lower RPM.

I can explain it to you, but I can't understand it for you. There's mountains of literature and even classes you can take.

Torque is what accelerates a vehicle. Period. End of discussion. POWER is what sustains or increases torque.

The post above that tries to equivocate the number of ignition events into some sorta wildly erroneous calculation about different motors at different rpms is just WRONG. The dyno measures the torque output to the rollers at whatever RPM the motor is operating at the time. There's no supplemental calculation. The RPM and ignition/combustion events required to PRODUCE THAT TORQUE are already incorporated into that value. (in essence, it couldn't produce that torque WITHOUT those cylinder events).

The reason that high horsepower, low torque motors can compete is called TORQUE MULTIPLICATION (i.e. a transmission).

High power, low torque motors (of any kind) can be made to be productive and useful through torque multiplication (which is why we use transmissions). However, transmissions have limits (i.e. the number of gears to keep the motor within the optimal operating rpm range).

This is why tractor trailers have 7,8,9 gears with a high low splitter, because TORQUE is what is required to accelerate a mass.

Horsepower is important, it helps maintain and keep that torque (and acceleration) going forward, but torque is what moves the vehicle. More torque, either by the motor's output or by artificially multiplying it with a mechanical gear is what accelerates it faster. Power is what helps to keep the torque and acceleration longer within the same gear.

 

[blakeman8192]

I can explain it to you, but I can't understand it for you. There's mountains of literature and even classes you can take.

Torque is what accelerates a vehicle. Period. End of discussion. POWER is what sustains or increases torque.

Newton figured this out over 300 years ago. Torque is a force. Power is force over time. Acceleration comes from power (work).

The post above that tries to equivocate the number of ignition events into some sorta wildly erroneous calculation about different motors at different rpms is just WRONG. The dyno measures the torque output to the rollers at whatever RPM the motor is operating at the time. There's no supplemental calculation. The RPM and ignition/combustion events required to PRODUCE THAT TORQUE are already incorporated into that value. (in essence, it couldn't produce that torque WITHOUT those cylinder events).

The dyno measures difference in barrel RPM over a fixed window of time to calculate the force required to achieve that speed delta for that known window of time. Engine RPM is then plugged in on the other side of the measurement to derive horsepower.

The reason that high horsepower, low torque motors can compete is called TORQUE MULTIPLICATION (i.e. a transmission).

High power, low torque motors (of any kind) can be made to be productive and useful through torque multiplication (which is why we use transmissions). However, transmissions have limits (i.e. the number of gears to keep the motor within the optimal operating rpm range).

This is why tractor trailers have 7,8,9 gears with a high low splitter, because TORQUE is what is required to accelerate a mass.

Again, power (work) is what accelerates a mass. Torque is a force, and power is force over time.

Horsepower is important, it helps maintain and keep that torque (and acceleration) going forward, but torque is what moves the vehicle. More torque, either by the motor's output or by artificially multiplying it with a mechanical gear is what accelerates it faster. Power is what helps to keep the torque and acceleration longer within the same gear.

Horsepower is entirely derived from torque and RPM, it has no influence on torque and it does not "keep torque going forward" (what do you even mean by that?). An engine makes a certain amount of force per combustion event (torque), and the rate at which it makes that force (RPM) is its total power output.

An engine that makes a shitload of torque running at 1RPM is able to get heavier loads moving in the first place, but is not going to do so quickly because it's only producing the force once per minute.

 

[Angrey]

I know you're not going to understand this, but here it is anyway

Microsoft PowerPoint - Area Under the Torque vs.ppt (tazcobra.com)

"Maximize the area under the torque vs. RPM curve and you Will maximize acceleration and final speed."

 

[robvas]

This is getting over-complicated. The whole point was you want to ride out the HP until it's less than what it would be when you shift.

Someone originally made a bad assumption looking at the original graph (saying you would not need to go past 6500).

It's that simple. Works for every car. You don't need math.

Erase the torque from the graph becuase it's just misleading. HP is the only one you care about for shift points in 1/4 mile racing

 

[blakeman8192]

I know you're not going to understand this, but here it is anyway

Microsoft PowerPoint - Area Under the Torque vs.ppt (tazcobra.com)

"Maximize the area under the torque vs. RPM curve and you Will maximize acceleration and final speed."

This document literally says "power, force times velocity" which is flat out wrong. Power is force over time. If I made a PowerPoint with fancy equations, would you take my word as gospel?

 

[Angrey]

This document literally says "power, force times velocity" which is flat out wrong. Power is force over time. If I made a PowerPoint with fancy equations, would you take my word as gospel?

Just stop. You're stomping around in puddles and you don't have the shoes for it.

Power is a unit of ENERGY/Time. In order to resolve units of ENERGY you need a force over a DISTANCE. So Power = force x distance/unit time (i.e. work/time). Hence, Power equal force-velocity.

Power can not be "force over time" you simpleton because power is energy/time and force is not energy.

I could walk you through basic physics and the different units and what they mean, but I'd be wasting key strokes.

power = force x velocity - Google Search

 

[blakeman8192]

Just stop. You're stomping around in puddles and you don't have the shoes for it.

Power is a unit of ENERGY/Time. In order to resolve units of ENERGY you need a force over a DISTANCE. So Power = force x distance/unit time (i.e. work/time). Hence, Power equal force-velocity.

Power can not be "force over time" you simpleton because power is energy/time and force is not energy.

I could walk you through basic physics and the different units and what they mean, but I'd be wasting key strokes.

power = force x velocity - Google Search

P = Fv is only in the case where a constant force is moving an object at a constant velocity. We are talking about acceleration.

If torque was the only factor in acceleration, then we wouldn't be measuring horsepower at all. If RPM was "baked in" to torque measurements, then how does the formula for horsepower, HP=TQ*RPM/5252, make any sense? Again, we just wouldn't be measuring horsepower at all. HP specifically involves RPM because the rate at which the engine produces power absolutely matters and torque doesn't capture that.

You claimed that gear ratios are the only thing that allows lower torque cars to accelerate at the same rate as higher torque cars. This is laughably wrong. Get on a bike and push the pedals as hard as you can, but only once every 5 seconds. Then push the pedals hard as you can, as fast as you can. That's why RPM matters, and why horsepower is a thing.

 

[Angrey]

P = Fv is only in the case where a constant force is moving an object at a constant velocity. We are talking about acceleration.

If torque was the only factor in acceleration, then we wouldn't be measuring horsepower at all. If RPM was "baked in" to torque measurements, then how does the formula for horsepower, HP=TQ*RPM/5252, make any sense? Again, we just wouldn't be measuring horsepower at all. HP specifically involves RPM because the rate at which the engine produces power absolutely matters and torque doesn't capture that.

You claimed that gear ratios are the only thing that allows lower torque cars to accelerate at the same rate as higher torque cars. This is laughably wrong. Get on a bike and push the pedals as hard as you can, but only once every 5 seconds. Then push the pedals hard as you can, as fast as you can. That's why RPM matters, and why horsepower is a thing.

I'm finished. This is my last post. If you don't understand it, I can't help you any more than I already have.

Force is not energy. You can not create work or kinetic energy WITHOUT a distance. Energy, applied over an amount of time becomes POWER.

So in order to create POWER, you must have a force, you must have a distance and you must have a time. You can not create power without all 3.

If you derive the constant 5252 and resolve the equation for horsepower, you'll answer your own question.

How do you convert engine torque to horsepower? | HowStuffWorks

 

[blakeman8192]

I'm finished. This is my last post. If you don't understand it, I can't help you any more than I already have.

Force is not energy. You can not create work or kinetic energy WITHOUT a distance. Energy, applied over an amount of time becomes POWER.

So in order to create POWER, you must have a force, you must have a distance and you must have a time. You can not create power without all 3.

If you derive the constant 5252 and resolve the equation for horsepower, you'll answer your own question.

How do you convert engine torque to horsepower? | HowStuffWorks

Please do let it be your last post then. The force with which the engine turns the crankshaft and the rate at which it does that both matter equally. The engine accelerates the car by converting fuel into mechanical energy, and the faster it is rotating, the faster it can do that.

 

[Angrey]

Please do let it be your last post then. The force with which the engine turns the crankshaft and the rate at which it does that both matter equally. The engine accelerates the car by converting fuel into mechanical energy, and the faster it is rotating, the faster it can do that.

Which has NOTHING to do with my original posit.

You're still lost on this idea of RPM.

The torque that the dyno is measuring at 7k rpm ALREADY TAKES INTO ACCOUNT THAT THE MOTOR IS SPINNING THAT FAST TO CREATE THE TORQUE AT THAT RPM.

In order to maintain 750 ft-lbs of torque to the wheels, the motor has to work HARDER AND HARDER AS THE SPEED/RPM increase to keep that level of acceleration/force constant.

Both numbers are tied by the 5252 constant. In order for the the torque to remain constant (and not drop) as rpm increases, the power to sustain that torque grows and grows.

Creating 750 ft-lbs of torque at 3k rpms requires less power than creating it at 7500 rpm. In either case, the car is ONLY accelerating forward at the constant acceleration.

You have a giant misconception of what the RPM and dyno values actually represent.

Whatever the motor is creating AT THAT RPM WITH THAT NUMBER OF COMBUSTION EVENTS is reflected on the torque readout. It ALREADY accounts for the fact that at 7K rpms, the motor is spinning 2 times faster than at 3500 rpms. At 3500 rpms, it takes LESS POWER to create a torque level.

Or put more so an idiot can understand it, if you want to keep the same torque:

HP1 = 200 hp.
TQ1 = 300 FT-lbs.
RPM = 3500 r/m

TQ2 = 300 FT-lbs
RPM2 = 7000 rpm

Therefore in order to maintain that same torque level, the motor has to work at twice the work to keep the same torque.

In either case, the car is being accelerated at the SAME acceleration rate, it just ha to work that much harder to propel it equally at the higher speed. And that's physically what you would expect. To continue to accelerate equally it gets more and more costly as speed increases.

300 ft-lbs of torque doesn't accelerate the car any faster at 3k rpms or 7k rpms.

I think you honestly need to do some studying on the nature of acceleration and how the value is exponential.

 

[blakeman8192]

Which has NOTHING to do with my original posit.

You're still lost on this idea of RPM.

The torque that the dyno is measuring at 7k rpm ALREADY TAKES INTO ACCOUNT THAT THE MOTOR IS SPINNING THAT FAST TO CREATE THE TORQUE AT THAT RPM.

In order to maintain 750 ft-lbs of torque to the wheels, the motor has to work HARDER AND HARDER AS THE SPEED/RPM increase to keep that level of acceleration/force constant.

Both numbers are tied by the 5252 constant. In order for the the torque to remain constant (and not drop) as rpm increases, the power to sustain that torque grows and grows.

Creating 750 ft-lbs of torque at 3k rpms requires less power than creating it at 7500 rpm. In either case, the car is ONLY accelerating forward at the constant acceleration.

You have a giant misconception of what the RPM and dyno values actually represent.

Whatever the motor is creating AT THAT RPM WITH THAT NUMBER OF COMBUSTION EVENTS is reflected on the torque readout. It ALREADY accounts for the fact that at 7K rpms, the motor is spinning 2 times faster than at 3500 rpms. At 3500 rpms, it takes LESS POWER to create a torque level.

Or put more so an idiot can understand it, if you want to keep the same torque:

HP1 = 200 hp.
TQ1 = 300 FT-lbs.
RPM = 3500 r/m

TQ2 = 300 FT-lbs
RPM2 = 7000 rpm

Therefore in order to maintain that same torque level, the motor has to work at twice the work to keep the same torque.

In either case, the car is being accelerated at the SAME acceleration rate, it just ha to work that much harder to propel it equally at the higher speed. And that's physically what you would expect. To continue to accelerate equally it gets more and more costly as speed increases.

300 ft-lbs of torque doesn't accelerate the car any faster at 3k rpms or 7k rpms.

I think you honestly need to do some studying on the nature of acceleration and how the value is exponential.

Nope, your entire premise is off because acceleration is relative. Consider an object in space, which will accelerate at a fixed rate with a fixed amount of thrust, no matter how fast it is going.

Engines make lower torque at higher RPM not because it is "harder" to make it faster spinning things spin faster; but because the engine is struggling to suck in air, struggling to mix air effectively with fuel due to the shorter time between combustion events, struggling to push out exhaust gases, and friction getting significantly higher overall.

Engines convert fuel into mechanical energy. When RPM rises, they can consume more fuel, and generate more mechanical energy. That is why we care about measuring horsepower in the first place.

You've said "this is my last post" multiple times now in this thread, and I recommend you stick to that. You're making a fool of yourself.

 

[Johnny maverick]

Um? There was a thread about top mount hellions in there somewhere wasn't there?

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