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PumaRacing

Blipping The Throttle

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Batfink

I thought only torque cannot be measured accurately as that can only be measured when a load is applied

An engine dyno places a known load on the engine to get a reading - how much energy the engine uses to overcome that friction means you get a torque figure.

 

as BHP = Torque (ft/lbs) x rpm/5252.

so as there will be no load on the engine no accurate bhp reading can be measured

the bhp is the same.

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hengti

here's my schoolboy bit

The Principle of Conservation of Energy : 'energy can be changed from one form to another, but it cannot be created or destroyed'

 

My best guess is ...

If the engine produces 150bhp at 6,000 rpm, it'll be doing that regardless of whether or not it's pulling a tinny french hatchback up a hill, or sat on some chav's drive.

 

This probably has everything to do with potential and kinetic energy - just can't be RS'd to reason it out right now

Edited by hengti

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M3Evo

Hmmmm. Been thinking about this all the way home.

 

Now, Power(P)=Torque(T)xAngular velocity(can't fint a theta so A will do) if I'm not mistaken?

 

So if we assumed that the engine suffered no frictional losses, the spinning parts wouldn't need any force to keep them moving (also neglecting acceleration of pistions and rods).

 

This being the case, once the engine had accelerated to our 6000rpm, it would need to make no power to stay at that speed (Newton 1 I think although I can't remember!) hence it wouldn't be making any power.

 

However, as there will be friction etc. the engine will need to make sufficient power to overcome said friction and no more.

 

Kinda makes sense that way to me as the pressure in the cylinder required to move the unloaded engine bits it pretty tiny, whereas the pressure in the cylinder required to move the loaded engine pulling the car up a hill is great, and as Torque in this case is the only independant variable, I reckon that it's making it's 150bhp on the hill, and bugger all on the drive.

 

I'll proudly stand by that untill the flames start licking around my ankles :lol:

Edited by M3Evo

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DrSeuss

At 6k rpm with a WOT the engine management is injecting 150bhp's worth of fuel, this energy goes into accelerating the minimal mass of the engine (as opposed to the total car weight). Hence the engine spins up very quickly.

 

An engine dyno produces a drag loss to hold the engine at a given rpm. Much like driving up a given incline/speed with a fixed throttle position will maintain a steady rpm. Drag and all other losses = power output.

 

Unless i've missed something this isn't a very interesting thought experiment, its just a waste of fuel.

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charlie_sav

It will make 150 bhp the energy is going into tuning the flywheel and other rotation parts which is the only load, so it will spin up very fast,

 

if you were in gear then the energy would go into turning the rotating parts and roataing the wheels, the load will be higher so the engine will spin up slower because of it,

 

another think to look at is if you were to look at the map on the ecu you would see it cut across from idle load 0 1000rpm to load site 15 speed site 6000rpm, the ecu is injecting the fuel and spark to make max power so the engine will make max power,

 

EDIT, by the time i had writen thay Drseuss had allready said the same. Doh.

 

But there is someone that lives very near me his main hobby early saturaday and sunday is to constantly rev his motor bike over and over again, Tuning it he says, he also used to do it with this escort cab, ment to be a turbo technics but no turbo insight but it very powerfull as it can wheel spin right down our road he says !! Really i said.

Edited by charlie_sav

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RichF
This being the case, once the engine had accelerated to our 6000rpm, it would need to make no power to stay at that speed

 

In your perpetual motion universe maybe :lol:

 

I interpreted the question to mean what power does the engine produce to achieve 6000rpm:

 

hold it there until the revs just hit 6k

 

Rather than what power is used to keep it at 6000rpm.

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PumaRacing

The stationary engine is producing the same power at all points in the rev range as it would do on the road or on a dyno. All the power is being used to accelerate the crank, flywheel and other internal components.

 

If the inertia of these components is known the the power can be calculated from the rate of acceleration. Inertia dynos do this instead of measuring torque and speed.

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petert

It's not using 150hp of fuel though. It's only using enough fuel to make enough power to turn the alternator, water pump, frictional losses etc. Whilst it has the ABILITY to make 150hp, it's not. A blip to 6000 rpm with this sort of load would only use 2ms (an educated guess) of injector duration, rather than 7.5ms.

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Martin@PRD
The stationary engine is producing the same power at all points in the rev range as it would do on the road or on a dyno. All the power is being used to accelerate the crank, flywheel and other internal components.

 

If the inertia of these components is known the the power can be calculated from the rate of acceleration. Inertia dynos do this instead of measuring torque and speed.

 

Is that the answer?, a bit blunt :lol:

 

I'm sorry but its sounds your on another ego trip again Dave, do you have a personailty?

Edited by Martin@PRD

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PumaRacing
It's not using 150hp of fuel though. It's only using enough fuel to make enough power to turn the alternator, water pump, frictional losses etc. Whilst it has the ABILITY to make 150hp, it's not. A blip to 6000 rpm with this sort of load would only use 2ms (an educated guess) of injector duration, rather than 7.5ms.

 

It is producing 150 bhp at WOT at 6k and it is using the same amount of fuel as it would any other time at WOT at 6k. The 'load' is the red herring in the question and the reason I asked it because the same topic is currently the subject of various misunderstandings on a usenet forum. The load is irrelevant. All that determines is the rate of acceleration and has no effect on the bhp being produced which is solely a function of throttle position and rpm.

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Martin@PRD

thats better, :lol:

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PumaRacing
Is that the answer?, a bit blunt :lol:

 

I'm sorry but its sounds your on another ego trip again Dave, do you have a personailty?

 

What the hell gives you the right to have a go at me for posing a question and then answering it in a perfectly normal way? You bloody well wouldn't do it face to face that's for damn sure.

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PumaRacing
The engine is producing full power. The revs build very fast due to the small amount of power required to spin up the flywheel. As the drive line is disengaged, all of the engine energy/power is going straight into kinetic energy now stored in the flywheel. Cylinder pressures therefore load are low as the piston is easily pushed down during the power stroke due to the lack of resistance.

 

?????????????????????????????????????????????????????????????????????

 

Ben

 

Started off well but then shot yourself down in flames. Cylinder pressures are the same as at any other time at WOT. The engine isn't affected by 'resistance'. All of its power is going into accelerating internal components in exactly the same way as all of the power would go into accelerating the car if it were on the road. In a way it's like having a very low first gear, say one that only goes to 5mph. The car accelerates very fast, grip notwithstanding, but for a very brief time. The fastest the engine can ever accelerate is when it's only driving its own mass which effectively is the same as being in a gear with an infinitely low ratio.

 

I don't think horsepower/torque is really the issue in this.

Ignoring the engine heat and friction losses, the engine is only producing as much power as is required to overcome the friction of the gearbox oil and shafts.

 

Kinda like what efy2bjg said, but it won't be putting out anywhere near full power, and it'll be using little fuel etc.

 

The engine would work much harder and produce much more power at full throttle, going up a steep hill.

 

Nope, it's working just as hard up a hill as it is accelerating hard on the flat or accelerating even faster when all it's doing is accelerating its own internal mass. All the external load does is determine the rate of acceleration. It can't affect the operation of the engine. WOT at 6k, however briefly, means 150 bhp whether the engine is driving a car, driving a dyno or just accelerating itself.

 

thats better, :lol:

 

As the person who got it completely wrong at the first attempt I'd suggest you avoid trying to be patronising.

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Martin@PRD

Love you too Dave, come and give me a hug!!!

 

To be honest Dave Im not the typical person to jump to conclusions like everyone else does with you, yes your a wizard with your theorys and deserve all the respect, but ease up on the god playing roll, life isnt all about advertising and money making.

 

I was generally interested in your red herring question which got us all thinking well.....everything, lol. But just to give a blunt answer and not explaining a little deeper after a lot of calories have been burnt is such a premature ejaculation!!!

 

Well we all get at each other for time to time. Thats life

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mfield
Probably completely wrong but,

 

Yes it still producing power

still 150bhp

The power is still going the same place as before

And really have not got a clue about load up a hill but would that be the same aswell ?

 

 

So was i nearly correct ?

:lol:

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RichF

Ha ha, it's like handing in a piece of homework ! :lol:

Edited by RichF

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boombang
It is producing 150 bhp at WOT at 6k and it is using the same amount of fuel as it would any other time at WOT at 6k. The 'load' is the red herring in the question and the reason I asked it because the same topic is currently the subject of various misunderstandings on a usenet forum. The load is irrelevant. All that determines is the rate of acceleration and has no effect on the bhp being produced which is solely a function of throttle position and rpm.

To take a car with no running load from 5.99''k to 6k it doesn't need WOT so why mention it at all?

 

In fact I can quite easily rev my car from idle RPM to redline without WOT.

 

Or am I missing the point?

 

or does WOT mean something other than wide open throttle in Davespeak?

 

 

just to add I can also blip the throttle without having to give it "WOT".

Edited by boombang

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M3Evo

Boo, I was very wrong :(

 

Still, gave me something to think about :)

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DaveW
Is that the answer?, a bit blunt :(

 

I'm sorry but its sounds your on another ego trip again Dave, do you have a personailty?

 

harsh mate :)

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Alex G
When the engine hits 6k with the car stationary and no load of any sort being applied is the engine producing any power?

 

:) erm..maybe im missing something but at that particular instance (WOT untill it hits 6k) its producing full power at maximum load???

 

but you say no load which makes me think you mean the instant you let go of the throttle and the engine begins to decelerate, so the only power being produced is due to the inertia of the engine components, which will be small)

Edited by Alex G

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petert
It is producing 150 bhp at WOT at 6k and it is using the same amount of fuel as it would any other time at WOT at 6k.

 

But if you blipped the throttle, it wouldn't be WOT at 6K (unless you're a numbskull and held it there). It would be WOT at say 2-3K, as it takes a huge gulp of air, then the load would come off and the revs would continue to rise, regardless now of throttle position, as it digests the fuel/air ingested earlier. Surely the "blip" is less than a second in time?

 

I wonder if Jonmugie, who has a Motec M48, would care to datalogue the event for us? I'd do it myself, except my race car is in bits. You could then plot throttle position, RPM, vacuum and injection pulse for that second.

 

Certainly one to think about.

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efy2bjg
Started off well but then shot yourself down in flames. Cylinder pressures are the same as at any other time at WOT. The engine isn't affected by 'resistance'.

 

 

Damn, Almost :)

 

So why do people speak of bedding in piston rings with lots of load to keep cylinder pressures high? or have i remembered this wrong? Also why does the engine sound completely different when under load?

 

Ben

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PumaRacing
But if you blipped the throttle, it wouldn't be WOT at 6K (unless you're a numbskull and held it there). It would be WOT at say 2-3K, as it takes a huge gulp of air, then the load would come off and the revs would continue to rise, regardless now of throttle position, as it digests the fuel/air ingested earlier. Surely the "blip" is less than a second in time?

 

I wonder if Jonmugie, who has a Motec M48, would care to datalogue the event for us? I'd do it myself, except my race car is in bits. You could then plot throttle position, RPM, vacuum and injection pulse for that second.

 

Certainly one to think about.

 

The question specifies WOT in that you hold the throttle to the floor until 6k and then release it quickly before the engine overspeeds. Engine speed won't continue to rise once the throttle is closed. Any air/fuel mix ingested at WOT will be used in one cycle. Sure you have to come off the pedal pretty sharpish before you hit the redline but that's easily enough done. Try it on the driveway sometime.

 

The point of the question is really to get people to appreciate that only throttle position and rpm determine power output. It's perhaps tempting to think that an engine 'works harder' on a dyno or going uphill but that's confusing external load with power output. As I posted on the usenet forum earlier today, someone once accused a rolling road operator I knew of blowing his engine up by 'loading it up too much'. Harder than it would have been loaded on the road. That's nonsense though because an engine is 'loaded' as much as it ever can be by the simple act of opening the throttle fully.

 

Damn, Almost :)

 

So why do people speak of bedding in piston rings with lots of load to keep cylinder pressures high? or have i remembered this wrong? Also why does the engine sound completely different when under load?

 

Ben

 

To bed in rings you ideally need short bursts of full throttle to generate high cylinder pressures and force the rings hard against the bore walls. The external load won't affect the cylinder pressures but it will affect how long and how quickly the car accelerates for. If you want to keep the rpm low but the cylinder pressure high for several seconds then you need a high gear or a hill. The load isn't to keep the cylinder pressures high, it's to stop the revs rising too fast.

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RichF
someone once accused a rolling road operator I knew of blowing his engine up by 'loading it up too much'.

 

Doesn't that depend on what part actually broke?

 

I mean, if the engine is trying to exert it's power on some components and you're restricting the movement of those components, (or components further down the chain e.g. the wheels) by applying a load/braking force then they are more likely to break aren't they ? To a lay-person (sp?) that may be interpreted as "you've blown my engine up".

 

Not saying the bloke is right, just trying to see why he might have said that.

Edited by RichF

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