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Today I measured an engine (4 cylinder) with no load.

RPM accelerates at approx 4.6 RPM per 1mS,

When accelerating the engine between 3400 rpm 5200 rpm it does it in 389ms.

Does anyone know how good or bad this is ?

Does anyone know how to work out the "1/2MV2 Theory"?

..


From the look of it he's trying to work out how quick a certain engine revs.

Obviously the weight of components in an engine and it's ignition curve will affect how quick it can rev. As for the "1/2MV2 Theory" I have no idea what it is but I'm sure an automotive book would have it. Maybe it's 1/2 x Mass x velocity sqr'd?

Hmmmm, Evil genius at work. Either that or they have intenet access at graylands now :) (joke)

Ok I'll play along

T = 2/3<1/2mv2>=?

As we know

PV=nRT --------- (1) (why did we use this equation for prove above?)

As we know.

n=N/NA (why we don’t put the P=F/A and volume formula)

Put in above equation (1)

PV= N/NA RT---------(2)

Where R/NA= k (k is Boltzman constant)

Equation 2 becomes

PV=NkT

Or

P=NkT/V------------(3) (why we found “P”)

As we know.

P=2/3*N/NA<1/2mv2>-------------(4) (Why we use this equation we can also use this formula)

Comparing 3 and 4. (why we compared)

NkT/V=2/3*N/NA<1/2mv2>.

We get.

kT=2/3<1/2mv2>

or

T=2/3k<1/2mv2>

T=constant<K.E>

T Directly Proportional to <K.E>

also

Defining gravitational potential energy

The change in gravitational potential energy of the object-earth system is just another name for the negative value of the work done on an object by the earth.

Total energy for object moving under gravity

Definition of gravitational potential energy allows us to write down a conservation law for the total energy

D (K+Ug)=0 => E=Ug+K=constant

½ mv2+mgh = constant

Crucial that U only depends on height!

Demo

A ball rolls down an inclined slope.

After reaching the bottom of the incline, its inertia keeps it moving.

It then rolls up an inclined slope. The inclination angles of the two slopes differ.

There is negligible friction.

Observed Result

The maximum heights achieved on the two slopes are the same.

Reason: Conservation of energy.

Conservative forces

Potential energies can only be defined for conservative forces

Those forces which do work on a object in moving it from one position to another in such a way that the work done is path independent.

eg. gravity, electric forces, spring forces

Not friction, air resistance

Pendulum demo

Energy (K+U) should be constant

If pendulum released with zero speed will return to same point (height) with zero speed (ignoring air drag, friction etc)

Stopped-pendulum demo

Another example: springs …

Force F=-kx (Hooke’s law)

(x extension)

W=-1/2kx2 (last time)

Therefore, can define elastic (spring) potential energy

U= 1/2kx2

(Horizontal) Spring

spring

frictionless table

x

1/2kx2+1/2mv2=constant

Motion ?

Imagine displacing mass from equilibrium to x=a – what happens ?

1/2ka2=1/2kx2+1/2mv2 or

1/2mv2= 1/2ka2-1/2kx2

what range does x lie in ?

when is v greatest ?

describe motion …

Graph

x

u(x)=1/2kx2

a

oscillates!

motion confined to

region below dotted

line

-a

Question

What is meaning of the dashed horizontal line in the graph?

Hint: What is conserved?

Choose a particluar location x between -a and +a.

Draw a vertical line through the x value.

What is the meaning of the intercept of that line with the graph?

What is the meaning of the distance from the intercept to the dashed line?

The graph can be used to get the speed v of the mass, at any location x.

How can this be done?

I would think that measuring an engine with no load would be impractical in a real world application. Isn't it dangerous for the engine as well?

The load you attach to an engine will change any of your results. Different loads = different results = what are you trying to accomplish?

........what are you trying to accomplish?

What is he trying to accomplish? And does it really belong here? This isn't a science think tank and the only connection this has with Z cars is that they also have an engine. I can put up with this as long as you all can put up with it. Rock on!

ok well ive got 2 engines of the same size but different mods

If i compare the noload acceleration rates i might have sum idear of power output.

I am recording anything i can so that when i put my copmuter (computer)

Into control the engine timing I can compare before and after effects.

note: engine is protected with my adjustible 2 stage rev limiter

ok so ive been doing little hooning

rev limit stop is set at 5500

rev limit soft is set at 5300

when traveling from sum traffic lights to 100 km/h sign

I have found taking off but taking good care of the clutch im doing 100 km/h.

I have found using launch control @3500 RPM im doing 110 km/h with tools+ passenger.

Cleaned out tools and passenger using launch control @3500 RPM im now doing 120 km/h to the same point.

so that 20 km/h is gained over the same distance by using the engines peak toque to start off and unloading extra mass, when i shift gears my foot does not leave the floor, its hard to get used to though, just change quickly not a slam

approx 100 kg?

car is 1060kg + 75kg

I have to go and get a rough distance measurment to work it out.

I think if you want to accelerate twice as fast you need 4 times the power

because you have to get half the distance in half the time.

thats how i understand it

Your useless ramblings have what to do with Classic Z cars? Other than the fact that Z cars have engines? You are clogging up this site with your theories, your attempts to justify something that no one is asking for, and in general, your crap. I DON'T KNOW WHY MIKE HASN'T KILLED THIS THREAD ALONG WITH YOUR OTHERS, AND I DON'T KNOW WHY HE HASN'T BANNED YOU SINCE YOU CONTRIBUTE NO Z RELATED CONTENT TO THIS SITE. I WONDER HOW MANY IGNORE LISTS YOU ARE ON. I WONDER WHAT YOU ARE ON.

DONE.

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