RE: Final EU vote on 2035 engine phaseout delayed

[911hope]

911hope said:

Pan Pan Pan said:

And if you stop living in cloud fantasy land, you might understand that it takes more energy to move 1814.36 kilograms, than it does to move 1360.77 kilograms. Perhaps you believe that changing the power source from ICE to electric makes 453.59 kilograms `magically' disappear?

You mention a mass increase of 33% between 2 cases, then ask if using an electric motor's well documented efficiency advantage over ICE ( about a factor of 3).

Have a go at the sums, you will find that you defeat your own argument.

If you can't, try dividing 1.33 by 3.

What does that answer tell you?

Was that too difficult?

[bigothunter]

GT9 said:

Pan Pan Pan said:

And if you stop living in cloud fantasy land, you might understand that it takes more energy to move 1814.36 kilograms, than it does to move 1360.77 kilograms. Perhaps you believe that changing the power source from ICE to electric makes 453.59 kilograms `magically' disappear?

Simple question.
Do you or do you not accept that kinetic energy is recoverable?

Who mentioned velocity? Pan only referred to the energy required to move 1814.36 kilograms or 1360.77 kilograms over the same distance.

Energy required per unit distance = A + Bv + Cv^2

Go sufficiently slowly and the aerodynamic term (Cv^2) tends to zero. Bv is virtually zero anyway. Which leaves the A term of tyre loss which is basically proportional to vertical load.

Vertical load using Pan's figures is 17798.9 N or 13349.2 N plus vehicle weight (assume 24525.0 N).

So if you travel below 5 mph transporting 4449.7 N greater load (W=mg) then approximately 11.7486% more energy is consumed. Do the same on steel rails and that figure will reduce below 1%.

Pan was correct all along


nb Assumes Pan was referring to units of Kilogram Mass not Kilogram Force and Powertrain Efficiency is disregarded. Also KERS is utilised on ICE Hybrids.

Edited by bigothunter on Sunday 2nd April 13:28

[bigothunter]

NMNeil said:

bigothunter said:

NMNeil said:

A gas powered turbine for electrical generation runs at around 60% efficiency, and with catalytic converters (which are unlikely to be removed unlike an ICE car) these emissions can be reduced to almost nothing.

What percentage of ICE cars have their catalysts removed? Absolutely miniscule I suggest.

Perhaps this trend is more common in the US than UK?

Maybe do a search on PH for 'decatting'.
And while waiting
https://www.avontuning.co.uk/services/advanced-opt...
https://www.japspeed.co.uk/product/exhausts/exhaus...

We agree - there is absolutely minimal occurrence of catalyst removal and no significant effect on emission levels. Assertion was erroneous. Case dismissed

[911hope]

bigothunter said:

Who mentioned velocity? Pan only referred to the energy required to move 1814.36 kilograms or 1360.77 kilograms over the same distance.

Energy required per unit distance = A + Bv + Cv^2

Go sufficiently slowly and the aerodynamic term (Cv^2) tends to zero. Bv is virtually zero anyway. Which leaves the A term of tyre loss which is basically proportional to vertical load.

Vertical load using Pan's figures is 17798.9 N or 13349.2 N plus vehicle weight (assume 24525.0 N).

So if you travel below 5 mph transporting 4449.7 N greater load (W=mg) then approximately 11.7486% more energy is consumed. Do the same on steel rails and that figure will reduce below 1%.

Pan was correct all along


nb Assumes Pan was referring to units of Kilogram Mass not Kilogram Force and Powertrain Efficiency is disregarded. Also KERS is utilised on ICE Hybrids.

Edited by bigothunter on Sunday 2nd April 13:28

You cannot use a term like "energy is consumer" and disregard drivetrain efficiency.

The original energy for the vehicle is chemical (hydrocarbon fuel and charged battery) by the time this has been converted to kinetic energy, the vast majority of the energy in the hydrocarbon fuel has been converted to heat and the tiny minority has been used to propel the load.

The exact opposite is the case for a battery/electric motor case.

You can try the simple arithmetic question I set for PPP yesterday. It will all become clear.

You really have made a mistake in backing PPP.

BTW. The unit for force is the Newton. Never Kg.

[NMNeil]

SpeckledJim said:

And that’s why a light ICE goes further on a unit of energy than a heavy EV does.

That’s right, isn’t it?

As we're talking about e-fuel, you need to factor in the energy used in it's production as well, so no, it isn't right.

[bigothunter]

911hope said:

You cannot use a term like "energy is consumer" and disregard drivetrain efficiency.

The original energy for the vehicle is chemical (hydrocarbon fuel and charged battery) by the time this has been converted to kinetic energy, the vast majority of the energy in the hydrocarbon fuel has been converted to heat and the tiny minority has been used to propel the load.

The exact opposite is the case for a battery/electric motor case.

You can try the simple arithmetic question I set for PPP yesterday. It will all become clear.

You really have made a mistake in backing PPP.

BTW. The unit for force is the Newton. Never Kg.

Energy consumed in propelling a vehicle along the road is overwhelming dominated by aerodynamic loads and tyre losses. It has nothing to do with drivetrain efficiency.

Unlike energy consumed by the propulsion unit which is hugely dependent upon whole drivetrain efficiency. The crucial point which some choose to ignore.

Kilogram-force (kgf or kgF) or kilopond (kp) is an established unit. During WW2, Germany measured the thrust of a rocket engines in kilograms-force. Kilogram-force remained the primary unit for thrust in the Russian space program until the late 1980s. In recent times, the unit has been deprecated.

Here's a reference which should improve your knowledge: https://en.wikipedia.org/wiki/Kilogram-force

Could be worth brushing-up on the meaning of irony too

[starsky67]

Even so and allowing you to ignore the impact of regenerative braking by asking that the whole journey is done at a constant speed, you still have the issue that the ICE power unit is around 3 x less efficient than the EV one.

That more than compensates for the effect of weight.

[911hope]

bigothunter said:

Energy consumed in propelling a vehicle along the road is overwhelming dominated by aerodynamic loads and tyre losses. It has nothing to do with drivetrain efficiency.

Unlike energy consumed by the propulsion unit which is hugely dependent upon whole drivetrain efficiency. The crucial point which some choose to ignore.

Kilogram-force (kgf or kgF) or kilopond (kp) is an established unit. During WW2, Germany measured the thrust of a rocket engines in kilograms-force. Kilogram-force remained the primary unit for thrust in the Russian space program until the late 1980s. In recent times, the unit has been deprecated.

Here's a reference which should improve your knowledge: https://en.wikipedia.org/wiki/Kilogram-force

Could be worth brushing-up on the meaning of irony too

In a thread about fuel and drivetrain rules and efficiencies, seems somewhat lacking to pretend these are not the most important things, and pretend they don't exist.

[bigothunter]

911hope said:

bigothunter said:

...energy consumed by the propulsion unit which is hugely dependent upon whole drivetrain efficiency. The crucial point which some choose to ignore.

In a thread about fuel and drivetrain rules and efficiencies, seems somewhat lacking to pretend these are not the most important things, and pretend they don't exist.

Not sure about 'rules' but otherwise agreed

[bigothunter]

starsky67 said:

Even so and allowing you to ignore the impact of regenerative braking by asking that the whole journey is done at a constant speed, you still have the issue that the ICE power unit is around 3 x less efficient than the EV one.

That more than compensates for the effect of weight.

ICE Hybrids utilise regenerative braking not just pure EVs.

Kinetic energy is a function of energy transfer during acceleration (positive or negative). Without losses (principally aerodynamic and tyre), no further energy input would be required to sustain perpetual motion. Holds true until hill climbing demands greater potential energy which can be partially recovered on the next downhill section.

Also rate of acceleration has no effect on the amount of energy consumed (ignoring any speed/load efficiency changes), which simply relates to the change in kinetic energy. Except going faster sooner slightly increases mean aero drag.

[500TORQUES]

bigothunter said:

Also rate of acceleration has no effect on the amount of energy consumed (ignoring any speed/load efficiency changes), which simply relates to the change in kinetic energy. Except going faster sooner slightly increases mean aero drag.

It bloody does, you need to use a lot more energy to accelerate faster.

[GT9]

500TORQUES said:

It bloody does, you need to use a lot more energy to accelerate faster.

Sorry 500 but he is correct.
What you meant to say was that it takes a lot more power.
Regardless of how long it takes to accelerate, the energy converted to kinetic energy is defined by the change in velocity alone.

[911hope]

GT9 said:

Sorry 500 but he is correct.
What you meant to say was that it takes a lot more power.
Regardless of how long it takes to accelerate, the energy converted to kinetic energy is defined by the change in velocity alone.

Integrate the area under the force/distance curve, and you will arrive at the total work done, to the load.

We will get some basic physics concepts accepted soon...I hope.

[500TORQUES]

GT9 said:

Sorry 500 but he is correct.
What you meant to say was that it takes a lot more power.
Regardless of how long it takes to accelerate, the energy converted to kinetic energy is defined by the change in velocity alone.

My fuel flow rate and torque output disagrees.

Accelerating at 1.2G uses more energy than accellerating at 0.2G.

[911hope]

500TORQUES said:

GT9 said:

Sorry 500 but he is correct.
What you meant to say was that it takes a lot more power.
Regardless of how long it takes to accelerate, the energy converted to kinetic energy is defined by the change in velocity alone.

My fuel flow rate and torque output disagrees.

Accelerating at 1.2G uses more energy than accellerating at 0.2G.

Yes but it takes 6 times longer to hit target speed at 0.2g

More force for less time vs. less force for more time.????????

If it says you have done 1.2g acceleration, consider it unlikely to be valid.

Edited by 911hope on Sunday 2nd April 17:19

[911hope]

500TORQUES said:

GT9 said:

Sorry 500 but he is correct.
What you meant to say was that it takes a lot more power.
Regardless of how long it takes to accelerate, the energy converted to kinetic energy is defined by the change in velocity alone.

My fuel flow rate and torque output disagrees.

Accelerating at 1.2G uses more energy than accellerating at 0.2G.

Torque isn't energy.
Torque is a force.
Energy is force integrated over distance.

[500TORQUES]

911hope said:

Yes but it takes 6 times longer to hit target speed at 0.2g

If it says you have done 1.2g acceleration, consider it unlikely to be valid.

By the time the car at 0.2G has got up to speed the other has long gone and is through multiple further brake and acceleration events.

1.2G is nothing extreme for acceleration in a high power and grip car, my road car can achieve that.

You use a lot more energy the faster the acceleration in the real world.

[500TORQUES]

911hope said:

Torque isn't energy.
Torque is a force.
Energy is force integrated over distance.

Fuel is energy.
You need more fuel the faster the acceleration.

[bigothunter]

GT9 said:

500TORQUES said:

It bloody does, you need to use a lot more energy to accelerate faster.

Sorry 500 but he is correct.
What you meant to say was that it takes a lot more power.
Regardless of how long it takes to accelerate, the energy converted to kinetic energy is defined by the change in velocity alone.

Years ago, Autocar ran acceleration versus fuel consumption tests on a regular petrol ICE car. They continued all tests over the same distance

Although final kinetic energy levels were identical, they discovered that accelerating fairly hard (without excessive revs) to 60mph actually used less fuel than accelerating gradually. Reason was that higher loads put the IC engine in a more efficient region of the BSFC map. Appearances can be deceptive.

Suspect conducting similar tests with an EV would demonstrate no perceptible difference in energy consumed.

[GT9]

500TORQUES said:

GT9 said:

Sorry 500 but he is correct.
What you meant to say was that it takes a lot more power.
Regardless of how long it takes to accelerate, the energy converted to kinetic energy is defined by the change in velocity alone.

My fuel flow rate and torque output disagrees.

Accelerating at 1.2G uses more energy than accellerating at 0.2G.

Energy is power multiplied by time.
Accelerate twice as fast and you need twice as much power but you get to the final speed twice as quickly, therefore the energy is the same.
Th equation for the actual amount of kinetic energy the car is carrying = 0.5 x m x v^2
m = mass, v = speed
This is really basic physics by the way.

Fuel flow is obviously also affected by the efficiency at which the engine can convert the fuel to energy, which varies with torque and engine speed.
During the acceleration run the car is also consuming energy to push it through the air, overcome tyre rolling resistance and power various systems around the car, so all that has to be factored in.