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Discussion Starter · #1 ·
Hello,

New guy here so please go easy on me, 2nd thread :)

Could someone please tell me how many watts a new Tesla PLAID would use in a 1/4 mile pass at full throttle.

I'm told it's 400 volts at 1,500 amps at full output but from there I'm lost on the calculation.

Do I break down KwH to minutes then to seconds? I think the car runs a low 9 second 1/4 mile.

Thank you!
 

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Hello,

New guy here so please go easy on me, 2nd thread :)

Could someone please tell me how many watts a new Tesla PLAID would use in a 1/4 mile pass at full throttle.

I'm told it's 400 volts at 1,500 amps at full output but from there I'm lost on the calculation.

Do I break down KwH to minutes then to seconds? I think the car runs a low 9 second 1/4 mile.

Thank you!
That would be max power - 600 kW - but it would NOT be able to use that at the start
The power will inevitably ramp up as the speed increases
Anyway call it 600 kW for 10 seconds
That is 6,000,000 Joules
Divide by 3600 to turn it into Watt hours - 1666 Watt hours - or 1.6 kWh
 

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Discussion Starter · #3 ·
That doesn't sound right, that would mean 2 of my electric bike batteries could power a Tesla PLAID from zero to 140mph in a 1/4 mile run?

I know it takes about 1/2 gallon of gas for a 1,000HP car to make a 1/4 mile run and isn't gas 33.7 KWH per gallon so about 17KWH for 1/2 Gallon of gas?

So your saying it just takes 1666 watt hours for a Tesla PLAID to make a o-140mph 1/4 mile run?

I think something is off somewhere?
 

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That doesn't sound right, that would mean 2 of my electric bike batteries could power a Tesla PLAID from zero to 140mph in a 1/4 mile run?

I know it takes about 1/2 gallon of gas for a 1,000HP car to make a 1/4 mile run and isn't gas 33.7 KWH per gallon so about 17KWH for 1/2 Gallon of gas?

So your saying it just takes 1666 watt hours for a Tesla PLAID to make a o-140mph 1/4 mile run?

I think something is off somewhere?
That is what the numbers say!
Your 17 kWh divided by the incredible inefficiency of a 1000 hp dragster will divide out to about 1.7 kWh

Another way of looking at it is the kinetic energy of the car
The model S is about 2000 kg - and 152 mph at the end of the 1/4 mile - 68 m/sec
Energy = 1/2 x 2000 x 68 x 68 = 4.6 Mega Joules

The power calculation gave 6 Mega Joules

When I take my machine to the drags it uses hardly any charge
 

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Discussion Starter · #5 ·
That is what the numbers say!
Your 17 kWh divided by the incredible inefficiency of a 1000 hp dragster will divide out to about 1.7 kWh

Another way of looking at it is the kinetic energy of the car
The model S is about 2000 kg - and 152 mph at the end of the 1/4 mile - 68 m/sec
Energy = 1/2 x 2000 x 68 x 68 = 4.6 Mega Joules

The power calculation gave 6 Mega Joules

When I take my machine to the drags it uses hardly any charge
Wow! Almost seems too good to be true, so for a fun 1/4 mile drag car I should just make five or six 2KWH batteries that can be removed and swapped to save weight and this would give me the five or six 1/4 passes.

Thanks you, really appreciate it!
 

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Wow! Almost seems too good to be true, so for a fun 1/4 mile drag car I should just make five or six 2KWH batteries that can be removed and swapped to save weight and this would give me the five or six 1/4 passes.

Thanks you, really appreciate it!
Not so fast!
That is the amount of energy you need
Now you have to think of POWER
Batteries have a "C" rating - C is how fast you can discharge the battery
5C means that you can discharge your 2 kWh battery at 5 x 2 = 10 kW

My 14 kWh battery (Chevy Volt) seems to survive a max discharge of 400 kW - 28C

The Tesla with a 100 kWh battery is only discharging at 6C

You can buy batteries for RC planes that claim 100C -

Then you get to the required voltage - for high speed you will need a high voltage - I am using 340 volts and my acceleration is dropping fast by the end of the 1/4 mile
 

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Discussion Starter · #7 ·
Not so fast!
That is the amount of energy you need
Now you have to think of POWER
Batteries have a "C" rating - C is how fast you can discharge the battery
5C means that you can discharge your 2 kWh battery at 5 x 2 = 10 kW

My 14 kWh battery (Chevy Volt) seems to survive a max discharge of 400 kW - 28C

The Tesla with a 100 kWh battery is only discharging at 6C

You can buy batteries for RC planes that claim 100C -

Then you get to the required voltage - for high speed you will need a high voltage - I am using 340 volts and my acceleration is dropping fast by the end of the 1/4 mile
Thanks for this, I knew some of it just not exactly yet if I will have to build my own battery packs or try to buy used ones from an EV and link them together for 400v but I understand this configuration will not be light.

What are your thoughts on me using 125x7 of these 200A Rated HEADWAY batteries or do you have a recommendation for the best cells so I can build smaller high discharge lightweight removable battery packs?


and thanks again for the help!
 

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Thanks for this, I knew some of it just not exactly yet if I will have to build my own battery packs or try to buy used ones from an EV and link them together for 400v but I understand this configuration will not be light.

What are your thoughts on me using 125x7 of these 200A Rated HEADWAY batteries or do you have a recommendation for the best cells so I can build smaller high discharge lightweight removable battery packs?


and thanks again for the help!
I started off with Headway - then I went to the batteries from a Chevy Volt
For normal road use you simply cannot beat the batteries from a production EV - they have invested huge amount of resources in getting them right

For ultimate performance the Lithium Poly cells the RC guys use are unbeatable - but they are also explodable!

The Headway will not be as dependable and cheap as the production EV cells
And will not be as powerful as the RC cells

Removable .... for road use and any decent life you really need liquid cooled - not so much for the cooling as to ensure that the cells are all at the same temperature

Headway - 127 x 7 = 889 cells x 360 gm = 320 kg
127 x 3,2 = 406 volts
7 x 200 amps = 1400 amps
1400 amps x 406 volts = 568 kW
In practice that will sag by at least 25% - so 426 kW

In comparison my 340 volt Volt pack is 133 kg - and I'm pulling 400 kW - OK that will sag to 320 kW - but I am drawing more than would be "advised"

Two of my packs would be more power than the Headways and still lighter
Two actual Volt packs would give you the 400 volts - and if you pulled 2400 amps that would only be abusing them the same as I do
 

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Best batteries for drag racing are NiCAD, imo. Or SuperCaps if you can afford them.

You initially asked about horsepower (or kW...they're interchangeable). Horsepower, in the absence of a transmission, merely determines top speed...

For your purposes, if you can deliver constant current (which is constant acceleration) you need 1/2 (horsepower)* (1/4 mile time) *746

joules of energy from the battery.
 

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You initially asked about horsepower (or kW...they're interchangeable). Horsepower, in the absence of a transmission, merely determines top speed...
No, power also determines (or limits) acceleration: the rate of change of kinetic energy is power.

For your purposes, if you can deliver constant current (which is constant acceleration) you need 1/2 (horsepower)* (1/4 mile time) *746

joules of energy from the battery.
Okay, except that if the motor is delivering constant torque the horsepower is not constant, so the power in that formula is the power at the 1/4 mile point - the highest level of power used during the run. The Tesla is power-limited before the end of a 1/4 mile run, so the delivered torque is not constant.

Also, constant acceleration and the corresponding constant torque corresponds to constant motor current, not battery current; I'm not sure that everyone understands that.
 

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Another way of looking at it is the kinetic energy of the car
The model S is about 2000 kg - and 152 mph at the end of the 1/4 mile - 68 m/sec
Energy = 1/2 x 2000 x 68 x 68 = 4.6 Mega Joules

The power calculation gave 6 Mega Joules
The energy calculation is correct. Well, the kinetic energy calculation is close, but any Model S has a mass over 2,000 kg, and the Plaid won't be any lighter.

And that's just the kinetic energy - there's also the energy used to move the vehicle against aero and rolling drag, although that's only about 300 Wh. Of course with slipping tires the energy loss to the tires is much higher, so add perhaps another kilowatt-hour.

The end result is that the sum of kinetic energy gained and drag losses agrees well with the assumption that the car is close to using 600 kW for the entire duration of the run; it's lower at the beginning, then reaches the motor maximum at some point when the speed multiplied by torque reaches that value, and stays there. And as Duncan already explained, it won't be 400 volts, since the battery won't be fully charged (a Tesla battery runs about 360 V nominal; 400 V is the maximum charging voltage) and because the voltage delivered will be lower than the resting voltage due to internal resistance (that's "sag").

Anyway... yes, less than two kilowatt-hours delivered by the battery for the whole run.
 

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No, power also determines (or limits) acceleration: the rate of change of kinetic energy is power.
I said constant acceleration -> constant torque -> constant current. Yes, power is an energy rate...so what?

You can accelerate a car with a 6V battery just as fast as a 500V battery if you can deliver the 1500 amps.

Power has nothing to do with acceleration in the absence of a transmission. ZERO.
 

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I said constant acceleration -> constant torque -> constant current. Yes, power is an energy rate...so what?
So the current multiplied by the voltage is power, and so the available power is limiting.

You can accelerate a car with a 6V battery just as fast as a 500V battery if you can deliver the 1500 amps.
No, you can't. Please remove all but one cell of the battery in any EV (so that battery current is unaffected by voltage and power are greatly reduced) and demonstrate your theory.

Current at the battery is irrelevant by itself, without voltage for context. The current from the battery is not the same as the current through the motor. Current through the motor determines torque, but how much torque depends on the motor design, and voltage is required to make that current flow (how much voltage depends on motor design and speed).

Power has nothing to do with acceleration in the absence of a transmission. ZERO.
The use (or not) of a transmission does not change this fundamental reality.

I get that people in this forum come from many backgrounds, all having their own strengths and weaknesses, and understanding basic physics is a common weakness.... but this is high school physics stuff. Please don't confuse new people with nonsense, and shouting the nonsense confidently just makes the situation worse.
 

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The energy calculation is correct. Well, the kinetic energy calculation is close, but any Model S has a mass over 2,000 kg, and the Plaid won't be any lighter.

And that's just the kinetic energy - there's also the energy used to move the vehicle against aero and rolling drag, although that's only about 300 Wh. Of course with slipping tires the energy loss to the tires is much higher, so add perhaps another kilowatt-hour.

The end result is that the sum of kinetic energy gained and drag losses agrees well with the assumption that the car is close to using 600 kW for the entire duration of the run; it's lower at the beginning, then reaches the motor maximum at some point when the speed multiplied by torque reaches that value, and stays there. And as Duncan already explained, it won't be 400 volts, since the battery won't be fully charged (a Tesla battery runs about 360 V nominal; 400 V is the maximum charging voltage) and because the voltage delivered will be lower than the resting voltage due to internal resistance (that's "sag").

Anyway... yes, less than two kilowatt-hours delivered by the battery for the whole run.
Are you high?

The car uses ZERO power at launch. A big fat ZERO. Not 600kW. ZERO. Omega is ZERO.

The power use rises linearly, assuming constant acceleration (force before the tires break traction) as torque remains CONSTANT, until maximum power is reached (torque will drop off in reality at higher speeds, speed will limit by pack voltage). The max speed of the car will be determined by its power/horsepower and that will be where the loads on the car equal the force at the tired (which comes from torque).

Duncan was correct in trying to determine Joules needed. Your 600kW number for the entire run is nonsense unless there's a CVT, which there isn't in a Tesla. And if you do assume constant acceleration , for the 1/4 mile, the POWER used per time curve is a straight line, starting at zero power, and ending, at the quarter mile time, at 600kW. The area under that curve is the energy needed in Joules. You can convert Joules to kWhr, but you gotta deliver the current that makes the torque that makes the acceleration and the voltage that makes the speed, and the max power.
 

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So the current multiplied by the voltage is power, and so the available power is limiting.


No, you can't. Please remove all but one cell of the battery in any EV (so that battery current is unaffected by voltage and power are greatly reduced) and demonstrate your theory.

Current at the battery is irrelevant by itself, without voltage for context. The current from the battery is not the same as the current through the motor. Current through the motor determines torque, but how much torque depends on the motor design, and voltage is required to make that current flow (how much voltage depends on motor design and speed).


The use (or not) of a transmission does not change this fundamental reality.

I get that people in this forum come from many backgrounds, all having their own strengths and weaknesses, and understanding basic physics is a common weakness.... but this is high school physics stuff. Please don't confuse new people with nonsense, and shouting the nonsense confidently just makes the situation worse.
The battery doesn't matter. The problem was framed as deliver 1500 amps.
 

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Are you high?

The car uses ZERO power at launch. A big fat ZERO. Not 600kW. ZERO. Omega is ZERO.
I am not high, and I did not at any point suggest that the car will use 600 kW at launch. It will deliver zero useful power at zero speed (but consume many kilowatts anyway due to electrical and motor losses), and what I mentioned was
"the assumption that the car is close to using 600 kW for the entire duration of the run; it's lower at the beginning, then reaches the motor maximum at some point when the speed multiplied by torque reaches that value, and stays there."​
That's ideally zero at zero speed, rising linearly to the maximum power point. Since the time spent at less than full power is a smal part of the run, a rough approximation of full power for the whole time isn't bad as a starting point.

Using "omega" for speed is just silly. That's the name of symbol used for angular velocity, not the name of a physical quantity. You're not convincing anyone that you have deep technical knowledge with nonsense jargon.

The power use rises linearly, assuming constant acceleration (force before the tires break traction) as torque remains CONSTANT, until maximum power is reached...
I guess what I wrote got across, because that's what I said.

The max speed of the car will be determined by its power/horsepower and that will be where the loads on the car equal the force at the [tires]...
Yes, which has nothing to do with the 1/4 mile acceleration run, which does not reach the car's top speed. The power available at high speed does limit the car's top speed; it also limits acceleration.

Duncan was correct in trying to determine Joules needed. Your 600kW number for the entire run is nonsense unless there's a CVT, which there isn't in a Tesla.
As I explained, the rough approximation of 600 kW for ten seconds is roughly comparable to results of Duncan's approach. No, there is no continuously variable transmission (CVT) but for much of the speed range of a Tesla the power is limited to a roughly constant value, as with almost any EV. That means with increasing speed the torque available decreases, just like a constant speed engine working through a CVT.

And if you do assume constant acceleration , for the 1/4 mile, the POWER used per time curve is a straight line, starting at zero power, and ending, at the quarter mile time, at 600kW. The area under that curve is the energy needed in Joules.
Yes, that's the basis of your 1/2*(peak power)*time calculation, and I agreed with that.

... you gotta deliver the current that makes the torque that makes the acceleration and the voltage that makes the speed, and the max power.
Obviously. You need to deliver the current the motor needs, the voltage the motor needs for that current to flow, and the electrical power that corresponds to those at every point during the run. When the power limit (imposed by the motor controller to protect the inverter and the battery) is reached, power becomes the limiting factor in acceleration. that's true for most of the 1/4 mile run.
 

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Here's a good writeup of what actually goes on in a Ludicrous Model S

Thanks for illustrating the reality of the dependence on power. From that article:
T= 1.76 Sec
Combined motor output power peaks at 680 hp as motor voltage waveforms reach 130 volts at 500 Hz. Again, that's an AC RMS voltage. Longitudinal acceleration drops to 1.00 g and continues falling from this point. The vehicle is traveling 51.0 mph.
So at 1.76 seconds into a 10.52 second run, at 51.0 MPH on its way to 125 MPH, the car hits the point that it is power limited (at only 680 HP or 507 kW). From that point on, it is limited by power, not peak motor torque, because the battery and controller can't provide enough power to push enough current to produce peak motor torque. Many people expect (and loudly insist that there will be) the same torque and acceleration for the whole run, but this shows roughly constant torque and acceleration only up to about 50 MPH, just as expected from the motor/controller performance data for Teslas and other typical EVs.

energy used during the run
first part: 1/2*(507 kW)*1.76 s = 892 kJ (actually much less because launch torque and current is less than peak)
remainder of run: 507 kW * (10.52 s - 1.76 s) = 4,441 kJ (or 1,233 Wh)
total: about two kilowatt-hours... but the point is that the majority of the power is consumed during the constant-power phase, which was why multiplying that constant power by the run time was a reasonable rough first approximation.
 
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