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Hello, long time lurker, not been active in years, but I'm thinking it might be time to do more.



What I'm trying to do is figure out what electric motor I would need to take the place of a Chevy V8/8 speed transmission that is capable of driving a 4000lb 2WD pickup through the quarter mile in 8 seconds. I can figure out what I need using ICE easily, but figuring this out for electrics is not so easy, at least for me. Any suggestions on this? Rear differential ratios, transmissions, any part of the driveline is open season for this post. Tire diameter is 30 inches. I'm thinking I would prefer an AC traction motor. The usage would be for drag racing, plus driving to and from the track.



Please assume I understand about losing weight and whatnot to improve times, so we need not go over that part of it. The weight listed is what I expect the truck to weigh with the ICE setup, and I figure this will change with electric. I'm not concerned about power supply at this point, only the drive section. I figure once I know what the driveline needs, I can work out batteries.



It would be great if there was a chart somewhere that said that an electric motor of this power rating would be the equivalent of an ICE of this HP/TQ rating.
 

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Power is power
If you need 400 hp - then you need 400 hp

The thing that confuses things is you can put more than "rated" power though an electric motor - just as you can through a dino burner
And the torque curve shape helps a LOT

Sounds like you need a LOT of power as in 1000 hp +

You can look for specialised drag racing motors - or you could take a Tesla motor and overun it a bit

Tesla's seem to run for hundreds of thousands of miles - to me that means that if you are not worried about that sort of reliability you could turn the wick up a bit

I'm using a 10 kw (13 hp) motor in my car - but I'm feeding it 400 Kw (544hp) and I'm nowhere near the times you want

I don't think a Tesla motor would take a 40:1 increase but a 2:1 increase would give you 1000 hp
 

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How much ICE horsepower would it take to make an 8-second 4000lb truck?

If you can build an electrical system with the same weight, and presuming doing so doesn't change the aerodynamic profile of the truck (stuff goes inside), then it should be the same power requirements for electric.

You'll of course have different torque curves than for gas, so it's different than a top speed calculation, but, you seem to know how to figure that stuff out. Transmission and gear ratios the same, it's no different than your ICE stuff, just look at the torque curve for a given motor to know what you'd need.

So, that's all you need to know about sizing the motor.

If you want to talk about batteries...

Unlike with gasoline, which is trivial to get the energy out of as fast as you want (it explodes, you could get all the energy out of a gas tank in about a 2 second fireball), batteries aren't just limited by energy, they're limited by how fast you can get at the energy. How fast you can get at it is called power, exactly the same kind of power as the power it takes to move the vehicle. Normally this is not an issue because you drive an electric car for hours before recharging. But in high power applications like drag racing you might not be able to get at the energy quick enough. The fastest you might be able to drain a battery regardless of size is, for example, 10 minutes.

That means if you want to drain it in 1 minute, you need a battery pack 10x the size, even if you don't need the extra energy, just because you need to get at the energy 10x as quickly. There's no gasoline equivalent, there's no such thing as "I need to have a gas tank 10x as big in order to use gas 10x as fast". I suppose you might say fuel line diameter or fuel pump flow rate, but, those are minor add-on components, not an intrinsic property of gasoline itself like is the case with the chemical reactions in batteries.

The energy required to move 4000lbs at a certain speed through air resistance for 1/4 mile is a quite small amount of energy, but, you'll need probably ballpark 50x the energy stored just to have enough power to do it.

I've heard Tesla dragracers say that their limit on speed isn't motor or controller, it's battery.

If there's a 75,000 kwh battery pack in a Tesla, and it maxes out at 500hp (666,000 watts), and if true that that is because of batteries, that means the cells are only capable of 9C discharge rates. There's often a tradeoff between optimizing for higher energy or higher power (same as a deep-storage vs. starting battery in lead acid). Drill batteries are capable of 20c, so, there are higher-power cells out there than those used in Teslas. It depends on whether similar cells are available or affordable, if you want to build your own pack, or if you want to find larger form factor sizes.

I suspect EV drag racers have a battery supplier that builds batteries more specifically for their use case, but I'm not in the scene and don't know. I don't imagine they'd be cheap either.
 

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Hi
To Matt's point about the batteries

It's a bit like the motors - the Tesla battery (for instance) can give about 800 hp (both motors) AND last 20 years and 300,000 miles

If you take MORE than that out - then it probably won't last 20 years and 300,000 miles!!

But an 8 second car using IC technology will also have a "limited" life

I'm using a Chevy Volt battery - designed for about 150 hp - and I'm pulling 500 hp
BUT I will not be surprised if it does not last for 200,000 miles
 

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Discussion Starter · #5 ·
This quickly turned into a battery discussion.



The build I have planned should have around 900 RWHP running through an 8 speed auto. I'll probably really be under 4000lbs, but that's what my target weight is.
 

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This quickly turned into a battery discussion.



The build I have planned should have around 900 RWHP running through an 8 speed auto. I'll probably really be under 4000lbs, but that's what my target weight is.
For AC motors, a dual core AMR motor can just about do 900 HP when driven by dual RMS PM250DZ inverters. I don't know of a single core motor--or a single inverter--available on the market that can output that much power.
 

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Moving the mass

4000 lbs truck with ET of 8 sec in 1320 ft (1/4 mile): the basic physics (kinematic equations) indicate 225 mph final speed with acceleration of 1.28g, and required power of 1540 hp. There are online drag racing calculators that will solve for this power also based upon weight and ET.

This does not include the power to overcome aero drag and rolling friction of the tires.

If you have the Cd drag coefficient and frontal surface area then the aero drag power can be computed.
 

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This does not include the power to overcome aero drag and rolling friction of the tires.
Holy crap. So just to accelerate 4000lbs to that top speed in that amount of time, just the kinetic energy component, requires 1500hp?

Supposing a coefficient of drag of 0.4 (ambitiously low) and a frontal area of 3m^2 (ambitiously low), the calculator I ran said it takes over 1000 hp extra just to overcome wind and rolling resistance.

Which, makes sense I suppose, trying to drive a brick of a truck 225 miles an hour.

Naturally that's only how much power it requires at top speed as it crosses the 1/4 line, it's logarithmically (?) lower towards the start line, but, still. That's a lot of power needed.
 

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4000 lbs truck with ET of 8 sec in 1320 ft (1/4 mile): the basic physics (kinematic equations) indicate 225 mph final speed with acceleration of 1.28g, and required power of 1540 hp.
Holy crap. So just to accelerate 4000lbs to that top speed in that amount of time, just the kinetic energy component, requires 1500hp?
A way to check the power required, assuming the ET and trap speed are correct:
4000 lb (1814 kg) at 225 mph (362 km/h or 100 m/s) is 9070 kJ of kinetic energy; over 8 seconds that's 1133 kW or 1520 hp.​

This does not include the power to overcome aero drag and rolling friction of the tires.

If you have the Cd drag coefficient and frontal surface area then the aero drag power can be computed.
Supposing a coefficient of drag of 0.4 (ambitiously low) and a frontal area of 3m^2 (ambitiously low), the calculator I ran said it takes over 1000 hp extra just to overcome wind and rolling resistance.

Which, makes sense I suppose, trying to drive a brick of a truck 225 miles an hour.

Naturally that's only how much power it requires at top speed as it crosses the 1/4 line, it's logarithmically (?) lower towards the start line, but, still. That's a lot of power needed.
Because the rolling drag power is proportional to speed and the aero drag power is proportional to the cube of speed, the vehicle's acceleration given a constant power source is much greater at low speeds, and the only reasonable way to find the performance for a constant power (or constant power requirement for a given ET) is by simulation (not by a formula). Kennybobby has done a lot of good work of this sort for previous drag racing discussions, such as Hypothetical Jr Comp Dragster.


Now we get to a problem with the numbers: an 8-second ET will not correspond to 225 mph, because that is based on a steady acceleration of 1.28 g, and the actual acceleration for an 8-second ET will be higher at low speed (when drag is low) and lower at high speed (when much of the power is being used up by drag). An 800-horsepower engine-driven car (which is not producing 800 hp most of the time) can do a quarter mile in 10.8 seconds, and only gets to 131 mph.

Doing the check as above:
4,469 lb (2,027 kg) at 131 mph (211 km/h or 58.6 m/s) is 3480 kJ of kinetic energy; over 10.8 seconds that's an average of only 322 kW or 432 hp. The rest of the power is going to overcoming drag.​

The target (as usual in drag competition) is the ET, not the trap speed. It would be helpful to run a simulation to find the constant power needed to get the car through the quarter in 8 seconds, with drag... knowing that the trap speed will be much lower than 225 mph.
 

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the only reasonable way to find the performance for a constant power (or constant power requirement for a given ET) is by simulation (not by a formula).
Shouldn't this be a relatively simple matter for calculus? I'm too rusty (and was never very good) to set it up but seems like this is exactly the situation calc was invented for.
 

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Here is a Tesla run with high current draw for about 6.5 seconds. Would have to integrate the speed curve to get distance. It appears that the launch mode holds the car at WOT for 1.5 seconds before release, and the current pulled back before he let off the throttle.

[edit: digitized the speed using https://apps.automeris.io/wpd/, then did piece-wise integration. 0 to 8 seconds distance was about 751 feet, so a little bit over half of a quarter mile (1/8th mile is 660 ft)]
 

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Shouldn't this be a relatively simple matter for calculus? I'm too rusty (and was never very good) to set it up but seems like this is exactly the situation calc was invented for.
If there were only one simple function, yes, but with aero drag being a function of speed, acceleration being a function of power and speed and aero drag, and time to run the 1/4 mile being an integral of speed (which is a function of acceleration) until enough distance accumulates, I doubt there is a reasonable solution... and I didn't even try. In general, anything can differentiated, but most real-world expressions of any useful complexity can't be integrated; in science exact solutions are rare, and numerical solutions are the norm.
 
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