# Understanding Formulas & how they apply to the EV World

P = I*V

The general application would be to say that for any given power requirement, if you increase voltage of system you would decrease the required flow of amps to produce said required power.

So, if you’re pulling less amps, would it follow, in the ev world, that you would need less AH in a higher voltage pack to get the same range as a lower voltage but higher AH pack? (assuming you were driving two of the same cars under same conditions but with different packs)

What other factors need to be considered?

yes

Peukert effect should be considered. Higher volts = less amps for the same power level. So if you use less current, and the batteries are only doing 1-2C discharge normally instead of 3-4C, you’ll be saving the batteries life, and actually get more useable Wh out of it.

P = Watts, but since we’re multiplying Ah times Volts, its Watt-Hours (Wh).

Also, higher volts (with a DC motor), the higher RPM, lower volts, lower RPM.

Higher current, more torque.

[QUOTE=frodus;5385]Also, higher volts (with a DC motor), the higher RPM, lower volts, lower RPM.

Higher current, more torque.[/QUOTE]

How does the controller play into this equation. I thought the controller was deciding what Volt/Amp combo to send to the motor. I guess the battery pack is the upper limit of voltage that it possible w/o a transformer. So higher the pack voltage, higher potential voltage to send to motor.

I’m guessing there is a good reason nobody puts a transformer in their systems. Would it be too big/heavy? I know almost nothing about transformers, except that they change I/O voltage.

Also, would it be a correct conclusion from the previous example, that the system with a lower voltage would produce more torque in any given situation because it is pulling more amps for said power requirement?
I suppose this might be advantageous if you do a lot of driving in hilly terrain?
Or is the controller smart enough to compensate for this too?

I think I have a ways to go before I can get to Peukert’s Law. But I’m getting there.
Thanks

P = Watts, but since we’re multiplying Ah times Volts, its Watt-Hours (Wh).

Also, frodus, how would one calculate the Ah savings for a higher voltage.

For a 4000lb car, drag-notwithstanding, what would be the difference in range be for example between a 144V/5568Ah pack(using 24x232Ah batts) or a 312V/4030Ah pack (using 26x155Ah batts)?

Is is as simple as doing a proportionality between the voltages?

In other words, 312/144 *4000 = 8666 Ah equivalent? That can’t be right. I know I’m missing at least one thing here.

From reading another post, I saw calculations like this:
144V * 232Ah = 33408Wh = 33.4kWh
312V * 155Ah = 48360Wh = 48.4kWh
But I’m not sure I understand what this means.

Also, frodus, how would one calculate the Ah savings for a higher voltage.
For the same power (say 7200W), you would either do 144V of 50Ah batteries, or 72V of 100Ah batteries and get the same power level. There’s no “savings”, but with higher Voltage, you’re using less amps, and keeping the peukert effect low… and it depends on battery and chemistry, all batteries are not created equal. I cannot answer the question. You have to have the peukert value of the battery, continuous amps useage and temperature among other things to calculate.

For a 4000lb car, drag-notwithstanding, what would be the difference in range be for example between a 144V/5568Ah pack(using 24x232Ah batts) or a 312V/4030Ah pack (using 26x155Ah batts)?
First, you’re not calculating Ah right. You don’t multiply Ah by batteries. If they’re in series, its still only the rating of one battery. For example. 12 batteries is 144V. Say they’re 50Ah, the pack is 144V 50Ah and 7200Wh.

144V at 232Ah is 33,408Wh
312V at 155Ah is 48,360Wh
Wh is range, so 312V would give higher range. Say the car has a cruising Wh/mile of 300Wh, the 144V would give only 80% of its total Watt hour rating for a range of 89 miles. The 312V would give 80% of its total WH rating for a range of 129 miles.

The big issue is that you’ll never fit 155Ah (or 232Ah) of batteries in a car/motorcycle or even a truck. Thats over 2k Lbs. Most i’ve seen in lead is about 25kWh

Is is as simple as doing a proportionality between the voltages?

P=VI

Figure out how far you want to go, then estimate a cruising Wh/mile based on what others have done, overshoot by 10-20% of their Wh/mile. Calculate the Power in Wh needed. Then chose a ratio of Volts and Amp-Hours to get the Wh you need. Keep in mind your speed requirement. If you go too low, you won’t be able to spin the motor fast enough to get to the speed you need, but if you go too high, you’ll have no torque/acceleration. You need the motor torque curve to calculate this.

In other words, 312/144 *4000 = 8666 Ah equivalent? That can’t be right. I know I’m missing at least one thing here.

where did you get that calculation? where’d the 4000 come from? LBS doesn’t equate anywhere in power calcs. P=V*I. Ah is calculated based on the battery itself.

From reading another post, I saw calculations like this:
144V * 232Ah = 33408Wh = 33.4kWh
312V * 155Ah = 48360Wh = 48.4kWh
But I’m not sure I understand what this means.

Wh means that its the amount that you can discharge per hour to 100% discharge. If you have 1kWh, and you discharge it at 1000W, you take 80% of that (because you don’t want to discharge lead below 80%), and you can run for 48 Minutes. (800W/1000W) * 60minutes = 48 Minutes.

If you have a 33.4kWh pack, its really 26726.4Wh. From that, if you were to cruise at 300Wh per mile, and go 45 miles in an hour, you’d be using a total of 13,500W per hour. At this rate, you could run for 118Minutes (almost 2 hours).

How does the controller play into this equation. I thought the controller was deciding what Volt/Amp combo to send to the motor.

The controller is a power converter, the batteries are still the limitation of how much power you can get.

I guess the battery pack is the upper limit of voltage that it possible w/o a transformer.

Transformers are AC, they won’t work with DC unless you use a switching power supply to bump the voltage up.)

So higher the pack voltage, higher potential voltage to send to motor.
Exactly

I’m guessing there is a good reason nobody puts a transformer in their systems. Would it be too big/heavy? I know almost nothing about transformers, except that they change I/O voltage.
They don’t use them because converting hundreds of amps of one voltage to another voltage requires a switch mode power supply. You can’t use a transformer with DC, it only works with changing voltage (like AC). It would cost more than the controller to “convert” to a higher voltage, and you’d decrease amps. Best thing to do is higher voltage with smaller Ah batteries for higher speed, but low end torque still suffers.

Also, would it be a correct conclusion from the previous example, that the system with a lower voltage would produce more torque in any given situation because it is pulling more amps for said power requirement?

Not a correct conclusion, Lower volts does not equal higher amps neccessarily. Your batteries still need to be able to handle it, the controller needs to be able to control it and provide current limiting to save the batteries. In the end, power in = power out + losses.

I think I have a ways to go before I can get to Peukert’s Law. But I’m getting there.

Basically, the higher amps you draw from the battery, the less total charge it will give. Kinda like gas engines, the heavier into the throttle you go, the less MPG you get. Let off the throttle and cruise, and you get further.

[QUOTE=frodus;5406]For the same power (say 7200W), you would either do 144V of 50Ah batteries, or 72V of 100Ah batteries and get the same power level. There’s no “savings”, but with higher Voltage, you’re using less amps, and keeping the peukert effect low… and it depends on battery and chemistry, all batteries are not created equal. I cannot answer the question. You have to have the peukert value of the battery, continuous amps useage and temperature among other things to calculate.
[/QUOTE]
Got it.

[QUOTE=frodus;5406]
First, you’re not calculating Ah right. You don’t multiply Ah by batteries. If they’re in series, its still only the rating of one battery. For example. 12 batteries is 144V. Say they’re 50Ah, the pack is 144V 50Ah and 7200Wh.
[/QUOTE]
I never knew this until right now.

[QUOTE=frodus;5406]
Wh is range, so 312V would give higher range. Say the car has a cruising Wh/mile of 300Wh, the 144V would give only 80% of its total Watt hour rating for a range of 89 miles. The 312V would give 80% of its total WH rating for a range of 129 miles.
[/QUOTE]
Makes sense, got it.

[QUOTE=frodus;5406]
The big issue is that you’ll never fit 155Ah (or 232Ah) of batteries in a car/motorcycle or even a truck. Thats over 2k Lbs. Most i’ve seen in lead is about 25kWh
[/QUOTE]
I’m going to have to disagree with you here, even if I end up looking the fool

[QUOTE=frodus;5406]
If you go too low, you won’t be able to spin the motor fast enough to get to the speed you need, but if you go too high, you’ll have no torque/acceleration. You need the motor torque curve to calculate this.
[/QUOTE]
Interesting

[QUOTE=frodus;5406]
where did you get that calculation? where’d the 4000 come from? LBS doesn’t equate anywhere in power calcs. P=V*I. Ah is calculated based on the battery itself.
[/QUOTE]
I screwed up, 4000 was an estimate of Ah, by adding them to 4030, not weight, but as you said the calculation was wrong because you don’t Ah when in series.

[QUOTE=frodus;5406]
Wh means that its the amount that you can discharge per hour to 100% discharge. If you have 1kWh, and you discharge it at 1000W, you take 80% of that (because you don’t want to discharge lead below 80%), and you can run for 48 Minutes. (800W/1000W) * 60minutes = 48 Minutes.

If you have a 33.4kWh pack, its really 26726.4Wh. From that, if you were to cruise at 300Wh per mile, and go 45 miles in an hour, you’d be using a total of 13,500W per hour. At this rate, you could run for 118Minutes (almost 2 hours).[/QUOTE]

My next step is to calculate what my specific Wh reqiurements will be for my conversion.

This really cleared up a lot for me, thanks.

Not a correct conclusion, Lower volts does not equal higher amps neccessarily. Your batteries still need to be able to handle it, the controller needs to be able to control it and provide current limiting to save the batteries. In the end, power in = power out + losses.

good explanation, thanks

Basically, the higher amps you draw from the battery, the less total charge it will give. Kinda like gas engines, the heavier into the throttle you go, the less MPG you get. Let off the throttle and cruise, and you get further.

okay, i can work with that.

[QUOTE=naiche;5409]I’m going to have to disagree with you here, even if I end up looking the fool
[/QUOTE]
Here are 2 examples:

255Ah battery:
Weight: 165.00 pounds
Length: 19.25 inches
Width: 10.94 inches
Height: 10.30 inches
Amps: 255.00

165lbs * 12 is 1980lbs, so it’d be CLOSE to that weight. Plus, they’re not “small” batteries.

163Ah battery:
Weight: 119.00 pounds
Length: 22.00 inches
Width: 4.90 inches
Height: 11.15 inches
Amps: 155.00

so 26 * 119 = 3094lbs

Again, not “small”.

I’ve seen large packs in other EV’s, but they were LiFePo, and over 25 grand. Oh yeah, one more thing. Once you get to a certain Ah rating in high capacity batteries, they’re very spendy.

Well it ain’t going to be light, but here are a couple other examples:

12V 155Ah @ 86.40lbs each
http://www.usbattery.com/usb_us12vxc_golg.html

26*86.40 = 2246lbs. for 312V system

8V 170Ah @ 64.25lbs. each
http://www.usbattery.com/usb_us8vgcxc_golf.html

24*64.25 = 1542lbs for 192V system

6V 232Ah @ 61.73lbs each
http://www.usbattery.com/usb_us2200xc_golf.html

24*61.73 = 1481.52

I’m converting a pickup and have figured out I can fit about 27 Max in the bed, and I figure I could get at least 4 under the hood as well to balance the weight a little. So it might be 22 in back, 4 up front.

It’s all about sacrifice with these designs. So I’m just going to have to make some hard decisions.

thats alot of weight…

I think anything over about 25kW or so is a bit overkill… you might just want to buy a hybrid it’ll definitely be a lead sled… 