# Battery math?



## kosstheory (May 26, 2008)

Where did this formula derive from?

range[miles]=250 x capacity[kWh] / (mass[lbs]^0.6)

I found it on the Wiki.

Specifically:

is range[miles] = the total range per charge?

where did the 250 come from and why is it used as a multiplier?

What is capacity[kilowatt hours], what capacity does it refer to?

why is the mass, which I assume to be the total mass of the object being moved carry an exponent of 0.6? What is the 0.6 exponent for?

Thanks


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## TheSGC (Nov 15, 2007)

The 250 Watt/Miles figure is average power per mile for small cars/sedans traveling at freeway speeds. 

There formula to derive it is [Volts * AMP Draw]/MPH. There is no definate way to determine your Watt/Mile until you actually build and drive you EV for a while and log your info.

The Capacity is the total capacity of your battery pack [Volts * Ah].

I have no idea what the exponent is for, but it's needed. I believe that came from one of the EV books out there and there is an explanation for it. Its only a general estimation and there are a few better ways to determine your range, like using EV Calculator. http://www.evconvert.com/tools/evcalc/


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## kosstheory (May 26, 2008)

TheSGC said:


> There is no definate way to determine your Watt/Mile until you actually build and drive you EV for a while and log your info.


Thanks for the link to the Calc tool. I'm going to see if I can contact the programmer to get my hands on the raw code.

I'm not looking for exact figures, and I don't think that's the intent of the wiki, because it's there to help people develop a concept for a vehicle, a theory on how best to build their vehicle. In fact, it's part of the battery sizing FAQ, so it would be impossible to keep a log if you haven't even sized the batteries yet.

So, I need to know how to plug in info that I do know, like the exact weight of my car with it's theoretical batteries, and its other theoretical equipment, so I can get an idea of what size my batteries need to be to optimize the system.

The calculator is useful, but you'd have to know the weight of the preset vehicles, which I'm guessing made the list, because there is imperical data available to extrapolate the calculations. The problem with this might be that the vehicles in question might suffer from design flaws. However, if the numbers are based purely on theory, there should be a way to crunch the same numbers for any vehicle.

I have a 67 Mustang Hardtop that will likely weigh just under 3000 lbs once the conversion is complete. I need to know how to figure out what I would need to get a 100 mile range with a top speed of 100 mph and an acceleration of 0-60 in under 10 seconds. I don't want to use generic values for things that can be better know like the kWh requirements to move a relatively aerodynamic vehicle that weighs ~3000 lbs at 70 MPH for roughly 80 miles. Using the 250 Watts/Mile value will lead to a less accurate theoretical model, and result in over/under-compensation in the design.

I know this is aiming high, but I need to start somewhere, and work my way back to what may be more reasonable. Goals are goals, whether or not you ever achieve them.


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## Dave_C (May 29, 2008)

kosstheory said:


> The calculator is useful, but you'd have to know the weight of the preset vehicles, which I'm guessing made the list...


You can choose a vehicle similar to yours and then click on "details" and edit the vales to better match your car.


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## favguy (May 2, 2008)

OK, first off, if you want to achieve 100 mph top speed and a 100 mile range, you will need large capacity lithium batts, with probably a higher voltage AC drive system, whats your budget, as that will easily cost you about $60K I reckon, if you're happy to have a 50 mile range with a top speed of say, 80ish and a cruising speed of 65ish, you can easily do this with a DC system and lead acid for about $12K


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## kosstheory (May 26, 2008)

favguy said:


> OK, first off, if you want to achieve 100 mph top speed and a 100 mile range, you will need large capacity lithium batts, with probably a higher voltage AC drive system, whats your budget, as that will easily cost you about $60K I reckon, if you're happy to have a 50 mile range with a top speed of say, 80ish and a cruising speed of 65ish, you can easily do this with a DC system and lead acid for about $12K


Well, what I really need are formulae tailored to my car, so I can do a proper analysis.

I can't have a budget until I know what I need to do this. At this point all I've got is a car and a dream.

I understand that what I'm shooting for will likely be expensive, the question is precisely how expensive does it have to be, and I can't know that without more accuracy.

Do you know where I can find less generalized formulae that I can use to determine my vehicles specific kWh needs at specific sustained speeds?


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## favguy (May 2, 2008)

I'm sorry I don't, but I'm relatively new to this myself! There are some on here who know much more, maybe they'll show up soon..


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## mattW (Sep 14, 2007)

Kosstheory if you can find out the rolling resistance (Crr) and CdA of your car I can work out how many kW it will take to maintain a certain speed and how many kWh it will take to go a certain distance at that speed... You'll just have to do the leg work to find me the numbers.


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## kosstheory (May 26, 2008)

mattW said:


> Kosstheory if you can find out the rolling resistance (Crr) and CdA of your car I can work out how many kW it will take to maintain a certain speed and how many kWh it will take to go a certain distance at that speed... You'll just have to do the leg work to find me the numbers.


That's what I'm talking about! Thanks! I'll get the info you require in short order.


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## mattW (Sep 14, 2007)

The equation, using all metric SI units is 
Pw=m x g x v x Crr (rolling resistance) + 0.6 x CdA x x v^3 (Air drag)

Power is in watts, m is mass in kg,g is acceleration due to gravity (9.8m/s/s), v is velocity in km/h, Crr is coefficent of rolling resistance, 0.6 is a constant (1/2 times the density of Air @~ 1.2kg/m^3), CdA is coefficient of drag times frontal area in m^2... i can do tha calculations if that is too intimidating lol


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## kosstheory (May 26, 2008)

mattW said:


> The equation, using all metric SI units is
> Pw=m x g x v x Crr (rolling resistance) + 0.6 x CdA x x v^3 (Air drag)
> 
> Power is in watts, m is mass in kg,g is acceleration due to gravity (9.8m/s/s), v is velocity in km/h, Crr is coefficent of rolling resistance, 0.6 is a constant (1/2 times the density of Air @~ 1.2kg/m^3), CdA is coefficient of drag times frontal area in m^2... i can do tha calculations if that is too intimidating lol


Can you check this? I re-written it to make sure I understand:

(Power in Watts) = ((Mass in kilograms)(Acceleration due to gravity)(Velocity in km/h)(Rolling Resistance)) + ((0.6)(Coefficient of drag)(Velocity³)

or with some of the variables filled in:

(Power in Watts) = ((1360 kg)(9.8 ms²)(160 km/h)(Rolling Resistance)) + ((0.6)(Coefficient of drag)(160 km/h³)

Let me know if I have it right so far.

Thanks again.


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## kosstheory (May 26, 2008)

found this gem on the net.


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## kosstheory (May 26, 2008)

mattW said:


> The equation, using all metric SI units is
> Pw=m x g x v x Crr (rolling resistance) + 0.6 x CdA *x x* v^3 (Air drag)
> 
> Power is in watts, m is mass in kg,g is acceleration due to gravity (9.8m/s/s), v is velocity in km/h, Crr is coefficent of rolling resistance, 0.6 is a constant (1/2 times the density of Air @~ 1.2kg/m^3), CdA is coefficient of drag times frontal area in m^2... i can do tha calculations if that is too intimidating lol


In the above formula what are the two consecutive multiplication operations for?

Just want to make sure I'm not missing something before I proceed with the calculations.


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## mattW (Sep 14, 2007)

Nah just a typo mate i was going to put 'rho' there for air density but I didn't have greek letters and figured it was just easier to use 0.6 for the constant rather than 1/2 x rho.
Your way of wording it looks good, though 160km/hr sounds a little hairy.


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## MrCrabs (Mar 7, 2008)

kosstheory said:


> Thanks for the link to the Calc tool. I'm going to see if I can contact the programmer to get my hands on the raw code.


http://www.geocities.com/CapeCanaveral/lab/8679/ev.html

This is Uve's page which has more information on the formulas used in the EV calculator. (The newer one is based off/is the same as Uve's calc)


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## kosstheory (May 26, 2008)

mattW said:


> The equation, using all metric SI units is
> Pw=m x g x v x Crr (rolling resistance) + 0.6 x CdA x x v^3 (Air drag)


Matt, I currently have Ford Design looking into the CdA number for me, but in the interim I have decided to go with a Cd of .5 and the frontal area is about 2.37 m².

I pluuged the Crr of .0062 and CdA of 1.1535 into the equation and I got some really high numbers.

I double checked my work, and then I decided to double check the equation. I found two sites that show basically the same equation for deriving drag using CdA, but both of them show the exponent of the velocity as being 2 instead of 3.


Here's what I found on Wikipedia.










where
_F_d is the force of drag, which is by definition the force component in the direction of the flow velocity,[1] _ρ_ is the mass density of the fluid, [2] _v_ is the velocity of the object relative to the fluid, _A_ is the reference area, and _Cd_ is the drag coefficient (a dimensionless constant, e.g. 0.25 to 0.45 for a car). 
Should your equation be:



mattW said:


> Pw=m x g x v x Crr (rolling resistance) + 0.6 x CdA x x v^2




Incidentally, when I resolved the equation using the square of the velocity instead of the velocity cubed the numbers obviously looked more in line.

Please check my work and let me know. I'd also like to know how the resulting power value is used to size the batteries. I'm guessing this is only good as an instantaneous value, and we're ultimately going to use it to determing the watt-hour requirement, but I don't know how to get from here to there. Would I multiply by the number of seconds in an hour?

Thanks for your time!


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## mattW (Sep 14, 2007)

Work is force x distance, power is work/time so power is force x distance/time. In the end you get force (as in your equation above) times velocity so the v^2 goes to a v^3 but everything else is the same. Bad CdA at high speed use monstrous amounts of power since it is a cubic function of velocity. That's why its better to drive 60 on the highway instead of 80.


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## MrCrabs (Mar 7, 2008)

Matt is correct, you can check with the units.
Power is measured in Watts.
1 W = (1 kg * 1 m^2 ) / s^3

In the equation the 0.6 constant (related to the density of air) gives us kg/m^3, the area term gives up m^2 and the velocity term gives us m^3/s^3
So kg/m^3 * m^2 * m^3/s^3 simplifies to kg * m^2 / s^3 which is a Watt



edit:
Why does the equation use velocity in km/h instead of m/s?


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## mattW (Sep 14, 2007)

Woops, it is meters per second actually, that makes a lot more sense because all the other units use meters and seconds, This is all looking back at my uni notes from three years ago so its a little rusty. The example they used was at 70km/h so I assumed that was the units but I guess they must have converted it to m/s before doing the calculations. Sorry kosstheory, hopefully that should clear things up a bit and give you better numbers to work with.


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## mattW (Sep 14, 2007)

If I take your numbers from the other post but plug it in m/s secong instead of Km/h you get:
_
Power in Watts = ((Mass in kg) (9.8m/s²) (Velocity in m/s) (Rolling Resistance)) + ((0.6465) (Coefficient of Drag) (Area in m²) (*Velocity^3)*)_

_Mass in Kg = 1367_
_Acceleration Due to Gravity = 9.8m/s²_
_Velocity= *37.8 m/s*_
_Rolling Resistance = .0062_
_Density of Air = .6465_
_Coefficient of Drag =.5_
_Area = 2.37 m²_

_Plugging all of the values in I get:_

_((1360.77 kg) (9.8 m/s²) (*37.8m/s*) (.0062)) + ((0.6465) (.5) (2.37 m²) (*37.8^3*))
(3125.3) + (41337)
44.5kW

That's with the correct velocity^3 in m/s for 136km/h. To do it for an hour would need a 44.5kWh pack which is $22,250 at $0.5/Wh.

Moral of the story, either drive slower or get a lower drag car or get ready to fork out some serious $$$
_


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## kosstheory (May 26, 2008)

mattW said:


> If I take your numbers from the other post but plug it in m/s secong instead of Km/h you get:
> 
> _Power in Watts = ((Mass in kg) (9.8m/s²) (Velocity in m/s) (Rolling Resistance)) + ((0.6465) (Coefficient of Drag) (Area in m²) (*Velocity^3)*)_
> 
> ...


I'm trying to present ev conversion as a viable alternative to using gasoline, and for most people 65 mph isn't fast enough, because they have to contend with traffic, and be able to react to change on the higway rapidly, whether that means slowing down or speeding up to get out of the way of potential problems. So, in reality The 80mph or even 100mph goal is based not on cruising speed but spurts of possible necessity. As my unnderstanding has developed, I get that jumping up to 80mph can be done as long as the motor can support that, but it will just drain the batteries faster. The cruising speed would be about 65 mph.

Now, as far as the equation, why does it differ, using V³ instead of V². The equation I found was supposedly already for high speed. I realize that you've been doing this for a minute, and probably have loads of emperical data to support the V³, but I need to understand why I'm using V³, especially when it makes such a big difference in the final numbers. And more importantly, when I'm considering forking over several thousand dollars based on it.

Thanks for all your help and your patience will such a pain in the arse noob!


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## mattW (Sep 14, 2007)

Using v^2 will give you the force, which you can use to work out the torque required at the wheel, but using v^3 will give you the power required at the wheel. The number for the aero drag with the v^2 forumula would actually be in Newtons not Watts. You are trying to work out the power not the force so you need to use the power equation not the force equation.

If its just for bursts of speed then you just need to make sure the motor's peak power is enough that after transmission losses you can reach those speeds, you just need to make sure that your cruising speed power requirements are a little less than the continuous rating of the motor...


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## MrCrabs (Mar 7, 2008)

See my above post in this thread where I mention units.
A Newton is 1 (kg * m)/s^2
A Watt is 1 (kg *m^2)/s^3
http://en.wikipedia.org/wiki/Newton
http://en.wikipedia.org/wiki/Watt

So the equation with the V^2 term returns the Force needed in Newtons.
The equation with the V^3 term returns the Power needed in Watts.

Using the Force, you could work back and figure out much force your motor needs to output to achieve that speed.
Using the Power, you can work back and figure out much battery energy storage you need.

http://en.wikipedia.org/wiki/Air_density


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## kosstheory (May 26, 2008)

MrCrabs said:


> See my above post in this thread where I mention units.
> A Newton is 1 (kg * m)/s^2
> A Watt is 1 (kg *m^2)/s^3
> http://en.wikipedia.org/wiki/Newton
> ...


So, how do we translate m²/s³, taken from the Watt equation you found, into our equation:

Power in Watts = ((Mass in kg) (9.8m/s²) (Velocity in m/s) (Rolling Resistance)) + ((0.6465) (Coefficient of Drag) (Area in m²) (*Velocity^3)*)

How do we account for the meters being squared, if we're just cubing the velocity as a whole in this equation?


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## mattW (Sep 14, 2007)

Well its a big tricky but you'll get there easily enough, the first half multiplies simply enough to give kg.m^2/s^3, for the second half you need to remember that the density of air is measured in kg/m^3, if you work out the equation as you've written it you get the second half as m^2.m^3/s^3 but you can take the m^3 and convert it to kg due to the units for density of air... so you get m^2.kg/s^3 = watts


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## kosstheory (May 26, 2008)

mattW said:


> Well its a big tricky but you'll get there easily enough, the first half multiplies simply enough to give kg.m^2/s^3, for the second half you need to remember that the density of air is measured in kg/m^3, if you work out the equation as you've written it you get the second half as m^2.m^3/s^3 but you can take the m^3 and convert it to kg due to the units for density of air... so you get m^2.kg/s^3 = watts


Thanks for straightening me out.

So, moving forward, using a mass of 1367 kg, and a speed of 65mph with a Cd of .5, a Crr of .0062, and an Area of 2.37m².

I can see that the power requirements would be about 21k, which should sustain my travel at 65mph for an hour, or 65 miles, right?

Assuming this is correct, doe sthis mean that I could travel twice that distance at half the speed? Say, 32.5 mph for 130 miles?

Thanks again


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## mattW (Sep 14, 2007)

kW are the power required to travel that speed, kWh is the energy, the product of power and time which is how you determine range. If you want a range of 65 miles at 65mph then 21kWh (with allowances for the battery type i.e DoD and peukert's effect but we'll ignore that for now) would do the job. If you want to travel 110 miles at the same speed you will need double the energy (42kWh) but will still consume the same amount of power at any instantaneous point on the journey.

Now that you have the formula worked out, you can work out the power requirements at the lower speed and then divide your planned energy (kWh) by the low speed power (kW) to get the time you can travel for. The range should more than double...


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## MrCrabs (Mar 7, 2008)

mattW, I assume this formula is for on the level.

Do you have a version handy for inclines or declines? I can work it out later in the week, but was curious if you had worked it out yet.


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## mattW (Sep 14, 2007)

I haven't yet, I was planning to next week some time. I'm planning to write a wiki on "How Much Power Do I Need?".. which will summarise this thread. If you want to work it out that would save me some time, its my exam week now, but I have holidays coming up


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## kosstheory (May 26, 2008)

MrCrabs said:


> mattW, I assume this formula is for on the level.
> 
> Do you have a version handy for inclines or declines? I can work it out later in the week, but was curious if you had worked it out yet.


Hadn't even thought of that...


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## kosstheory (May 26, 2008)

mattW said:


> kW are the power required to travel that speed, kWh is the energy, the product of power and time which is how you determine range. If you want a range of 65 miles at 65mph then 21kWh (with allowances for the battery type i.e DoD and peukert's effect but we'll ignore that for now) would do the job. If you want to travel 110 miles at the same speed you will need double the energy (42kWh) but will still consume the same amount of power at any instantaneous point on the journey.
> 
> Now that you have the formula worked out, you can work out the power requirements at the lower speed and then divide your planned energy (kWh) by the low speed power (kW) to get the time you can travel for. The range should more than double...


How does it work out that the resulting power in watts can be sustained for 1 hour? I'm not sure if I'm wording this question right. Maybe I'm missing something simple?

Does this sum up the reason why this works out? (Taken from wikipedia entry on Watts)

_Confusion of watts and __watt-hours_
_Power__ and __energy__ are frequently confused in the general media. A watt is one 1 joule of energy per second. So watts multiplied by a period of time equals energy. For example, if a 100 watt light bulb is turned on for one hour, then an amount of energy is used corresponding to 100 watts of power being generated for a time period of one hour, i.e. 100 watts times one hour, i.e. 0.1 kilowatt-hour._

Are we actually multiplying the Power in Watts by 1 hour to result in a Watt-Hour value? In the example given on the Wiki page the Watt is said to be 1 joule of energy per second. If this is true then why isn't 3,600 used as a multiplier when figuring Watt-Hours?


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## mattW (Sep 14, 2007)

The reason is because an hour is a much more useful unit than a second for measuring energy. Its ok to multiply by hours because the units are in hours, you just need to remember that when you convert it back into Joules. You could measure energy in watt-seconds but most peoples numbers would be really big. There is an explanatory example on the page:

(1 kW·h)(1000 W/kW)(3600 s/h) = 3,600,000 W·s = 3,600,000 J = 3.6 MJ.

You learn something everyday too, I thought power was instantaneous but it is Joules per second so I guess its not.


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## mattW (Sep 14, 2007)

Ok the wiki article about this is now here...


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## gerd1022 (Jun 9, 2008)

i think the power eq. for hill climbing should be:

mass*g*sin(arctan(%grade/100))*velocity

m*g*sin(theta) is the vertical force to push the car up the hill.

force times velocity is power.

i think thats right.


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## MrCrabs (Mar 7, 2008)

Well I'm out of town on a business trip this week, doing a chiller water plant startup. Tonight I have more of the program to finish up for tomorrow's continued startup.
Ive been limiting myself to just an hour or so of diyelectriccar.com reading (thats how addicted I am, I should really be doing more work at the hotel each night, but I can't leave this site alone  )
If I get bored on Sunday this weekend I might start to work it out and put it into an excel spreadsheet. It looks like gerd's equations are on the right track.
'
Am I correct in thinking the rolling resistance will go down when you are going up hill, however there will be a much larger increase in force needed to overcome gravity? I was planning on modeling that to have the most realistic calculation possible. (even tho the reduction in RR will be almost negligible compared to the increase due to gravity)


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## TheSGC (Nov 15, 2007)

I just wanted to add my research with the EV Calculator. http://www.evconvert.com/tools/evcalc/

After about 6 months of comparing my own calculations, real world EV data from the EV Album and various people, estimates from EV of America and a few others, I have found a combo that seems to give the best estimations using that EV Calculator. 

Using the EV Calculator, input your car's data, select your motor, battery and controller from the list. Then in the last section, use 80% as DoD, 0.6 as the incline and 0.0015 as the rolling resistance. Now this seems to be dead on for small/midsized cars with a moderate hilly area. You can compare it yourself to some of the cars on the EV Album. I uses something the lines of 20-30 cars that are similar in specs/size to my Civic to come up with these numbers.


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## kosstheory (May 26, 2008)

Ok, so at 80% DoD, which I understand to be reasonable for lead acid batteries, my Ev will require about 32kWh of energy to accomplish my goal of about 100 miles per charge with ~60% at ~60 mph and ~40% at ~35mph.

This means that using an Azure Dynamics #AC55 AC motor, which should be more efficient than a DC motor, I would need ~18 X 18V Batteries with a 104Ah rating. 18 of them to get 312V to support the operating voltage of the motor (312/18=17.333), and 104Amp-Hours because 32k divided by 312 equals about 104, right?

Where can I find the best source for the lightest, least expensive Lead Acid or Lithium Ion batteries that fit these specs?


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## kosstheory (May 26, 2008)

Ok...Spot check...Calling all gurus...

I'm now leaning towards using a 144V-192V DC motor like the FB1-4001 or the WarP 9".

I decided that the additional cost to build a 312v batt system just can't be resolved. Lithium Ions are too expensive, so the additional weight of lead acid batteries, ~2500 lbs, is just unreasonable.

I did manage to find a relatively inexpensive battery that may be sufficient, at least on paper. The Interstate Battery SRM-4D, which is supposed to have 234Ah at 12V. I would only need 12 of these to get 144V, and the Amp Hour rating is actually a little higher than my original estimated requirement of 220Ah, giving a 33.696kWh energy rating for my battery storage. I originally estimated that I would need 32kWh given a 80% DoD.

My question is, has anyone here done any business with Interstate BAtteries, and how do their batteries stack up against others? The SRM-4D is supposed to be a true industrial application, deep cycle battery that's built to last.

A local distributor is willing to let them go for only $178.00 per batt, and that includes the core charge and a 12 month warranty +30 month pro-rate.

What do you guys think?


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## mattW (Sep 14, 2007)

I have no experience with them, but it sounds like a good price. I am jealous of the US battery market. What do you mean by this line;

"my Ev will require about 32kWh of energy to accomplish my goal of about 100 miles per charge with ~60% at ~60 mph and ~40% at ~35mph."

It didn't really make sense to me?


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## kosstheory (May 26, 2008)

mattW said:


> "my Ev will require about 32kWh of energy to accomplish my goal of about 100 miles per charge with ~60% at ~60 mph and ~40% at ~35mph."
> 
> It didn't really make sense to me?


Using the formula that you gave me to determine power, and energy requirements I came to the conclusion that it should be possible to go about 60 miles at about 60 MPH and about 40 miles at about 35 MPH with a 32kWh battery store.

Did you mean that you didn't understand my usage of the "~" sign? or are you saying that there is something wrong with my reasoning?


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