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Mathematical exercises on Ford's EV (out of main)

BioSehnsucht

Model 3 LR
Apr 1, 2016
1,791
4,877
DFW, TX
Wow, the upcoming Ford "Mustang inspired SUV" is going to A) be a massive guzzler, B) be expensive, and C) be super-heavy

Electrify America NEWSROOM

"Ford will be releasing its all-new, Mustang-inspired electric SUV, that aims to achieve a targeted EPA-estimated range of 300 miles. The vehicle’s 150kW charging ability will allow very quick charging speeds on Electrify America’s 150 to 350kW network. Ford estimates that its all-new, all-electric Mustang-inspired SUV will be capable to charge up to 47 miles in 10 minutes using Electrify America DC fast chargers,"

Let's do some math:
  • 150000W / (47mi/(1/6 hour) * 6) = 532 Wh/mi
  • 300mi * 0,532 kWh/mi = 160kWh usable. Probably 170-180kWh nominal.
  • Charge rate: 282mph / 453kph

532Wh/mi? Ford hasn't learned a d*mned thing about the importance of streamlining to making a good EV.

I'm not sure that first equation works as stated? As stated I don't get a sensical much less identical result - 47 / 0.1666... (1/6 being 0.166...) is ~282, then times six is 1692, divided into 150k is 88.65 ... ??

Though I can arrive at the same end results differently... 150kW * 1/6h (10 minutes) = 25kWh, 25kWh / 47 miles = 532Wh/mi.

I was thinking maybe it was supposed to be 47% range not 47 miles in 10 minutes, but then I did the math. 47% of 300 miles in 10 minutes would be improbably good and I have a hard time believing it, as it would come out to 177 Wh/mi. Somehow I doubt Ford managed to build a CUV with 30% less Wh/mi than the Model 3...

I suspect the reality is it can only hit the 150kW for a minute or two and then tapers hard and the charge speed is garbage. If we instead it had bad but not as bad Wh/mi (let's say something around the EQC's 355 Wh/mi) it might have actually a 100~110kW pack, and maybe it averages 100kW charging in that 10 minutes (which means that if it started at 150kW and dropped down to something like 88 kWh - and at < 20% SoC!). I don't even want to think about the taper over a whole 90% charge.

So while I doubt it's efficiency is quite as bad as 532Wh/mi, I also expect it to have really terrible charging rates. Perhaps insufficient pack thermal management to sustain high C rate? So basically no redeeming qualities no matter how you slice it.
 

StealthP3D

Well-Known Member
Dec 12, 2018
9,939
81,962
Maple Falls, WA
I'm not sure that first equation works as stated? As stated I don't get a sensical much less identical result - 47 / 0.1666... (1/6 being 0.166...) is ~282, then times six is 1692, divided into 150k is 88.65 ... ??

You are actually doubting KarenRei's basic EV electrical math? :eek:

What's next? Wondering if the sun will rise tomorrow? ;)
 

Carl Raymond

Active Member
Oct 18, 2018
1,565
13,270
NSW, Australia
You are actually doubting KarenRei's basic EV electrical math? :eek:

What's next? Wondering if the sun will rise tomorrow? ;)

Her answer is right, but there’s a “ * 6 ” in her working that oughtn’t be there.
I can see the train of thought. To divide by one sixth, you multiply by 6 instead. She’s shown both. One or t’other.
 
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StealthP3D

Well-Known Member
Dec 12, 2018
9,939
81,962
Maple Falls, WA
Her answer is right, but there’s a “ * 6 ” in her working that oughtn’t be there.
I can see the train of thought. To divide by one sixth, you multiply by 6 instead. She’s shown both. One or t’other.

I see. It didn't even occur to me to check her "work". Somehow she came up with the right answer.
 

KarenRei

ᴉǝɹuǝɹɐʞ
Jul 18, 2017
9,619
104,589
Iceland
Her answer is right, but there’s a “ * 6 ” in her working that oughtn’t be there.
I can see the train of thought. To divide by one sixth, you multiply by 6 instead. She’s shown both. One or t’other.

It's the standard way to cancel terms. E.g.

X foo/bar * Y bar/baz = X*Y foo/baz

In this case, it's:

X mi/(1/6 hour) * 6 (1/6 hour)/hour = X*6 mi/hour

Note that each term remains self-consistent (e.g. there's six "1/6th hour units" (e.g. 10 minutes) per hour, and 6*(1/6) hours = 1 hour) and that the numerator of the second term cancels the denominator of the first.

These sort of equations are common in physics and chemistry. :)
 

KarenRei

ᴉǝɹuǝɹɐʞ
Jul 18, 2017
9,619
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Also, there's no reason such a large pack shouldn't be able to maintain 150kW for 10 minutes. That'd actually be an even worse admission :)
 
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mongo

Well-Known Member
May 3, 2017
13,461
42,017
Michigan
It's the standard way to cancel terms. E.g.

X foo/bar * Y bar/baz = X*Y foo/baz

In this case, it's:

X mi/(1/6 hour) * 6 (1/6 hour)/hour = X*6 mi/hour

Note that each term remains self-consistent (e.g. there's six "1/6th hour units" (e.g. 10 minutes) per hour, and 6*(1/6) hours = 1 hour) and that the numerator of the second term cancels the denominator of the first.

These sort of equations are common in physics and chemistry. :)
Yes, unit cancellation is common. But you dropped a term or were doing work inline.

As written
150000W / (47mi/(1/6 hour) * 6)
As intended?
150000W/ (47mi/(1/6 hour) * ((1/(6 hour))* 6 hour))


Yes, foo/baz is standard, but I have never seen foo/foo or baz/baz.

The item causing confusion:
"
Let's do some math:

150000W / (47mi/(1/6 hour) * 6) = 532 Wh/mi
"
Note that the orignal story had
"Ford will be releasing its all-new, Mustang-inspired electric SUV, that aims to achieve a targeted EPA-estimated range of 300 miles. The vehicle’s 150kW charging ability will allow very quick charging speeds on Electrify America’s 150 to 350kW network. Ford estimates that its all-new, all-electric Mustang-inspired SUV will be capable to charge up to 47 miles in 10 minutes using Electrify America DC fast chargers,"

So, to be easily followable (as much as unformatted triple demonator text can be), one could potentially start with:
150000W / (47mi/10min)
However, jumping straight to this would not be untoward:
150000W / (47mi/(10min*(1hr/60min)))=532Wh/mile. Which clearly shows the minutes to hours conversion.

Or one could also add an intermediate step showing the fractional conversion:
150000W / (47mi/(1/6hour)
=150000W / (6x47mi/hour)

6 (1/6 hour)/hour

Adding a 1/6×6 hr/hr is valid; however, that was not what was written:

150000W / (47mi/(1/6 hour) * 6)
= 532 Wh/mi
Is equivalent to
150000W / (6*47mi(1/6 hour)
Or
150000W / (6*6*47mi/hour)
= 88.7Wh/mi

Even volating the OoO and taking it as:
150000W / (47mi/((1/6 hour) * 6))
Or
150000W/(47mi/hour) is incorrect
It would have needed to be: (with or without extra hr/hr units)
150000W/( (47mi/(1/6hr)) *(1/6* 6))
 

KarenRei

ᴉǝɹuǝɹɐʞ
Jul 18, 2017
9,619
104,589
Iceland
Yes, unit cancellation is common. But you dropped a term or were doing work inline.

As written
150000W / (47mi/(1/6 hour) * 6)
As intended?
150000W/ (47mi/(1/6 hour) * ((1/(6 hour))* 6 hour))


Yes, foo/baz is standard, but I have never seen foo/foo or baz/baz.

The item causing confusion:
"
Let's do some math:

150000W / (47mi/(1/6 hour) * 6) = 532 Wh/mi
"
Note that the orignal story had
"Ford will be releasing its all-new, Mustang-inspired electric SUV, that aims to achieve a targeted EPA-estimated range of 300 miles. The vehicle’s 150kW charging ability will allow very quick charging speeds on Electrify America’s 150 to 350kW network. Ford estimates that its all-new, all-electric Mustang-inspired SUV will be capable to charge up to 47 miles in 10 minutes using Electrify America DC fast chargers,"

So, to be easily followable (as much as unformatted triple demonator text can be), one could potentially start with:
150000W / (47mi/10min)
However, jumping straight to this would not be untoward:
150000W / (47mi/(10min*(1hr/60min)))=532Wh/mile. Which clearly shows the minutes to hours conversion.

Or one could also add an intermediate step showing the fractional conversion:
150000W / (47mi/(1/6hour)
=150000W / (6x47mi/hour)



Adding a 1/6×6 hr/hr is valid; however, that was not what was written:

= 532 Wh/mi
Is equivalent to
150000W / (6*47mi(1/6 hour)
Or
150000W / (6*6*47mi/hour)
= 88.7Wh/mi

Even volating the OoO and taking it as:
150000W / (47mi/((1/6 hour) * 6))
Or
150000W/(47mi/hour) is incorrect
It would have needed to be: (with or without extra hr/hr units)
150000W/( (47mi/(1/6hr)) *(1/6* 6))

This is a long way of writing: you left off the units on that one term.

To which the answer is: yes, because I thought it was obvious, and that writing out a 6*(1/6 hour)/hour was more confusing than just writing 6. I guess not.

ED: To Daveepisilon, who disagreed with this: Take the final element in Mongo's post:

It would have needed to be: (with or without extra hr/hr units)
150000W/( (47mi/(1/6hr)) *(1/6* 6))

Put the necessary units in that he left out:
150000W/( (47mi/(1/6hr)) * ((1/6 hr)* 6)/hr)

Move the "times 6" to the left side of the numerator:
150000W/( (47mi/(1/6hr)) * 6*(1/6 hr)/hr)

Now omit the conversion term, because I felt it was bloody obvious how you convert from 1/6th hours to whole hours, and figured more people would be confused by seeing it than who would nitpick over its absence:
150000W/( (47mi/(1/6hr)) * 6)

Look familiar? There's no math error, I literally just left off writing the conversion term because I never could have imagined that people would go off on this bizarre tangent over something so obvious.
 
Last edited:

KarenRei

ᴉǝɹuǝɹɐʞ
Jul 18, 2017
9,619
104,589
Iceland
yes, it's obviously a mistake in that line. I find this website to be quite useful for unit analysis and it even can convert to Wh/mi. 150000W / (47mi/(1/6 hour) ) to Wh/mi - Wolfram|Alpha

Which shows that my answer is right. Why are we even still talking about this? I simply left off writing the units on the 6 to try to make things less confusing because I figured that people would inherently understand that there's six "1/6th hours" in an hour.

A decision which inadvertently set off this increasingly weird conversation.
 
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davepsilon

Member
Sep 29, 2019
183
657
Massachusetts
This is a long way of writing: you left off the units on that one term.

To which the answer is: yes, because I thought it was obvious, and that writing out a 6*(1/6 hour)/hour was more confusing than just writing 6. I guess not.

No, it's a simple typo. The line should be either
150000[W] / (47[mi] * 6 [hr^-1])
or
150000[W] / (47[mi]/(1/6 hr) )

including both in the line gives the wrong answer: 89 Wh/mi

As an aside when you say you wrote 6 instead of 6*(1/6) to reduce confusion that's where the error is introduced. (6 * (1/6)) = 1 not 6
 
Last edited:
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KarenRei

ᴉǝɹuǝɹɐʞ
Jul 18, 2017
9,619
104,589
Iceland
No, it's a simple typo. The line should be either
150000[W] / (47[mi] * 6 [hr^-1])
or
150000[W] / (47[mi]/(1/6 hr) )

including both in the line gives the wrong answer: 89 Wh/mi

No. (1/6 hr) is units; you don't multiply by units, you cancel them until you get your final units. I simply - and now for, what, the 4th time that I have to repeat it? - omitted spelling out the conversion term on the 6*(1/6 hr)/hr conversion for what I thought was clarity. The omitted (1/6 hr) numerator cancels the (1/6hr) denominator.

Again, I thought this was bloody obvious. Why do we have to keep going over this? No matter how many times you try to insist that I was doing something else, the actual answer will continue to be: I omitted the unit conversion because I thought it was obvious, and never imagined that people would rail against me for doing so. My answer was correct (as Wolfram confirms), as was the formula was correct apart from deliberately omitting the unit conversion terms to try to avoid confusion. Based on the notion that I (mistakenly) thought that even an idiot could tell that if you're working in units of miles per 1/6th hours, the conversion is just to multiply by six.
 

mongo

Well-Known Member
May 3, 2017
13,461
42,017
Michigan
Look familiar? There's no math error, I literally just left off writing the conversion term because I never could have imagined that people would go off on this bizarre tangent over something so obvious.

No. (1/6 hr) is units; you don't multiply by units, you cancel them until you get your final units. I simply - and now for, what, the 4th time that I have to repeat it? - omitted spelling out the conversion term on the 6*(1/6 hr)/hr conversion for what I thought was clarity. The omitted (1/6 hr) numerator cancels the (1/6hr) denominator.

Again, I thought this was bloody obvious. Why do we have to keep going over this? No matter how many times you try to insist that I was doing something else, the actual answer will continue to be: I omitted the unit conversion because I thought it was obvious, and never imagined that people would rail against me for doing so. My answer was correct (as Wolfram confirms), as was the formula was correct apart from deliberately omitting the unit conversion terms to try to avoid confusion. Based on the notion that I (mistakenly) thought that even an idiot could tell that if you're working in units of miles per 1/6th hours, the conversion is just to multiply by six.

No need to imply anyone is an idiot.
I have never seen the unit (1/6hr), nor do I expect anyone else has. The Ford statement made no mention of 1/6 hour units so I am unclear how anyone would make that leap. When the average person sees (1/6) in an equation, they will assume those are numbers. If one wants to remove confusion, might I suggest writing it out (sixths of an hour). Or else use common units like 10 minutes which is what the original statement had.
 

davepsilon

Member
Sep 29, 2019
183
657
Massachusetts
No. (1/6 hr) is units; you don't multiply by units, you cancel them until you get your final units. I simply - and now for, what, the 4th time that I have to repeat it? - omitted spelling out the conversion term on the 6*(1/6 hr)/hr conversion for what I thought was clarity. The omitted (1/6 hr) numerator cancels the (1/6hr) denominator.
...
of miles per 1/6th hours

haha, I see what you were doing now. It's really confusing to mix in fractions with traditional units to define new units.
 
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