# Watts per mile climbing a 6% grade?



## Zak650 (Sep 20, 2008)

Hi,

In my area there is a 3 mile long 6% grade. I'm looking for feedback from people that can tell me their watts per mile are while climbing a similar long steep hill. 

What does your car weigh?
What is your pack voltage?
State the % grade
State the amperage
State motor brand/type
Length of grade?

Thanks,
Zak


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## Duncan (Dec 8, 2008)

Hi Zak

_In my area there is a 3 mile long 6% grade. I'm looking for feedback from people that can tell me their watts per mile are while climbing a similar long steep hill. _

Watts per mile is a nonsense measure - I assume you mean Watt hours per mile
The other thing you could mean is 
How many more watts do I need to climb the hill - and that is speed dependent

the contribution from the grade is easy

6% 0f 1 mile is 105 yards = 93.4 meters

Energy needed is mass of vehicle (Kg) x 9.81 x 93.4(m) = Joules - 1 Watt hour = 3,600 Joules

This give the amount of extra watt hours needed per mile (over your needs on a flat road)

So if your car weighs 1000Kg you will need 

1000 x 9.81 x 93.4 = 916,254 Joules - 254 watt hours per mile *EXTRA*

If you are interested in the number of watts (power required) then add speed into the equation

30 mph - 1 mile = 2 minutes = 120 seconds 

916,254 Joules divided by 120 seconds = 7635 watts = 7.635 Kw


60 mph - 1 mile = 1 minutes = 60 seconds 

916,254 Joules divided by 60 seconds = 15,270 watts = 15.27 Kw

substitute your cars weight and you can see how much more power you will need

You can use the extra power needed and your battery voltage to calculate the extra battery current you will draw

Motor current is a bit more complex


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## tomofreno (Mar 3, 2009)

For my car, 2260 lb, 115V pack, level ground/6% grade:

30 mph 45 mph 60 mph
3.8/15.8 HP 8.2/26.2 HP 15.5/39.6 HP
31/130A 66/210A 126/320A battery current

HPEVS AC50 motor, 3 phase AC, so the currents are rms, per phase.
The above numbers are calculated, but the spreadsheet has been checked against performance over a range of conditions and agrees fairly well. My controller may well overheat if I tried it at 60 mph. I know from experience it can do 5% at 40 mph for 6 miles in 91 F ambient no problem. Also 4.5% at 45 mph for 19 miles in 90 F ambient.




Zak650 said:


> Hi,
> 
> In my area there is a 3 mile long 6% grade. I'm looking for feedback from people that can tell me their watts per mile are while climbing a similar long steep hill.
> 
> ...


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## bjfreeman (Dec 7, 2011)

Duncan said:


> Hi Zak
> 
> 
> 
> ...


could you explain this
my understanding is
1% grade is 1 foot in one mile or 
0.3048 meters in 1.609344 kilometers or
0.0003048 kilometers in 1.609344 kilometers


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## bjfreeman (Dec 7, 2011)

Zak650 said:


> Hi,
> 
> In my area there is a 3 mile long 6% grade. I'm looking for feedback from people that can tell me their watts per mile are while climbing a similar long steep hill.
> 
> ...


the top level formula I use is:
Rolling Resistance (hp) + Aerodynamic Drag (hp) + Hill Climbing ((hp) + Acceleration (Hp)


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## Duncan (Dec 8, 2008)

_*my understanding is
1% grade is 1 foot in one mile or


*_1% grade is 1 ft in 100 feet _*
*_1 ft in 1 mile is 0.019% grade_*

*_6% is a steep gradient 

6 ft per mile - you would notice it on a bicycle but not in a car _*
*_


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## Zak650 (Sep 20, 2008)

Hi all,

Thanks for the info and the formulas you have provided. I am almost certain the % grade would mean the rise in elevation over the distance traveled. For example a 10% grade would have a rise of 528 feet per mile of straight level travel. A 100% grade would be 5280 feet rise per mile of level travel. In the latter case I hope the folks in front of you have a very powerful engine and everyone's brakes work well.

Zak


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## bjfreeman (Dec 7, 2011)

Zak650 said:


> Hi all,
> 
> Thanks for the info and the formulas you have provided. I am almost certain the % grade would mean the rise in elevation over the distance traveled. For example a 10% grade would have a rise of 528 feet per mile of straight level travel. A 100% grade would be 5280 feet rise per mile of level travel. In the latter case I hope the folks in front of you have a very powerful engine and everyone's brakes work well.
> 
> Zak


the USA standard is 
1% grade is 1 foot in one mile or 
0.3048 meters in 1.609344 kilometers or
0.0003048 kilometers in 1.609344 kilometers

so 6% would be 6 feet or ~1.8 metre


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## Duncan (Dec 8, 2008)

bjfreeman said:


> the USA standard is
> 1% grade is 1 foot in one mile or
> 0.3048 meters in 1.609344 kilometers or
> 0.0003048 kilometers in 1.609344 kilometers
> ...



COBBLERS the USA standard is in percentage

if you were correct a 25% grade (and we do have some) would be a 1320% grade - COBBLERS


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## bjfreeman (Dec 7, 2011)

just some numbers having todo with climb.
a 25% is a 12 degree hill.
you can rule of thumb use 1HP per %


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## bjfreeman (Dec 7, 2011)

Duncan said:


> COBBLERS the USA standard is in percentage
> 
> if you were correct a 25% grade (and we do have some) would be a 1320% grade - COBBLERS


Yes the USA is stated in % but the physical definition is 1 foot in one mile or 
0.3048 meters in 1.609344 kilometers



what does the 1320% in physical terms.


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## bjfreeman (Dec 7, 2011)

bjfreeman said:


> just some numbers having todo with climb.
> a 25% or 25Ft is a 45 degree hill.
> you can rule of thumb use 1HP per %


I stand corrected by wiki
http://en.wikipedia.org/wiki/Grade_(slope)


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## major (Apr 4, 2008)

bjfreeman said:


> just some numbers having todo with climb.
> a 25% is a 12 degree hill.
> you can rule of thumb use 1HP per %


"1HP per %" regardless of vehicle mass  And one foot per mile is 1%  You sure can come up with strange figures.


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## major (Apr 4, 2008)

bjfreeman said:


> what does the 1320% in physical terms.


It means for every horizontal mile travelled you have gone up 13.2 miles in elevation, or up an 85.7º slope angle (nearly vertical).


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## Duncan (Dec 8, 2008)

major said:


> It means for every horizontal mile travelled you have gone up 13.2 miles in elevation, or up an 85.7º slope angle (nearly vertical).



Hi Major

the 1320% was what a 25% slope would be in BJ's weird 1ft per mile = 1% system

I wonder if the 1ft per mile = 1% is an old railway standard??


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## bjfreeman (Dec 7, 2011)

major said:


> "1HP per %" regardless of vehicle mass  And one foot per mile is 1%  You sure can come up with strange figures.


that is 1HP added.
using the top formula
Rolling Resistance (hp) + Aerodynamic Drag (hp)
+ Hill Climbing ((hp) + Acceleration (Hp)
and I gave a link to wiki that shows how us uses it and how europe shows it and how they are both the same but not as %


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## tomofreno (Mar 3, 2009)

Duncan said:


> Hi Major
> 
> the 1320% was what a 25% slope would be in BJ's weird 1ft per mile = 1% system
> 
> I wonder if the 1ft per mile = 1% is an old railway standard??


My understanding is that the standard for freight railroad grades in the U.S. is less than 3%.


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## tomofreno (Mar 3, 2009)

bjfreeman said:


> that is 1HP added.
> using the top formula
> Rolling Resistance (hp) + Aerodynamic Drag (hp)
> + Hill Climbing ((hp) + Acceleration (Hp)
> and I gave a link to wiki that shows how us uses it and how europe shows it and how they are both the same but not as %


You are missing a parenthesis in your link. I didn't see anything in the wiki about 1 HP added per percent increase in grade, could have missed it. Seems this cannot be correct since the added force due to a grade is the product of the vehicle weight and the sine of the angle of the grade. Power is the (inner) product of force and velocity, so the required additional power would be different for different vehicle speeds.


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## major (Apr 4, 2008)

bjfreeman said:


> that is 1HP added.


Yes, I am aware of "added". But nevertheless, that power is very dependent on the mass because it is the rate of change of potential energy. So it cannot be the same "1HP per %" for a 40,000 lb truck and a 2500 lb car assuming the speeds are the same.



> and I gave a link to wiki that shows how us uses it and how europe shows it and how they are both the same but not as %


I just read your reference link again and I see this: 


> In the U.S., this percentage "grade" is the most commonly used unit for communicating slopes in transportation (streets, roads, highways and rail tracks), surveying, construction, and civil engineering.


Since the discussion was about road grade, it is commonly expressed in percentage in this country. I live in the flattest county in the country so don't see many grade signs. But I have done many RV miles out West and remember all those roads and highways to and from the passes posted with percentage grade signs.

BTW, the typical elevation change in this county is just about 1 foot per mile


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## bjfreeman (Dec 7, 2011)

tomofreno said:


> You are missing a parenthesis in your link. I didn't see anything in the wiki about 1 HP added per percent increase in grade, could have missed it. Seems this cannot be correct since the added force due to a grade is the product of the vehicle weight and the sine of the angle of the grade. Power is the (inner) product of force and velocity, so the required additional power would be different for different vehicle speeds.


the 1 HP as I stated originally is a rule of thumb not calculated. Like ball park.
the link was to define the grade.

A 6% grade will require you to take 6 times the car weight rolling resistance to calculate the Hp required. it takes 6 to 8 Hp for every 1000 pounds of car, Thus a 4000 pound car would require 4x6Hp x 2 or 48 Hp to push it 50 MPH up a 2%, assuming that the Hp for Aerodynamic Drag and Acceleration is zero.


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## tomofreno (Mar 3, 2009)

I can understand your motivation in constructing rules of thumb since you said you don't do well with math. I am trying to see how you arrived at the rule of thumb. Your estimation of the additional power required for a 2% grade doesn't seem to use vehicle speed other than stating it is at 50 mph. Are you saying this rule of thumb works for that speed, and there would be a different one for 30 mph?



bjfreeman said:


> the 1 HP as I stated originally is a rule of thumb not calculated. Like ball park.
> the link was to define the grade.
> 
> A 6% grade will require you to take 6 times the car weight rolling resistance to calculate the Hp required. it takes 6 to 8 Hp for every 1000 pounds of car, Thus a 4000 pound car would require 4x6Hp x 2 or 48 Hp to push it 50 MPH up a 2%, assuming that the Hp for Aerodynamic Drag and Acceleration is zero.


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## tomofreno (Mar 3, 2009)

Here is a graph of additional required HP to move a 1000 lb vehicle up 3 different percent grades at speeds from 10 to 60 mph, using an additional force equal the product of vehicle weight and sine of the incline angle (angle calculated from percent grade, e.g. 2% grade is 1.1 degree). At 50 mph on a 2% grade, a 4000 lb vehicle would require about 4*6.7 HP = 27 HP additional, where 6.7 HP is the additional power required for a 1000 lb vehicle. An 8% grade would require about 107 HP additional for a 4000 lb vehicle at 50 mph. Unless I've made a mistake, your estimate seems to be high.
View attachment Additional power required to climb a grade.pdf

Edit: Maybe I should explain...I calculated the additional required wheel torque from the above additional force. The additional required power is the product of this torque and the angular velocity of the wheels. I divided this power by 0.9 assuming 90% drive train efficiency.



bjfreeman said:


> the 1 HP as I stated originally is a rule of thumb not calculated. Like ball park.
> the link was to define the grade.
> 
> A 6% grade will require you to take 6 times the car weight rolling resistance to calculate the Hp required. it takes 6 to 8 Hp for every 1000 pounds of car, Thus a 4000 pound car would require 4x6Hp x 2 or 48 Hp to push it 50 MPH up a 2%, assuming that the Hp for Aerodynamic Drag and Acceleration is zero.


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## tomofreno (Mar 3, 2009)

tomofreno said:


> Here is a graph of additional required HP to move a 1000 lb vehicle up 3 different percent grades at speeds from 10 to 60 mph, using an additional force equal the product of vehicle weight and sine of the incline angle (angle calculated from percent grade, e.g. 2% grade is 1.1 degree)...


Got distracted there for a while. To continue on...here are some rules of thumb:
1) At 30 mph on a 4% grade 8 additional HP are required for 1000 lb vehicle weight, and increasing speed by 10 mph increases the required power by 2.67 HP.
2) Doubling the grade doubles the required additional power.


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## bjfreeman (Dec 7, 2011)

does your calculation for Rolling Resistance ?


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## Zak650 (Sep 20, 2008)

It seems to me that with an accurate weight of your vehicle and driver, a GPS to measure distance and elevation, two different downhill constant slope sections of straight road you should be able to take readings and arrive at rolling resistance of your particular car and tires and it's particular aerodynamic coeffient of drag.

DON'T CAUSE AN ACCIDENT ARRIVING AT THESE MEASUREMENTS!

road section 1: find the down hill speed it maintains in nuetral, GPS should tell you the change in distance and elevation. The GPS measurements do not have to be taken while moving. It can be recorded on the side of the road.

find the same for section 2.

The greater difference in slope the more accurate the outcome will be. Zero wind speed would also be helpful.

Since the over all drag minus aerodynamics is a constant curve and aerodynamic drag changes with the square of speed it seems that both of these curves could be plotted.


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## bjfreeman (Dec 7, 2011)

Zak650 said:


> It seems to me that with an accurate weight of your vehicle and driver, a GPS to measure distance and elevation, two different downhill constant slope sections of straight road you should be able to take readings and arrive at rolling resistance of your particular car and tires and it's particular aerodynamic coeffient of drag.
> 
> DON'T CAUSE AN ACCIDENT ARRIVING AT THESE MEASUREMENTS!
> 
> ...


rolling resistance=Crr * WeightKG * Gravitational acceleration 
Resulting units = Newtons 
Crr: Coefficient of rolling resistance (rolling drag). A value of 0.006 to 0.010 represents \low rolling resistance tires on a smooth surface; 0.010 to 0.015 represents ordinary car tires on concrete (see Wikipedia for other sample values). 
_*newton*_ is the unit for force _joules_ is the unit for work done by definition, work done = force X distance so multiply _newton_ by metre to get _joules_ 
*metre* (*meter* in the US) is the base unit of length Since 1983, it is defined as the length of the path travelled by light in vacuum in 1 ⁄ 299,792,458 metres (approximately 186,282 miles) of a second. 
Universal Gravitational Constant(G) = 6.6726 x 10-11N-m2/kg2conventional standard value of exactly 9.80665 m/s2 (approx. 32.174 ft/s) 

*Aerodynamic Drag: *

http://en.wikipedia.org/wiki/Automotive_aerodynamics
Aero drag formula: 0.5*rho * V^2 * Cd * A 
Resulting units = Newtons 
Cd: Coefficient of drag. If you don't know your vehicle's Cd, you can try to find it online (e.g. see: the large list of Cd values at Wikipedia, and the even bigger list of figures from the Mayfield Company, or the Nology Vehicle Specifications (pdf)). You can also determine your drag coefficient through coastdown testing. 



this is a function of the speed squared and the frontal area. Your drag goes up exponentially. Meaning that if you drive a very aerodynamic shaped vehicle the drag may be about .7 Hp at 25 MPH. If you drive the vehicle at 50 MPH the drag is increased to a little under 3 Hp. Most older cars have a drag coefficient less that the more aerodynamic cars of today so it would generally mean that you can figure a drag of 1 Hp at 25 MPH and something over 4 Hp at speeds over 50 MPH. 
Note: except for the newer bus the frontal is flat square so has the most Drag.

Note these converted to Hp are additive in the top level formula I posted


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## tomofreno (Mar 3, 2009)

bjfreeman said:


> does your calculation for Rolling Resistance ?


 If you are asking me if my calculation included rolling resistance the answer is no. Only the additional force required for a grade, no drag force either.


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## major (Apr 4, 2008)

bjfreeman said:


> and I gave a link to wiki that shows how us uses it and how europe shows it and how they are both the same *but not as %*


You tell us here that grade is not expressed as percent (%), but in the post quoted below you use percent to describe grade. What gives bj 


bjfreeman said:


> the one factor not talked about is the extra weight that would effect the need for more power to pull the trailer, especially up a *15% grade* (mountain passes)


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## bjfreeman (Dec 7, 2011)

major said:


> You tell us here that grade is not expressed as percent (%), but in the post quoted below you use percent to describe grade. What gives bj


the European use a ratio. USA use %. It was explained in the link I gave.


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## major (Apr 4, 2008)

bjfreeman said:


> the European use a ratio. USA use %. It was explained in the link I gave.


So that is contrary to your statement.


bjfreeman said:


> ....and I gave a link to wiki that shows how *us* uses it and how *europe* shows it and how they *are both the same* but not as %


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## bjfreeman (Dec 7, 2011)

major said:


> So that is contrary to your statement.


I consider that you way of looking at things.
I see no difference.

now if you had said in your view that is your opinion then I would agree.


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## major (Apr 4, 2008)

bjfreeman said:


> I consider that you way of looking at things.
> I see no difference.
> 
> now if you had said in your view that is your opinion then I would agree.


bj,

First you said Europe and US "are both the same" but "not as %".

Then you say Europe uses "ratio" and US uses "%".

I say those two statements are a contradiction. First they "are" then they "aren't". So you can call it my opinion; I call it reading what you wrote.

major


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## Brute Force (Aug 28, 2010)

BJ,

Horsepower required to push a given aerodynamic shape is proportional to the cube of the speed. So if it takes 1 HP to push a vehicle 25 MPH, it will take 8 HP to push it 50 MPH.


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## bjfreeman (Dec 7, 2011)

major said:


> bj,
> 
> First you said Europe and US "are both the same" but "not as %".
> 
> ...


but you did not read where I corrected my self, again you selective reading.


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## bjfreeman (Dec 7, 2011)

Brute Force said:


> BJ,
> 
> Horsepower required to push a given aerodynamic shape is proportional to the cube of the speed. So if it takes 1 HP to push a vehicle 25 MPH, it will take 8 HP to push it 50 MPH.


thanks, will update my data sheet.


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## Zak650 (Sep 20, 2008)

Really if it takes 2hp to drive something 25mph then doubling the speed to 50mph cubes the power required to 8hp. 

1hp in the previous post is a bad number to use since 1 cubed is still 1


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## Ziggythewiz (May 16, 2010)

Really it's only cubed if using an equation with a cube in it. As you just demonstrated, if the value is 1, it will not increase, if it is less than one it will decrease. You can't make a rule that is based on a formula without including the variables and units from that formula.


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## tomofreno (Mar 3, 2009)

Zak650 said:


> Really if it takes 2hp to drive something 25mph then doubling the speed to 50mph cubes the power required to 8hp.
> 
> 1hp in the previous post is a bad number to use since 1 cubed is still 1


 The drag force goes as vehicle speed squared. The product of this force and vehicle speed is the power or rate of work done against the drag force, and goes as the SPEED cubed. If this were the only force on the car then the power required when vehicle speed is doubled from some speed v would be (2v)^3, or 8 times greater than that at v, whether v is 1 mph or some other speed. 

The rolling resistance force is larger than the drag force up to about 45 mph for most cars, so the above relation does not hold over most of the speeds typically driven (Edit: meaning the speed will increase by less than a factor of 8)


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## Brute Force (Aug 28, 2010)

I don't make the rules, I just follow them. 

For the particular theoretical vehicle in the above example:

Horsepower Required = Proportionality Constant x Speed ^3

Where the Proportionality Constant is .000064 HP per MPH ^ 3
and Speed is in MPH

So:

Horsepower Required = .000064 x 25 x 25 x 25 = 1

And

Horsepower Required = .000064 x 50 x 50 x 50 = 8

So the example cited holds true. (Note: I did leave the units out of the equations for the sake of brevity)

If you want a more rigorous treatment of the subject, I'll happily post a full derivation from first principles with all the proper dimensional analysis.


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