Going up steep hills- downshift or floor it?

[redpoint5]

Another thing I have pondered with regards to hill climbs and efficiency is speed. Consider that it takes energy just to maintain position on a hill. You could burn fuel just sitting in one place at a standstill. Let's say it takes 1hp to just sit there and not move. If you double the horsepower by giving more throttle to the engine, then you are spending 1hp to merely maintain your position on the hill, and 1hp is spent actually climbing it. That would be a very slight throttle opening, and half of the power would be wasted to just maintaining hill position. If you increased throttle opening even more and now were developing 10hp, 9 of those would be spent climbing the hill, and only 1 maintaining. In this example, increasing power from 2hp to 10hp increased hill climbing efficiency from 50% to 90%.

Since just spending time on a hill without climbing it consumes fuel, it follows that getting to the top of the hill quickly reduces the amount of fuel wasted just maintaining position. Getting to the top in 1min instead of 2min reduces this wasted "maintenance" fuel expenditure by half.

The only factor that makes it more efficient to go slower, and this is significant, is wind resistance. There is some sweet spot of speed where getting to the top quickly is balanced by the exponential forces of wind resistance. Finding this ideal speed for various hills seems a daunting task for this limited mind to figure out.

[jtbo]

Originally Posted by redpoint5

Another thing I have pondered with regards to hill climbs and efficiency is speed. Consider that it takes energy just to maintain position on a hill. You could burn fuel just sitting in one place at a standstill. Let's say it takes 1hp to just sit there and not move. If you double the horsepower by giving more throttle to the engine, then you are spending 1hp to merely maintain your position on the hill, and 1hp is spent actually climbing it. That would be a very slight throttle opening, and half of the power would be wasted to just maintaining hill position. If you increased throttle opening even more and now were developing 10hp, 9 of those would be spent climbing the hill, and only 1 maintaining. In this example, increasing power from 2hp to 10hp increased hill climbing efficiency from 50% to 90%.

Since just spending time on a hill without climbing it consumes fuel, it follows that getting to the top of the hill quickly reduces the amount of fuel wasted just maintaining position. Getting to the top in 1min instead of 2min reduces this wasted "maintenance" fuel expenditure by half.

The only factor that makes it more efficient to go slower, and this is significant, is wind resistance. There is some sweet spot of speed where getting to the top quickly is balanced by the exponential forces of wind resistance. Finding this ideal speed for various hills seems a daunting task for this limited mind to figure out.

For some hills different tricks are needed than for other hills, letting speed go down is good for some hills, but one need to let speed go enough down that next downhill will again increase speed with neutral coasting so that next uphill is started again with enough speed, works well when there are continuously hills, but if there is just one hill it can be different story.

I wonder if hill would be endless and I will use only tiny bit more power than driving on flat ground, it would mean that my car would stop and would not even stay on hill, but would start going backwards, how much power one would need to stay at hill?

Even with lowest gear at steep hill, one needs to ask more power from engine so I doubt that it is just 1hp for staying on hill, even 10hp might be bit too little? But hp is bit difficult here as it is torque from wheels multiplied by gear ratios that would be holding car on place at steep hill, bit complicated for my tired brain at this moment, but it is quite a lot of force that 1000kg or more weighting vehicle is making with gravity at 5% slope, I think it is possible to calculate how much power is actually needed.

[redpoint5]

Originally Posted by jtbo

... works well when there are continuously hills, but if there is just one hill it can be different story.

... how much power one would need to stay at hill?

Even with lowest gear at steep hill, one needs to ask more power from engine so I doubt that it is just 1hp for staying on hill, even 10hp might be bit too little? ...but it is quite a lot of force that 1000kg or more weighting vehicle is making with gravity at 5% slope, I think it is possible to calculate how much power is actually needed.

For the purposes of my example, imagine a very long and continuous hill. I'm used to driving up mountains around here, and it's not uncommon to travel many miles of continuous 5% grade. I can engine-off coast for about 10 miles when I'm descending Mt. Hood!

The 1hp example used above was an arbitrary number I choose for ease of math. Of course, the amount of power required depends on the weight of the vehicle and the angle of the slope. Calculating these requirements is possible, but difficult considering the varying slopes and weights of vehicles. I'd be curious to know the power requirement to hold a 3000lb vehicle on a 5% slope, as a baseline.

My Subaru Legacy averaged 27mpg, but on days that I traveled to the mountains, I could achieve 33mpg. I would drive in the highest gear possible while giving about 85% throttle.

[jtbo]

Originally Posted by redpoint5

For the purposes of my example, imagine a very long and continuous hill. I'm used to driving up mountains around here, and it's not uncommon to travel many miles of continuous 5% grade. I can engine-off coast for about 10 miles when I'm descending Mt. Hood!

The 1hp example used above was an arbitrary number I choose for ease of math. Of course, the amount of power required depends on the weight of the vehicle and the angle of the slope. Calculating these requirements is possible, but difficult considering the varying slopes and weights of vehicles. I'd be curious to know the power requirement to hold a 3000lb vehicle on a 5% slope, as a baseline.

My Subaru Legacy averaged 27mpg, but on days that I traveled to the mountains, I could achieve 33mpg. I would drive in the highest gear possible while giving about 85% throttle.

Force required to stay at hill is certainly interesting, I found this thread, which might be helpful for math part:
Work to push a sled up a hill: Friction, inclines,force,work

It is about sledge pushing up to slope, but from my understanding it is same for car, only base parameters need to be changed?
Perhaps members with stronger math skills can help with that?

From my understanding it would just need to put to excel or another spreadsheet program and input slope, weight and rolling resistance that should include any losses of drivetrain to be able to make fits for all calculator that outputs needed power for staying on hills, there one can even workout forces and thus power needed for different approaches of hill climbing.

I posted altitude profile of my shopping route earlier, it has highest climb of 7%, luckily not many miles, but there is good long climbing too, I'm getting to 1200rpm if I keep top gear and something like 80-90% throttle pedal down and there I need to shift down already, even turbodiesel has good torque and at low rpm it is not enough for these hills.

Because of no throttle butterfly I get better efficiency at low throttle than your gasoline Subaru, which make it possible to use low throttle and let speed decrease with top gear and use lower gear then to climb at slow speed low throttle. Good part of diesels is that you are having intake open to outside all the time, which helps greatly with low throttle efficiency.

However I'm not using really high rpm, just around 1500 with 4th gear, on steeper hills, speed can go below 60kph, so I save from air resistance enough even I use bit more diesel with lower gear.

But I'm not sure if it still is the best method, even I'm getting same or perhaps bit lower overall consumption at these hills than at level ground, maybe getting air resistance to enough low level might help with climbing at bigger speed and top gear, but this around 1300kg weight is bit of burden at longer and steeper hills. Easiest way to help with that is to get rid of driver's extra weight

[kingsway]

Could it be as simple as using a Scanguage/Ultraguage and going up the hill using various speeds and gears and identifying a sweet-spot that is a good compromise of rate of consumption and time elapsed?

[jtbo]

Originally Posted by kingsway

Could it be as simple as using a Scanguage/Ultraguage and going up the hill using various speeds and gears and identifying a sweet-spot that is a good compromise of rate of consumption and time elapsed?

If you have scan gauge, that should be able to tell it, also consumption over distance would work.



But if no Scan gauge, one have to figure it out with math or something.

Now I did read somewhere here that it is same amount of power needed to climb despite which gear you have selected.

That is not entirely so, it is true at the wheels, but lower gear has lower gear ratio and because of that engine needs to produce less power to achieve same amount of power at driven wheels.
Think about driving up a hill with bicycle on top gear, you need more power to overcome ratio. Or think about lifting something heavy, with 2:1 ratio you can lift more than with 1:1 ratio.

Then one must apply distance/time to that too also fuel used per power unit and amount of power used so one can calculate how much power is used for each second or each distance unit, then difference between gears and time and it should come clear when to use which tactic.

I think it should be possible to make a calculator that calculates it all, but it would need gear ratios and BSFC map to work. Lot of other parameters too, but it should not be impossible.

edit: Ah, it might be possible to get around of BSFC map issue, perhaps bit inacurate, but still, one just would need to estimate a bit.

Turbo diesel motor can be up to 40% effective, at part load that will drop, but not as much as with gasoline engine, I haven't found part throttle efficiency yet, but my guess is somewhere around 30%?
Gasoline engine was 26% and under 20 at part throttle?

One can use efficiency estimate to multiply consumption, it should give pointers to right direction then.

[IamIan]

Originally Posted by Tesla

only one concept can be right

I don't think it's that simple.

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If you look at the BSFC for a given engine ... there is an example posted above ... the best energy efficiency point will have areas around it ... above , bellow, left , right ... which is a result of a long list of factors.

This has two effects from the Engine's perspective:

#1> The same load at lower RPMs can sometimes be less energy efficient than the same load at higher RPMs ... than past a certain point it reverses and higher RPMs at the same load become less efficient... lower RPMs or higher RPMs either one may be good or bad depending on where you are on the BSFC chart.

#2> More load at the same RPMs when you are load bellow the best BSFC point is better for efficiency ... but more load at the same RPMs when you are load above the best BSFC point is less efficient... lower Load or Higher Load either one may be good or bad depending on where you are on the BSFC chart.

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The user's control of the current desired engine load is a function of throttle ... current vehicle external forces , being external are not under the user's control ... although they can be predicted.

The user's control of current desired engine RPM is a function of Gear Choice and current vehicle speed... current vehicle speed is a function of previous choices of both Gear + Throttle.

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From a Vehicle travel perspective.
Slower always takes less energy to travel the same distance.

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In the real world those different effects combine together.

So for example ...

If reducing your vehicle speed reduces your energy per mile needs by 10% , but it moves you ( due to gear and/or throttle ) out of the BSFC best point so that you are ~11% less engine efficient than you are a net of ~1% less or lower MPG.

If increasing your vehicle speed increases your energy per mile needs by 10% , but moves you ( due to gear or throttle ) closer / more into the best BSFC point so that you are ~11% more engine efficient than you are a net of ~1% more or higher MPG.

etc ... etc ... they combine many different ways.

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It is a % difference in current % ... so there is a ~11% difference between 30% and ~33.7% ... just like there is a ~11% difference between 25% and ~28%.

It is not just a difference in % ... so you don't need as big of a jump as 20% to 31% would be ... in order to get ~11% difference ... a 20% to 31% jump would be ~55% difference...

If my gear & throttle choices allow me to jump from 20% to ~31% Engine Efficiency , than I can either increase my MPG by ~55% at the same vehicle energy per mile needs ... or I could get the same MPG at ~55% more vehicle energy per mile needs... or somewhere between the two.

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All the above still applies equally weather you are going up hill , down hill, or on flat level ground.

[redpoint5]

Originally Posted by IamIan

From a Vehicle travel perspective.
Slower always takes less energy to travel the same distance.

Good post, but I disagree with this. If you are engine on coasting down a hill, then slower does not take less energy to travel the same distance. Also, going slower up a hill does not always take less energy. If you go 1mph up a steep hill, enormous amounts of energy will be wasted just maintaining position on the hill.

There is a balance of engine efficiency, speed, wind resistance, and rolling resistance.

[jtbo]

Also remember that it is efficiency per produced power unit and used fuel to produce that power unit, so less power units is better than more efficiency after some point.

It is same amounts of energy needed to hold position of hill despite moving speed, there is of course minimum amount of energy that can be produced, so often it is possible to use higher gear and still produce only same amount of energy only distance covered per time unit is greater.

At some point more power is needed and from there one can find speed for that certain slope and wind condition which results lowest consumption to get up on that specific hill.

So highest possible gear with least possible power produced should result lowest possible fuel consumption.

What I have seen those BSFC maps, one need to be at edges to really get big change to fuel consumption, something that is really hard to happen in reality, but 1 power unit less is already making quite bit of effect to fuel consumption.

If it is 250 grams for 1hp, then going from 10hp to 9hp will result drop of 2500g to 2250g, to get that much less by better BSFC position you should end up to 225g position, which still is possible, however change of 2 or 3hp would already cause rather large difference that would be quite hard to overcome by using most optimal spot from BSFC map.

I still think that it depends from situations, there can't be general rule because there are so many variables.

[IamIan]

Originally Posted by redpoint5

There is a balance of engine efficiency, speed, wind resistance, and rolling resistance.

This part I agree with.

A few notes:
Speed is the only user controllable variable for wind resistance, or rolling resistance... for a given vehicle.

Sense Rolling resistance increases linearly with speed it does not itself directly change the net vehicle energy needed per mile ... going 2x as fast uses 2x the power for rolling resistance but gets there in 1/2 the time ... resulting in the same energy for rolling resistance for any given distance for any given speed ... for a given vehicle weight and Crr for the tires.

So it is the exponential wind resistance effects from speed that always result in less energy for a given distance at a slower speed than a faster speed.

Example Aerodynamics:
40 MPH requires 4x the power of 20MPH ... but only covers the same distance in 1/2 the time ... resulting in 2x more energy spent ( even if it was potential energy ) to travel the same distance.
60MPH requires 9x the power of 20MPH ... but only covers the same distance in 1/3 the time ... resulting in 3x more energy spent ( even if it was potential energy ) to travel the same distance.

Faster always requires more energy to move the vehicle at that speed and slower always requires less energy to move the vehicle at that speed... no mater where you get the energy from.

Originally Posted by redpoint5

Good post, but I disagree with this. If you are engine on coasting down a hill, then slower does not take less energy to travel the same distance. Also, going slower up a hill does not always take less energy. If you go 1mph up a steep hill, enormous amounts of energy will be wasted just maintaining position on the hill.

I'm sorry I was not more clear ... there are multiple interconnected aspects ... but from a vehicle perspective , yes slower is always less energy needed for that given speed ... weather or not it is more energy efficient application for the ICE is a separate issue ... the separate issues that are related to the net result is what I was trying to describe.

Even in your example ... The slower vehicle speed up or down the hill will use less energy to travel the same distance at a slower speed than it will at a higher speed ... going down a hill is just converting the potential energy you stored when you went up the will ... even if you timed it perfectly to come to a near complete stop at the crest of the hill ... going down or up the hill does not change the exponential effects of wind resistance with speed.

In your example you had the engine on while coasting down the hill ... so it becomes a question of while coasting down that hill what is the energy efficiency of the ICE converting fuel energy to useful mechanical energy to move the vehicle? ... that effects the ICE energy efficiency , but not the energy needed for the vehicle to travel a given speed.

There is a set amount of energy needed to lift a specific mass / weight of object up a given hill / distance ... how fast you do it does not effect the amount of energy needed ... just the amount of power needed ... but if you spend 2x the power you will get done in 1/2 the time and you still end up spending the same amount of energy.

All the energy spent fighting gravity to go up the hill is stored as potential energy at 100% efficiency ... Rolling resistance up or down the hill are separate ... Aerodynamic losses up or down the hill are separate ... ICE operating efficiency up or down the hill are separate.

These separate issues each contribute to the net result ... but they each have their own specific directions / functions.

I hope I was a bit more clear this time.