Does Driving Uphill at Constant Speed Use More Fuel Than on Flat Terrain?

[genjutsu]

Hi everyone, so one day I came upon a very confusing and complicated question. I was pondering about stuff and then while watching a car it just hit me and puzzled me. So basically what I'm confused about is that do you use the same amount of gas while going at constant 100mph vs constant 10mph going up a hill. In real life for me, I find myself feeling that I have to push down more on my gas pedal for 100mph than 10mph. However, I did some calculation trying to see if it would help.

Values:
mass=4.37kg
gravity=9.8m/s^2
coefficient between rubber (wheel) and dry concrete (road) =0.6 (source: http://hypertextbook.com/facts/2006/MatthewMichaels.shtml)
Angle of hill=30 degrees
Assuming air resistance and any other resistance to be the same

So going up the hill at a 30 degree angle would have friction and gravity pulling u down. I calculated that Force of gravity is 35,561N and Force of friction is 9,239N. Combined total of 42,801N force that you will need to match in order for your car to keep going at a net 0 force. When I got this it still wasn't really clear. If I was going at a constant speed 100mph then it would be 0 acceleration so that would mean the car is not using any more force than going at constant 10mph. So would that be true that if I go constant 140mph to get to my destination I would burn the same amount of gas as going constant 10mph? If that's true then wouldn't I reach my destination 14 times faster and still burn the same amount of gas?

But how come I always feel that I'm using more gas when i go uphill? Is it just my imagination?

Thank you for everyone who can help. Appreciate it.

 

[Integral]

...
Assuming air resistance and any other resistance to be the same

...

NO.
Air resistance is strongly dependent on speed.

 

[Cleonis]

But how come I always feel that I'm using more gas when i go uphill? Is it just my imagination?

Think of it this way:
Imagine a very shallow hill, and you are coasting (freewheeling) downhill. If the downhill slope would be very steep you would go faster and faster all the time, but let's say this descent is so gradual that you actually don't speed up.

What is happening is that on the downhill slope gravity is helping you, maintaining your velocity. Without the downhill slope friction would drain kinetic energy, and you come to a halt.

Conversely, going uphill it's the other way round. Going uphill gravity is joining up with friction to slow you down.
If you're going 60 mph on a level road, and when starting a hill you want to sustain that 60mph then your engine needs to put out more power.

On a level road gravity is affecting you only indirectly, the rolling resistance of the tires comes from the weight of the car deforming the tires, but other than that there is no effect from gravity on a level road. Uphill or downhill gravity does become a factor. To calculate you decompose gravity into a component perpendicular to the local surface and a component parallel to the local surface, and the parallel component is affecting your velocity.


So the following reasoning doesn't add up: 'When I sustain 60 mph going from a level road to a climbing road I don't accelerate therefore it should not require more power from the engine.'

There's something counter-intuitive there.
To focus on the counter-intuitive element, think of an elevator. Let's say that elevator is in an amusement park, and it's a scary ride, capable of free fall.
During the phase of free fall an accelerometer you are carrying with you will read a G-count of zero G.

- During deceleration at the end of the ride the elevator will have to pull more than 1 G, in order to come to a stop.
- During an ascent with constant velocity an onboard accelerometer will read 1 G
- During a descent with constant velocity an onboard accelerometer will read 1 G

However, for the elevator engines ascent and descent are very different of course. Let's say the elevator has a full passenger load.
Going up or going down the tension in the elevator cables is the same, but going up the engines have to put out power, and on the descent you can reverse the power flow. Some types of electromotor can also serve as a generator when mechanically driven. Going down the engines can be switched to regenerative braking.