A Theory Into Why It Should Be Cheaper Driving North

This post has nothing to do with IT, just happened to have been a curiosity conjured during my travels up North and back down South on various IT projects.


The Earth is not a perfect sphere, it is a spheroid that bulges out at the equator – the Earth's equatorial radius is greater than the Earth's polar radius. From high-school physics we know that Potential Energy = mass x gravity x height and so it follows that we might expect the potential energy of an object on the Earth's surface (sea-level) at the equator, to be greater than the potential energy of an object on the Earth's surface closer to the poles, since we can think of sea-level at the equator as being higher (further away from the Earth's core/center of mass) than sea-level close to the poles.

Application of the Hypothesis

If I travel from London to Glasgow achieving an MPG of 50 (Diesel), by how much would I expect the MPG to be affected on the drive back from Glasgow to London (because of the need to burn more fuel to acquire the additional potential energy)?

This application is based on a complete fantasy scenario where there are no traffic problems, the road is upon a perfectly flat spheroidal Earth (it could be argued that even with undulations in the carriageway, would still need to acquire more potential energy on the drive to London,) and I travel from sea-level in London to sea-level in Glasgow, and is really more of a mathematical exercise that attempts to calculate if there would be any noticeable difference. Apologies in advance for any flaws in the calculations!

The Mathematics
An old copy of Maple 7 was used for the calculations, and some of the lines below in red represent the Maple Execution Group Inputs with some formulas in blue.


Latitudes in degrees North:

The Earth's equatorial radius a and polar radius b in metres:

Mass of the automobile in kg:

Calorific value of diesel in J/kg:

Density of petroleum diesel in kg/l:

Litres in a UK gallon:

Distance London to Glasgow in miles:


Radians as a function of Degrees:
Earth's gravity (in ms-2) as a function of Radians:
Radius (in metres) at a given geodetic Latitude as a function of Radians (or distance from the Earth's center to a point on the spheroid surface):
f:=phi->sqrt( ( (a^2*cos(phi))^2 + (b^2*sin(phi))^2 ) / ( (a*cos(phi))^2 + (b*sin(phi))^2 ) );
PotentialEnergy in Joules with mass (in kg) gravity (in ms-2) and height (in m):

The Calculations:

GlasgowLatitudeRadians = 0.9751154533
LondonLatitudeRadians = 0.8991430162
GlasgowGravity = 9.818471842 ms-2
LondonGravity = 9.816854446 ms-2
*Notice that the gravity in Glasgow worked out as very slightly stronger!
GlasgowRadius = 6363522.841 m
LondonRadius = 6365075.641 m
And Potential Energy for the 1000kg automobile:
GlasgowPotentialEnergy = 62480069830 J
LondonPotentialEnergy = 62485021110 J
And the potential energy difference for LondonPE minus GlasgowPE:
PEDifference = 62485021110 - 62480069830 = 4951280 J
Kilos of diesel required:
KilosOfDiesel = 4951280 / 45300000 = 0.1092997793
Litres of diesel required:
LitresOfDiesel = KilosOfDiesel / 0.832 = 0.1313699271
Gallons of diesel required:
GallonsOfDiesel = LitresOfDiesel / 4.54609188 = 0.02889733216

The Result:

A journey from London to Glasgow of 405.1 miles at 50 MPG uses:
GallonsToGlasgow = 405.1/50 = 8.102
To get back to London requires an additional 0.02889733216 gallons of diesel making the MPG:
MPGtoLondon = 405.1/(8.102+0.02889733216)
= 49.82229924


The difference would be barely noticeable!