Hello all;
I don't remember where but I remember reading a calculation about elevation change's effect on your consumption. I think it was a Bjorn video but can't find it now. This is a very simplified and noob calculation (I'm not an engineer so this is all 'internet knowledge') but I would like to know if it more or less works as I have a 260km road trip coming up this weekend including a bit of a climb with a BEV BMW i3 towards the end. That is a pure 19kWh usable energy, pathetic but what we have.
So it goes;
'weight of the car' x 'climbing altitude' x 'gravity constant' gives you, in joules, energy required for that climb. Divide it by 3,600,000 you get kWh. Also you add to that your regular flat road consumption for the lenght of the road spent climbing. For example with my upcoming roadtrip in the BMW i3;
Car weighs 1300kg incl. me.
I will climb 1200m in 60km
Gravity is 9,8m/s
So 1300 *1200 * 9,8 = 15,28 million joules required.
Let's say the car is 85% efficient so I need 18million joules. In other words 5kWh.
Lenght of the climb is 60km. And my flat road consumption would be 150wh/km. So 9 additional kWh needed for that.
9+5 = 14kWh. + 10% error margin = 15,5kWh needed for the trip. So around 75-80% charge should be doable.
On the way back I will have 15 million joules of potential energy. If I can harvest it with regen only at 70% efficiency I get 10,5 million joules, a.k.a 3kWh. So I should gain around 15% SoC on the way back with i3's tiny battery.
Anyone want to chime in? Is the calculation making any sense to base a roadtrip on? Thoughts appreciated.
I don't remember where but I remember reading a calculation about elevation change's effect on your consumption. I think it was a Bjorn video but can't find it now. This is a very simplified and noob calculation (I'm not an engineer so this is all 'internet knowledge') but I would like to know if it more or less works as I have a 260km road trip coming up this weekend including a bit of a climb with a BEV BMW i3 towards the end. That is a pure 19kWh usable energy, pathetic but what we have.
So it goes;
'weight of the car' x 'climbing altitude' x 'gravity constant' gives you, in joules, energy required for that climb. Divide it by 3,600,000 you get kWh. Also you add to that your regular flat road consumption for the lenght of the road spent climbing. For example with my upcoming roadtrip in the BMW i3;
Car weighs 1300kg incl. me.
I will climb 1200m in 60km
Gravity is 9,8m/s
So 1300 *1200 * 9,8 = 15,28 million joules required.
Let's say the car is 85% efficient so I need 18million joules. In other words 5kWh.
Lenght of the climb is 60km. And my flat road consumption would be 150wh/km. So 9 additional kWh needed for that.
9+5 = 14kWh. + 10% error margin = 15,5kWh needed for the trip. So around 75-80% charge should be doable.
On the way back I will have 15 million joules of potential energy. If I can harvest it with regen only at 70% efficiency I get 10,5 million joules, a.k.a 3kWh. So I should gain around 15% SoC on the way back with i3's tiny battery.
Anyone want to chime in? Is the calculation making any sense to base a roadtrip on? Thoughts appreciated.