Quite a lot of momentum is lost when my trains go uphill. However, when first launching OpenTTD, the initial screen shows trains that lose no speed while ascending. Aside from toggling some setting to make the "physics" unrealistic, is there something I can be doing to my trains to minimize the amount of impact they suffer from going up an incline?
If you really need to climb the hill (and can't go through or around it) then the ways to get a train to maintain speed as much as possible up-hill are:
All these assume you have realistic train acceleration enabled (and you really should - the original acceleration model is really slow on all hills, where realistic attempts to simulate momentum).
Also, if you want to tweak things a little bit in to your favour, you can change the calculated steepness of slopes if need be.
First, you could make a tunnel through the hill if really need it to be fast, and the faster that it's going when it hits the hill, the faster that it will get over the hill, and the less that it will slow down.
The most direct answer is that you should buy trains with higher horsepower so that they can power through it better, and the fewer cars that are hooked up the less that it has to pull. If you're having trouble finding powerful enough trains, try electric rails, those have a lot higher power than most diesel or steam trains.
Another way to make trains climb a hill faster is by setting a lower weights for. This can be done in the advanced settings.
If you use
Base resistance of rail is 35N per metric ton. Add 100N for each % of grade. So a 7-tile train with 7% train incline setting would encounter 135N per metric ton of weight if it had one incline tile and 6 flat tiles under it. Multiply by the speed in m/s to get the amount of power needed.
Note that trains that are very small thus encounter steeper grades. It can sometimes also be useful to have one or two empty cars behind a very heavy locomotive to distribute the weight of the train more evenly. If most of the weight of a train is on a slope, the 'effective slope' the whole train encounters is higher. The formula for effective slope is
Where Si here is the slope of the i-th train car, and wi is the weight of the i-th train car.
Let's do an example. We have a 'Willis 2-8-0' and want to know how many cars of coal it can haul up a 1% incline at a speed of 80 kph (or 50 mph, 22+2/9 m/s). In order to make the math somewhat manageable we gloss over the complexity of the diffrent weights of the cars and assume a 'constant' slope instead of discrete tiles. In reality the speed will wobble up and down a bit around 50 mph as the lighter respectively heavier parts of the train are on an incline tile.
Its power is 820kW, which corresponds to 820 / (22+2/9) = 36.9kN of TE at the indicated speed. (Note, if a locomotive's max TE is lower than the figure computed, you use that figure instead). 36900 / 135 = 273.3t. Subtract the weight of the locomotive itself (145t) to be left with 118t of pulling power, which is about 3 cars worth of coal (rounded up from 2.6736).
In a typical game scenario on a typical map, hills are not very tall (often not more than a few tiles). With
Be very careful about exceeding the TE limit of a train when you can't guarantee a running start. If a train ahead of your train has to slow down, your train may have to stop on an incline and accellerate from zero on that same incline later. If it does not have the TE to pull itself up, it will climb very slowly at the minimum speed, which is less than 1 kph. The same applies to a train that has an unfortunate breakdown on an incline. In our example the 'willis' has a max. TE of 430kN, enough for this not to be an issue. It may be a slow start, but it will get itself up 1% hills with up to 3185t of weight.
You can find the 'max TE speed' by dividing 'Power' by 'TE'. This is the highest speed a locomotive achieves maximum pull power at. For the willis it's about 6.86 km/h, for example.
Note that at very high speeds (above around 150 km/h) these formulas don't apply anymore because the 'air resistance' term becomes significant. You would need to correct for quadratic air resistance. For the exact formulas, you can visit the openTTD wiki.
The extra force needed for air drag is given by:
Here v is your speed in m/s, n is the length of the train, and the force is given in N. (to use kN like the rest of the post divide it by 1,000 afterwards). Usually for freight trains this is a very small number compared to rolling resistance. For fast passenger trains it's worth considering.