Tracked-type tractors and bulldozers are very slow engineering vehicles. Their top speed is less than 15 km/h and when pushing it is even lower (< 5 km/h). For example: the Caterpillar D9 top speed in forward gear is 11.0 km/h while in reverse 13.5 km/h, while in 1st gear it is 3.5 km/h.

From elementary physics,

$$ P = \frac{dW}{dt} = \frac{d}{dt} \int_{\Delta t} \mathbf{F} \cdot \mathbf {v} \, dt = \mathbf{F} \cdot \mathbf {v}, $$

which means that for a given output $F = P/v$, that is: to increase the force one pays with decreasing speed.

I know that a vehicle drag force or pushing force is expressed in terms of torque (in units such as newton$\times$meters) but I don't know how to relate that to the previous formula other than putting the torque $$\tau = |\mathbf{r} \times \mathbf{F}| = r \cdot F = r \cdot P/v$$ which says that in order to increase torque one needs to decrease speed.

However, I wonder why can't we engineer an engine with a planetary shift that has both low gears for the maximal force at low speed while having also high gears for a reasonable speed (> 30 km/h) for better mobility.

Tracked-type tractors and bulldozers are very slow engineering vehicles. Their top speed is less than 15 km/h and when pushing it is even lower (< 5 km/h). For example: the Caterpillar D9's top speed in forward gear is 11.0 km/h while in reverse 13.5 km/h, while in 1st gear it is 3.5 km/h.

From elementary physics,

$$ P = \frac{dW}{dt} = \frac{d}{dt} \int_{\Delta t} \mathbf{F} \cdot \mathbf {v} \, dt = \mathbf{F} \cdot \mathbf {v}, $$

which means that for a given output $F = P/v$, that is: to increase the force one pays with decreasing speed.

I know that a vehicle drag force or pushing force is expressed in terms of torque (in units such as newton$\times$meters), but I don't know how to relate that to the previous formula other than putting the torque $$\tau = |\mathbf{r} \times \mathbf{F}| = r \cdot F = r \cdot P/v$$ which says that in order to increase torque one needs to decrease speed.

However, why can't we engineer an engine with a planetary shift that has both low gears for the maximal force at low speed while having also high gears for a reasonable speed (> 30 km/h) for better mobility?

Tracked-type tractors and bulldozers are very slow engineering vehicles. Their top speed is less than 15 km/h and when pushing it is even lower (< 5 km/h). For example: the Caterpillar D9 top speed in forward gear is 11.0 km/h while in reverse 13.5 km/h, while in 1st gear it is 3.5 km/h.

From elementary physics,

$$ P = \frac{dW}{dt} = \frac{d}{dt} \int_{\Delta t} \mathbf{F} \cdot \mathbf {v} \, dt = \mathbf{F} \cdot \mathbf {v}, $$

which means that for a given output $F = P/v$, that is: to increase the force one pays with decreasing speed.

I know that a vehicle drag force or pushing force is expressed in terms of torque (in units such as Newton$\times$Meters) but I don't know how to relate that to the previous formula other than putting the torque $$\tau = |\mathbf{r} \times \mathbf{F}| = r \cdot F = r \cdot P/v$$ which says that in order to increase torque one needs to decrease speed.

However, I wonder why can't we engineer an engine with a planetary shift that has both low gears for the maximal force at low speed while having also high gears for a reasonable speed (> 30 km/h) for better mobility.

Tracked-type tractors and bulldozers are very slow engineering vehicles. Their top speed is less than 15 km/h and when pushing it is even lower (< 5 km/h). For example: the Caterpillar D9 top speed in forward gear is 11.0 km/h while in reverse 13.5 km/h, while in 1st gear it is 3.5 km/h.

From elementary physics,

$$ P = \frac{dW}{dt} = \frac{d}{dt} \int_{\Delta t} \mathbf{F} \cdot \mathbf {v} \, dt = \mathbf{F} \cdot \mathbf {v}, $$

which means that for a given output $F = P/v$, that is: to increase the force one pays with decreasing speed.

I know that a vehicle drag force or pushing force is expressed in terms of torque (in units such as newton$\times$meters) but I don't know how to relate that to the previous formula other than putting the torque $$\tau = |\mathbf{r} \times \mathbf{F}| = r \cdot F = r \cdot P/v$$ which says that in order to increase torque one needs to decrease speed.

However, I wonder why can't we engineer an engine with a planetary shift that has both low gears for the maximal force at low speed while having also high gears for a reasonable speed (> 30 km/h) for better mobility.

Tracked-type tractors and bulldozers are very slow engineering vehicles. Their top speed is less than 15 km/h and when pushing it is even lower (< 5 km/h). For example: the Caterpillar D9 top speed in forward gear is 11.0 km/h while in reverse 13.5 km/h, while in 1st gear it is 3.5 km/h.

From elementary physics,

$$ P = \frac{dW}{dt} = \frac{d}{dt} \int_{\Delta t} \mathbf{F} \cdot \mathbf {v} \, dt = \mathbf{F} \cdot \mathbf {v}, $$

which means that for a given output $F = P/v$, that is: to increase the force one pays with decreasing speed.

I know that a vehicle drag force or pushing force is expressed in terms of torque (in units such as Newton$\times$Meters) but I don't know how to relate that to the previous formula other than putting the torque $$\tau = |\mathbf{F} \times \mathbf{r}| = F \cdot r = r \cdot P/v$$ which says that in order to increase torque one needs to decrease speed.

However, I wonder why can't we engineer an engine with a planetary shift that has both low gears for the maximal force at low speed while having also high gears for a reasonable speed (> 30 km/h) for better mobility.

Tracked-type tractors and bulldozers are very slow engineering vehicles. Their top speed is less than 15 km/h and when pushing it is even lower (< 5 km/h). For example: the Caterpillar D9 top speed in forward gear is 11.0 km/h while in reverse 13.5 km/h, while in 1st gear it is 3.5 km/h.

From elementary physics,

$$ P = \frac{dW}{dt} = \frac{d}{dt} \int_{\Delta t} \mathbf{F} \cdot \mathbf {v} \, dt = \mathbf{F} \cdot \mathbf {v}, $$

which means that for a given output $F = P/v$, that is: to increase the force one pays with decreasing speed.

I know that a vehicle drag force or pushing force is expressed in terms of torque (in units such as Newton$\times$Meters) but I don't know how to relate that to the previous formula other than putting the torque $$\tau = |\mathbf{r} \times \mathbf{F}| = r \cdot F = r \cdot P/v$$ which says that in order to increase torque one needs to decrease speed.

However, I wonder why can't we engineer an engine with a planetary shift that has both low gears for the maximal force at low speed while having also high gears for a reasonable speed (> 30 km/h) for better mobility.