**Wheel torque** can be calculated function of **engine torque** if the parameters and status of the transmission are known. In this tutorial, we are going to calculate the wheel torque and force for a given:

- engine torque
- gear ratio (of the engaged gear)
- final drive ratio (at the differential)
- (free static) wheel radius

Also, we are going to assume that there is **no slip** in the clutch or torque converter, the engine being mechanically linked to the wheels.

This method can be applied to any powertrain architecture (front-wheel drive or rear-wheel drive) but, for an easier understanding of the components, we are going to use a **read-wheel drive (RWD) powertrain**.

As depicted in the image above, the engine is the source of torque. The gearbox is connected to the engine through the clutch (on a manual transmissions) or torque converter (on an automatic transmissions). We consider that there is absolutely no slip in the clutch (fully closed) or in the torque converter (lock-up clutch closed). In this case the engine torque *T _{e} [Nm]* is equal with the clutch/torque converter torque

*T*.

_{c}[Nm]Further, the **engine torque** is transmitted through the gearbox, where is multiplied with the gear ratio of the engaged gear *i _{x} [-]* and outputs the

**gearbox torque**

*T*.

_{g}[Nm]The propeller shaft is transmitting the torque to the rear axle, where is multiplied with the final drive gear ratio *i _{0} [-]*. This gives the

**torque at the differential**

*T*.

_{d}[Nm]If the vehicle is driven on a straight line, the torque at the differential is equally split between the left wheel *T _{lw} [Nm]* and the right wheel

*T*.

_{rw}[Nm]The sum of the left and right wheel torque gives the **torque at the differential**.

Replacing (2) in (3) in (4) gives the **mathematical expression of the wheel torque function of the engine torque**, for a given gearbox ratio *i _{x}* and a final drive ratio

*i*.

_{0}The **formula of the wheel torque** (6) applies to a vehicle which is driven on a straight line, where the left wheel torque is equal with the right wheel torque.

From mechanics (static), we know that the **torque** is the product between a **force** and its **lever arm length**. In our case, the wheel torque is applied in the wheel hub (center) and the lever arm is the wheel radius *r _{w} [m]*. For this example we assume that both wheel have the same radius

*r*.

_{w}The same principle applies to the **right wheel torque**.

Assuming that both left and right wheel torque and radius are equal, we can write a generic expression of the wheel force *F _{w} [N]*, function of wheel torque

*T*and wheel radius

_{w}[Nm]*r*.

_{w}[m]From (10) we can extract the formula of the **wheel force** function of the **wheel torque** and **wheel radius**.

Replacing (6) in (10) will give the mathematical expression of the **wheel force** function of **engine torque**, gearbox **gear ratio**, **final drive ratio** and **wheel radius**.

**Example 1**. Calculate the **wheel torque** and **force** for a vehicle with the following parameters:

- engine torque, T
_{e}= 150 Nm - gearbox (1
^{st}) gear ratio, i_{x}= 4.171 - final drive ratio, i
_{0}= 3.460 - tire size marking 225/55R17

**Step 1**. Calculate the (free static) **wheel radius** from the tire size marking. The method for calculating the wheel radius is described in the article How to calculate wheel radius. The calculated wheel radius is r_{w} = 0.33965 m.

**Step 2**. Calculate the **wheel torque** using equation (6).

**Step 3**. Calculate the **wheel force** using equation (11).

**Example 2**. For a given gearbox, with multiple gears (gear ratios), we can calculate the **wheel torque** and **force** for each gear. Let’s calculate the **wheel torque** and force for a vehicle with the following parameters:

- engine torque, T
_{e}= 150 Nm - wheel radius, r
_{w}= 0.33965 m

The gearbox is automatic (ZF6HP26), with the following gear ratios and final drive ratio.

Gear # | Gear ratio symbol | Gear ratio |

1 | i_{1} | 4.171 |

2 | i_{2} | 2.340 |

3 | i_{3} | 1.521 |

4 | i_{4} | 1.143 |

5 | i_{5} | 0.867 |

6 | i_{6} | 0.691 |

Final drive | i_{0} | 3.460 |

To speed up calculations, we can use a Scilab script.

clc // Input data Te = 150; ix = [4.171 2.340 1.521 1.143 0.867 0.691]; i0 = 3.460; rw = 0.33965; // Wheel torque and force calculation Tw = (ix .* i0 .* Te)/2; Fw = Tw ./ rw; // Display results mprintf("\n%s\t\t%s\t\t%s\t\t%s\n","Gear","ix [-]","Tw [Nm]","Fw [N]") for i=1:length(ix) mprintf("%d\t\t%.3f\t\t%.2f\t\t%.2f\n",i,ix(i),Tw(i),Fw(i)); end

Executing the above script will output the following results in the Scilab console:

Gear ix [-] Tw [Nm] Fw [N] 1 4.171 1082.37 3186.73 2 2.340 607.23 1787.81 3 1.521 394.70 1162.08 4 1.143 296.61 873.28 5 0.867 224.99 662.41 6 0.691 179.31 527.94

**Example 3**. For our third example we are going to use the full load torque curve of an engine and calculate the **wheel torque and force** (traction) in each gear. Calculate the wheel torque and force (traction) for a vehicle with the following parameters:

- engine torque, T
_{e}= 150 Nm - wheel radius, r
_{w}= 0.33965 m - the gear ratios of ZF6HP26 (see
**Example 2**)

The engine torque at full load is given by the following parameters:

N_{e} [rpm] | 800 | 1312 | 1800 | 2276 | 2800 | 3316 | 3806 | 4300 | 4770 | 5300 | 5800 | 6300 |

T_{e} [Nm] | 116 | 135 | 148 | 157 | 165 | 172 | 178 | 184 | 188 | 187 | 183 | 171 |

_{e}is

**engine speed**and T

_{e}is

**engine torque**.

Since we need to perform a lot of calculations, we’ll use a Scilab script to calculate the **wheel torque** and force curves for each gear. The results are going to be plotted in a graphical window.

clc // Input data Ne = [800 1312 1800 2276 2800 3316 3806 4300 4770 5300 5800 6300]; Te = [116 135 148 157 165 172 178 184 188 187 183 171]; ix = [4.171 2.340 1.521 1.143 0.867 0.691]; i0 = 3.460; rw = 0.33965; // Plot engine torque figure(1) hf = gcf(); hf.background = 8; plot(Ne,Te,"LineWidth",2) xgrid() xlabel("Engine speed [rpm]","FontSize",3) ylabel("Engine torque [Nm]","FontSize",3) title("x-engineer.org","Color","blue","FontSize",2) // Calculate wheel torque and force for i = 1:length(ix) for j = 1:length(Te) Tw(i,j) = (ix(i) .* i0 .* Te(j))/2; Fw(i,j) = Tw(i,j) ./ rw; end end // Plot wheel torque and force figure(2) hf = gcf(); hf.background = 8; plot(Ne,Tw,"LineWidth",2) xgrid() xlabel("Engine speed [rpm]","FontSize",3) ylabel("Wheel torque [Nm]","FontSize",3) title("x-engineer.org","Color","blue","FontSize",2) legend("1st gear","2nd gear","3rd gear","4th gear","5th gear","6th gear",2) figure(3) hf = gcf(); hf.background = 8; plot(Ne,Fw,"LineWidth",2) xgrid() xlabel("Engine speed [rpm]","FontSize",3) ylabel("Wheel force [N]","FontSize",3) title("x-engineer.org","Color","blue","FontSize",2) legend("1st gear","2nd gear","3rd gear","4th gear","5th gear","6th gear",2)

Executing the script will output the following graphical windows.

The same method can be applied for an **electric vehicle**, the engine torque being replaced by the **motor torque**.

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## saalai thenagan R

what if Fwl is not equal to Fwr?

## Anthony Stark

Then it’s a bit more complicated. If the differential is open (without slip control) then both wheels will have the minimum force of the two. For example, if left wheel contact force is 1000 N, the same amount will be available to the right wheel.