It is a Two Wheel Drive rover, meaning it will have a small electric motor on each of its two traction wheels and it will have a third traction-less wheel on the back, just to balance it out.

It will be able to steer like tanks do, by changing the speed of each motor accordingly.

The simple way to look at this is by considering some basic combination of speeds :

1) The two wheels are turning in the same direction (forward) and with the same speed -> the rover will drive on a straight line, moving forward

2) The wheels are turning with opposite directions, but the same speed -> the rover will turn on itself, around the middle point between the two wheels

3) One wheel is turning, the other one is still -> The rover will rotate and the center of rotation will be the still wheel.

It is easier to do that using vector match which is quite intuitive, I am saying this because the target audience of this article includes very young makers that eventually did not study it yet... don't get scared :)

If we represent the velocity as an arrow starting from the center of the wheel and pointing to the direction in which the rover is moving, we can use some basic geometry to figure stuff out.

As you can see in the example above the right wheel is rotating roughly twice as fast as the left one, this is why the V2 arrow is about twice as long as the V1 one.

This means that in the same amount of time the right wheel will travel twice as the distance traveled by the left one.Now, if we connect the base of the arrows and extend the line, then we connect the top of the arrows and we also extend that line, we discover they eventually touch themselves in a point.

That point (the red triangle) is the center of rotation, in fact if V1 = V2 those lines will never touch which means that the center of rotation does not exist... and that is because there is no rotation at all.

If V1 = V2 the rover would drive in a straight line.

The same concept works for about any combination of velocities.

In this second example V1 and V2 have different direction and modulus (modulus just means how "long" a vector is), can we compute the position of the center of rotation?

Indeed we can, as previously said, basic geometry comes in help.

We want to find X, it is the distance between the left wheel and the center of rotation, you can easily see that we can then relate it to any point we consider the center of the rover, like the middle point between the two wheels.

The green and red triangles are similar, so we can state that :

**X : V1 = (D + X) : V2**

Note that we use (D + X) or (D - X) depending on which way the X arrow is pointing.

Resolving :

X = V1/V2 * (D + X)

X = V1/V2 * D + V1/V2 * X

X (1-V1/V2)) = V1/V2 * D

**V1/V2 * D**

**X = ------------------**

**1-V1/V2**

Now say we want to relate everything to the middle point of the axis connecting the two wheels :

**V1/V2 * D**

**X = ---------------- + D/2**

**1-V1/V2**

Let's assume D = 10 cm, V1 = -50 rpm , V2 = 100 rpm.

**.66 cm**

**The coordinate system we are using has the center in the middle point of the axis connecting the wheels and positive X facing left.**

Knowing the center of rotation is key to be able to plot maneuvers, i.e. if we need to drive around an obstacle, but it is also important to calculate the position of the robot at any time.

In fact we might try to drive the two motors with a predictable velocity, eventually equal, but for various reasons one might turn a bit faster than the other forcing our rover into a undesired shallow turn and by consequence missing the predicted trajectory.

That's not really a huge issue provided we can timely apply corrective actions, and this becomes possible if we also

**measure the real velocity**of the wheels.

A typical way to measure the rotation speed of something is to attach an encoder, a small and simple device that sends an impulse every known number of degrees of rotation.

This is typically achieved using a small plastic wheel with a number of holes in it

The U shaped thingy visible in the bottom right is an opto-coupler, on one side of the U there is an LED and on the other side a photo-transistor.

When light generated by the LED hits the transistor, current is allowed to flow through it and that can be detected as a logic 1 on the pins of a MCU.

If the plastic wheel is placed in the U, when it turns it will generate a 0 level each time there is no hole between the LED and the transistor and a 1 each time a hole is between the two.

Finally, if we know how many holes (

**N**) are in the wheel , we can easily compute how many degrees there are between each hole.

deg = 360 / N

And if we count how many times a 0-1 pulse happens on the selected pin of the MCU , we can also compute how many revolution the traction wheel performed:

rev = pulses * 360 / N

The schematics to connect the encoder to the Arduino are pretty simple :

the green triangle represents the LED, the red / white crossed circle represents the transistor.

There is a specific reason why we selected pin 2 on the Arduino, but we will dig into that detail later.

At this point we ASSUME we can activate the two electric motors and we know how we can use a "feedback loop" to measure how much they really moved.

Next step will be to see the details on how to activate the motors.

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