# How to calculate the tangential velocity?

I am trying to measure the tangential velocity from an Inertial Measurement Unit (IMU) sensor. I read from an article that they calculate the tangential velocity according to the explanation below:

$$v = R \cdot \omega \quad (1)$$

Where $$v$$ is the tangential speed, $$R$$ the radius of the curve and $$\omega$$ the angular speed.

The centripetal acceleration is given by:

$$a = R \cdot \omega^2 \quad (2)$$

Reformulating $$(2)$$ and substituting in $$(1)$$, I obtained:

$$v = \frac{a}{\omega} \quad (3)$$

Both $$a$$ and $$\omega$$ are given by the IMU. However, when $$\omega$$ tends to zero the tangential velocity tends to infinite.

I am using the tangential velocity to calculate the speed of an object:

$$v = \sqrt{v_x^2+v_y^2}$$

where $$v_x$$ is the linear velocity and $$v_y$$ is the tangential velocity.

What does it mean physically when the $$\omega$$ tends to zero? The speed tends to infinity in math, but it is not true physically.

Is it correct to calculate the tangential velocity according to the expression $$(3)$$?

Here is the link to the aforementioned article.

• Hello! It is preferable to use MathJax (LaTeX) to display formulas. You can find a tutorial at MathJax basic tutorial and quick reference. Please edit your question accordingly. Thanks! Apr 22, 2021 at 16:20
• Find $R = a/\omega^2$ for when it is spinning, and then use this value at low speeds. Apr 23, 2021 at 13:13

I can't access freely to the PDF document you cite in your question, so I cannot give a complete answer to your question. However, here are some preliminary remarks:

1. The relationships you mention in your question are limited to a uniform circular motion (on a straight line, you have $$a$$ finite and $$w = 0$$).

2. In theory, you don't need $$\omega$$, you can integrate (numerically) $$a$$ over time

$$a = dv / dt$$,

so

$$v = v_{0} + \int^{t1}_{to} a \ dt$$

But that implies monitoring the acceleration from the beginning of the trip to the finish, without interruption... (in which case $$v_{0} = 0$$).

1. An easier way to get an estimate of a vehicle's speed is to check the change of GPS coordinates at regular intervals and calculating an average speed based on the distance traveled during time intervals.

Here is a paper which proposes combining both approaches: https://www.researchgate.net/publication/260725857_Vehicle_speed_estimation_using_GPSRISS_Reduced_Inertial_Sensor_System

• Here's a freely accessible version of the article. Apr 22, 2021 at 16:55
• Thanks for the article. I will have a look and try to come up with a more complete answer. Apr 22, 2021 at 17:42
• As I stated earlier, the relationship $v=ωR$ is only valid when there is a circular motion, (i.e. when the vehicle turns): The Document you cite states the same thing: "Sensing Turns:When a vehicle makes a turn, it experiences a centripetal force, which is related to its speed..." Apr 22, 2021 at 20:23

It doesn't make sense that you have a value for the centripetal acceleration $$a$$ while $$\omega$$ tends to zero. Those two follow each other.

Physically, if you have a centripetal acceleration $$a$$ causing you to move around a curve, then you also must have a nonzero angular speed $$\omega$$ around that curve. Otherwise you wouldn't be moving around the curve and then there wouldn't be a centripetal acceleration in the first place.

Mathematically, if you do have an $$\omega$$ tending to zero, then $$a$$ will as well tend to zero. You can see that from your equation $$(2)$$.

$$\quad\omega\to 0 \quad\Rightarrow \quad a=R\omega^2\to 0$$

And then you can't conclude that the tangential speed grows to infinity anymore since not only the denominator in your equation $$(3)$$, but also the numerator simultaneously tend towards zero.

$$\omega\to 0 \quad\land \quad a\to 0 \quad \Rightarrow \quad v=\frac a \omega\to\,?$$

Notice that in equation 2, as the angular velocity goes to zero, the centripetal acceleration also goes to zero (as does the tangential velocity). Also in circular motion, the linear speed is the magnitude of the tangential velocity (unless the radius is also changing). The components, $$v_x$$ and $$v_y$$, both change as a function of time (unless you are letting the, xy, coordinate system rotate).