The space diagram, velocity vector diagram and the acceleration vector diagram are the important tools for analyzing a mechanism.

With the help of the following example, i will explain the steps required to draw a typical velocity vector diagram.

**Example**:

The input data for the above four bar link mechanism is as below:

OA=50 mm

AB=150 mm

X=50 mm

Theta = 45 Degree

Speed of the link OA, Noa=5000 rpm

Upon completion of the velocity vector diagram, we will be able to analyze the four bar link to find out the absolute velocity of the **piston** **B**** **as well as the relative velocity of the piston **B** with respect to the joint **A.**

**Required**** ****steps**** ****to**** ****produce**** ****the**** ****velocity**** ****diagram:**

**Step-1:**** ****Draw**** ****the**** ****space**** ****diagram:**** **The space diagram is the “scaled” representation of the mechanism at the desired orientation of the links. We will use **0.1** **scale**** ****factor**** **here**.**** **So, the link OA will be represented by a **5**** ****mm** line starting from the point **O** at an angle of **45**** ****degree**** (**because theta=45 degree in our case) with horizontal.

At the end of the first step, you should get a line like **OA**** **(as shown above).

**Step-2:**** ****Space**** ****diagram**** continued****:** Next draw the link AB starting from the point A of the already drawn line OA in such a way that the **length**** ****of**** ****AB=**** ****15**** ****mm**** **(remember we took a scale factor of 0.1mm). Also ensure that the distance between the two horizontal lines passing from the points **O and B respectively is 5 mm (because X=50 mm and scale factor=0.1). **

At the end of the **step-2,**** **you will get a space diagram consisting of the two lines OA and AB as shown above.

**Step-3:**** ****calculating**** ****linear**** ****velocity**** ****of**** ****joint**** ****A:**** **

We know the speed of the link OA **=5000**** ****rpm=(5000*2*3.14/60)**** ****rad/s**** ****=523.33**** ****rad/s**** **

We also know the length of OA=**50**** ****mm.**

Further, to convert the angular velocity to the linear velocity, the following equation needs to be used:

**Linear**** ****velocity**** ****=**** ****Angular**** ****velocity**** ******* **Length**** ****of**** ****the**** ****link**

So, linear velocity of joint A with respect to O, **Voa** =**523.33*50**** ****mm/s**** ****=**** ****26166.67**** ****mm/s**

**Step-4:**** ****Start**** ****drawing**** ****the**** ****velocity**** ****vector**** ****diagram:**** **The linear velocity of the joint A will be perpendicular to the link OA and will start from the point O of the space diagram we have already drawn.

We will use a **scale**** ****factor**** ****of**** ****0.001** for drawing the velocity vector. Hence, a line of length **26.166**** ****mm** and perpendicular to the line OA of space diagram will represent the velocity vector **Voa**** ****of**** ****value**** ****26166.67**** ****mm/s.**

After finishing the Step-4, you should get the velocity vector O’A’ as shown above. Please observe the space diagram in the black colored line and velocity vector diagram in magenta colored line.

**Step-5:**** ****Velocity**** ****vector**** ****diagram**** continued****:**** **The next thing we have to draw is the velocity vector

for the link **AB. **

We know two things about the velocity vector (we will represent it by **A’B’)**, one, it will start from the point **A’** of the velocity vector diagram and it will be perpendicular to the orientation of the link **AB **of the space diagram.

The thing we don’t know about the **A’B’ **is its value (or the linear velocity of the joint **B** with respect to the joint **A). **

So, for the time being we will just start the velocity vector **A’B’** from the point **A’** and to the direction perpendicular to the link **AB** and the velocity vector diagram at this stage will look like below:

**Step-6 : Velocity vector diagram continued: **Now we know the direction and the starting point of the link **A’B’ **, we have to find the other end of the vector. To find out the location of the other end, we will use the property of the joint **B ** that it will always have the motion only in horizontal direction and will pass through the point **O’**(*because the joint ***B*** is connected to the piston and the piston always oscillate in horizontal direction, and if you are wondering why the horizontal line should pass through the point ***O’*** then the reason is by joining the horizontal line with the point ***O’ ***will give the velocity of the joint ***B ***with respect to the joint ** O*). Now we have end up with the final velocity vector diagram of the mechanism:

**Step-7: Calculate the unknown velocity values: **Now if we measure the length of the line **A’B’ **and factored the value with the scale factor (*0.001 in our case*) , we will get the linear velocity of the joint **B ** with respect to the joint **A.**

We are getting the length of the line **A’B’** as **28.5878 mm. **So for getting the velocity value of the joint **B ** with respect to the joint **A **we have to divide it by the scale factor 0.001 and viola*!* We got the velocity value as **28587.8 mm/s**.

Similarly we got the velocity of the piston (joint **B)** of the four bar linkage mechanism with respect to ground (or the joint **O) **as **31.4587/0.001 = 31458.7 mm/s.**

easy expain thanks mech guru.

pls how do i determine the sense of an angular velocity,from the velocity diagram whether clockwise or anti clock

wise

I’m so glad for having understand this chapter in a fraction of second… thanks a lot

I’m so glad for having understand this chapter in a fraction of second… thanks a lot