# No. Of Instantaneous centres

I’m reading “Theory of Machines and Mechanisms” by Joseph Shigley.

Page No. 118 - The author states that A mechanism has as many instant centres as there are ways of pairing the link numbers. Thus, for an n-link Mechanism there are N = n(n-1)/2 instantaneous centres.

Now consider the following image having 3 links (one of them being the fixed reference frame), two of them being connected by a revolute joint.

The way I see it, there should be 2 instantaneous centres. One, point A itself about which each link can rotate. Second, some instantaneous centre in the plane about which this linkage system, as a whole, describes rotational motion at any given instant. But by the formula the no. Should be 3.

So, where am I going wrong ?

• Could you upload a somewhat clearer picture of the mechanism you consider? What part exactly is the third link, what does the dashed circle do? What is the point I? – OpticalResonator Mar 4 '18 at 21:54
• A 3link mechanism with 2 revolute joints is a double pendulum? – joojaa Mar 4 '18 at 23:43
• I is the other instantaneous centre. The dashed circle envelopes the 2 links together and shows how this system as a whole can itself exhibit mixed motions (rotational + translational) in the fixed reference frame which is the 3rd link. So, I should turn out to be the instantaneous centre for this motion as a whole of the linkage system. – RedHelmet Mar 5 '18 at 1:02
• Ah, I get it now. How are the links 1 and 2 connected to the reference frame? – OpticalResonator Mar 5 '18 at 9:32
• @OpticalResonator they’re not. It’s customary to consider the fixed reference frame as a link anyhow. For example, a 4 bar linkage system with 1 of the links fixed has 4 links including the fixed link. Now consider this very same 4 bar linkage system, except now its able to move freely about in the plane. So, now there are 4 links + 1 fixed reference frame (acting as a link)....giving a total of 5 links. – RedHelmet Mar 5 '18 at 10:50