Missing technique used to find acceleration of a 4-bar mechanism?

[MechE]

Hi there!
There is a technique that is being used by my lecturer for mechanisms course at uni and i could not understand how this technique is manipulated.I wish i would know this methods name but no chance!.
This technique is being used to find normal acceleration of link in 4-bar mechanism.Manipulation:If you know the velocity of a link then draw a semi circle whose diameter is link's length and draw another circle whose radius is velocity vector and intersect these two circles then you obtain the normal acceleration of link.
What is the name of this method?
How can i find more examples on this method?
Thank you very much!

 

[haruspex]

Where is the second circle centred in relation to the first?

 

[MechE]

Where is the second circle centred in relation to the first?

First circle centered at midpoint of link
Second circle is centered at the rotating tip of link.
(Assume link is grounded at one point.And second point is rotating at a constant angular velocity-it is a 4-bar mechanism)
I will upload a picture of it.

 

[haruspex]

First circle centered at midpoint of link
Second circle is centered at the rotating tip of link.
(Assume link is grounded at one point.And second point is rotating at a constant angular velocity-it is a 4-bar mechanism)
I will upload a picture of it.

Very cute.
It's just an application of Pythagoras. Let the bar be AB, length 2r. Let C be the centre of the bar, P be the point where the arcs intersect and N where the perpendicular from P to the bar meets the bar. Then v = BP, CB=CP = r, BN = a. Applying Pythagoras to CNP, (r-a)²+PN²=r². Applying it to BNP, a²+PN²=v². Subtract one equation from the other and you end up with a = v²/2r.
When using this drawing technique, you have to scale velocity to distance in such a way as to ensure the curves intersect. That same scaling applies to the ratio of acceleration to velocity, enabling you to interpret the result.

 

[MechE]

Very cute.
It's just an application of Pythagoras. Let the bar be AB, length 2r. Let C be the centre of the bar, P be the point where the arcs intersect and N where the perpendicular from P to the bar meets the bar. Then v = BP, CB=CP = r, BN = a. Applying Pythagoras to CNP, (r-a)²+PN²=r². Applying it to BNP, a²+PN²=v². Subtract one equation from the other and you end up with a = v²/2r.
When using this drawing technique, you have to scale velocity to distance in such a way as to ensure the curves intersect. That same scaling applies to the ratio of acceleration to velocity, enabling you to interpret the result.

Thank you for explanation.This was what i was looking for.