Can Someone Explain Step 4 in the Buckingham Pi Theorem Homework?

[hotjohn]

Homework Statement

http://www-mdp.eng.cam.ac.uk/web/li...mal_dvd_only/aero/fprops/dimension/node9.html

can somoene expalin about step 4 in the first photo attached ?
What does it mean by each group has all the repeating variables and non-repeating variable ?

Homework Equations

The Attempt at a Solution

As in the second photo , the repeating variable is M , L , and T . Each Fd , D , V , µ and ρ have M, L, T . So , what where is the non-repeating factor ?
In the first phto, we can see that the author group ( D, V , ρ and F ) into π1 ,
(D , V , µ and ρ ) into π2 . I didnt see any non-repeating factor into both π1 and π2

 

[haruspex]

As in the second photo , the repeating variable is M , L , and T .

No, in both images the repeating variables are D , V, and ρ.
Each Pi group must contain all repeating variables and exactly one non-repeating. Each Pi group must use a different non-repeating variable, hence the name.
The two non repeating vars are F and mu, so the two groups are D, V, rho, F and D, V, rho, mu.

 

[hotjohn]

No, in both images the repeating variables are D , V, and ρ.
Each Pi group must contain all repeating variables and exactly one non-repeating. Each Pi group must use a different non-repeating variable, hence the name.
The two non repeating vars are F and mu, so the two groups are D, V, rho, F and D, V, rho, mu.

Why the repeating variables are D , V, and ρ ? how to see it ? for three of them , they contain L , L/ T and M / ( L^3) respectively...

 

[haruspex]

Why the repeating variables are D , V, and ρ ? how to see it ? for three of them , they contain L , L/ T and M / ( L^3) respectively...

Not sure what it is you are asking how to see.
If you mean, how to choose them, it's like picking a basis for a vector space. Since the only fundamental dimensions present are M, L and T, the space is three dimensional. A basis is therefore three linearly independent vectors. Linearly independent here means that you cannot construct a dimensionless quantity from them, except trivially.
Since M only occurs in rho, you cannot get rid of it again by bringing in a combination of the D and V. So any dimensionless construct from them cannot involve rho. That leaves D and V. T occurs in V but not in D, so there is no nontrivial combination of D and V that is dimensionless.

 

[hotjohn]

Not sure what it is you are asking how to see.
If you mean, how to choose them, it's like picking a basis for a vector space. Since the only fundamental dimensions present are M, L and T, the space is three dimensional. A basis is therefore three linearly independent vectors. Linearly independent here means that you cannot construct a dimensionless quantity from them, except trivially.
Since M only occurs in rho, you cannot get rid of it again by bringing in a combination of the D and V. So any dimensionless construct from them cannot involve rho. That leaves D and V. T occurs in V but not in D, so there is no nontrivial combination of D and V that is dimensionless.

do u mean we choose D , V, and ρ becasue they are the simplest quantity , that we cannot derive from any other physical unit ?

 

[haruspex]

do u mean we choose D , V, and ρ becasue they are the simplest quantity , that we cannot derive from any other physical unit ?

Not sure if they're the simplest. In the worked example you posted (ac.png) any three of D, V, rho, mu will do. None of them can be dimensionally constructed from the other three.