# Choosing repeating variable in pi Buckingham theorem

## Homework Statement

why we cant form the pi group by using repeating variable of (μ, ρ , v) or (D, v , μ ) ?
http://www-mdp.eng.cam.ac.uk/web/li...mal_dvd_only/aero/fprops/dimension/node9.html
http://www.efm.leeds.ac.uk/CIVE/CIVE1400/Section5/dimensional_analysis.htm

## The Attempt at a Solution

according to the principle , when we choose the repeating variable , it cant form the dimensionless group , right ? for (μ, ρ , v) , i have (M/LT)(M/(L^3))(L/T) , i got (M^2)(L^-3)(T^-2)
for (D, v , μ ) , i have (L)(L/T)(M/LT) , i got ML(T^-2) , both group are not dimensionless, why is it a must to form group by using (D, ρ , v) ??

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Orodruin
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In general, it is perfectly fine to choose any group which is dimensionally independent. But be aware that it is not sufficient to check the product of the variables to conclude this as you seem to be doing.

Edit: Just one additional thing. Although you are free to pick any dimensionally independent group, some might work better or worse for your purposes. The theorem does not tell you which is the best choice.

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In general, it is perfectly fine to choose any group which is dimensionally independent. But be aware that it is not sufficient to check the product of the variables to conclude this as you seem to be doing.

Edit: Just one additional thing. Although you are free to pick any dimensionally independent group, some might work better or worse for your purposes. The theorem does not tell you which is the best choice.
so , we have to try and error to get the correct ans ?

haruspex
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well, after reading the thread , i still cant understand why we cant choose other combination of repeating factor other than (D, ρ , v) ? can you explain further?
As I wrote in the final post on the thread, any three of D, ρ, v or μ could have been used.

As I wrote in the final post on the thread, any three of D, ρ, v or μ could have been used.
As I wrote in the final post on the thread, any three of D, ρ, v or μ could have been used.
do you mean no matter which combination of D, ρ, v or μ , i would get the same answer too?

haruspex
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do you mean no matter which combination of D, ρ, v or μ , i would get the same answer too?
Try it.

Try it.
taking ρ, v and μ as repeating variables , i form 2 pi group , but i cant get the ans .... why ?

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Orodruin
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You will not get the same ##\pi##s, but the relations will be equivalent.

haruspex
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taking ρ, v and μ as repeating variables , i form 2 pi group , but i cant get the ans .... why ?
You left something out of the L equation in the left hand column.

• foo9008
You left something out of the L equation in the left hand column.
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what equation ? can you point out please ?
ok , i have done the correction , and my ans is Fρ / (μ^2) = f(ρvD / μ) , but my ans is not consistent with the author's working which is
μ/ ρvD = f ( F / ρ(v^2)(D^2) )

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You left something out of the L equation in the left hand column.
i have also tried out forming pi group by taking the repeating variable as ρ , μ , D , the ans that i got is same as post #13 .
however , when i choose repeating variable as v , μ , D , my ans is different ( i got (FD/ vμ ) = f(vDρ/μ) instead of (Fρ/ μ^2 ) = f( vDρ/μ) ) . why is it so ?

haruspex
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ok , i have done the correction , and my ans is Fρ / (μ^2) = f(ρvD / μ) , but my ans is not consistent with the author's working which is
μ
/ ρvD = f ( F / ρ(v^2)(D^2) )
It is consistent.
Let's write x=μ/ ρvD and y=Fρ / (μ^2). You got y=f(x), the author got x=f(y/x2) for some other function f. These are effectively the same.

• foo9008
It is consistent.
Let's write x=μ/ ρvD and y=Fρ / (μ^2). You got y=f(x), the author got x=f(y/x2) for some other function f. These are effectively the same.
sorry , i dont understand . do you mean from the author's working , i can remove the f( ) , and make μ/ ρvD = F / ρ(D^2)(V^2) , so eventually , i will get μDV = F ? for pi group with (ρ , V , μ , F) and (ρ , V , μ , D) , i will get Fρ/ (μ^2) = f( ρvD / μ) , after rearranging , i will also get μDV = F

haruspex
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sorry , i dont understand . do you mean from the author's working , i can remove the f( ) , and make μ/ ρvD = F / ρ(D^2)(V^2) , so eventually , i will get μDV = F ? for pi group with (ρ , V , μ , F) and (ρ , V , μ , D) , i will get Fρ/ (μ^2) = f( ρvD / μ) , after rearranging , i will also get μDV = F
No, I mean that saying y is a function of x is, generally speaking, equivalent to saying x is a function of y/x2.
Examples:
Y=2x is the same as x= 2/(y/x2)
Y=x4 is the same as x=(y/x2)1/2.

But it doesn't always work. There is no way to express y=x2 in the form x=f(y/x2)
Translating that back, your form can express F=AρV2D2 but the author's cannot. Likewise, the author's can express ρvD=Aμ, but yours cannot. That is a limitation of writing things in the form y=f(x); it cannot express a vertical line, i.e. x is constant. That problem would be solved by writing instead f(x,y)=0.

• foo9008
No, I mean that saying y is a function of x is, generally speaking, equivalent to saying x is a function of y/x2.
Examples:
Y=2x is the same as x= 2/(y/x2)
Y=x4 is the same as x=(y/x2)1/2.

But it doesn't always work. There is no way to express y=x2 in the form x=f(y/x2)
Translating that back, your form can express F=AρV2D2 but the author's cannot. Likewise, the author's can express ρvD=Aμ, but yours cannot. That is a limitation of writing things in the form y=f(x); it cannot express a vertical line, i.e. x is constant. That problem would be solved by writing instead f(x,y)=0.
So, my working in post 13 is correct??

No, I mean that saying y is a function of x is, generally speaking, equivalent to saying x is a function of y/x2.
Examples:
Y=2x is the same as x= 2/(y/x2)
Y=x4 is the same as x=(y/x2)1/2.

But it doesn't always work. There is no way to express y=x2 in the form x=f(y/x2)
Translating that back, your form can express F=AρV2D2 but the author's cannot. Likewise, the author's can express ρvD=Aμ, but yours cannot. That is a limitation of writing things in the form y=f(x); it cannot express a vertical line, i.e. x is constant. That problem would be solved by writing instead f(x,y)=0.
in my working , i only show that F(ρ)/ (μ^2) = f(ρvD / μ) , where is F=AρV2D2 ?
can you point out which part of the author's working show ρvD=Aμ ? i didnt see it

haruspex
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in my working , i only show that F(ρ)/ (μ^2) = f(ρvD / μ) , where is F=AρV2D2 ?
can you point out which part of the author's working show ρvD=Aμ ? i didnt see it
No, you misunderstand. F=AρV2D2 is merely an example of Fρ/ (μ^2) = f(ρvD / μ) . Specifically, it is if the function f is f(x)=Ax2.

haruspex
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So, my working in post 13 is correct??
Yes.

No, you misunderstand. F=AρV2D2 is merely an example of Fρ/ (μ^2) = f(ρvD / μ) . Specifically, it is if the function f is f(x)=Ax2.
No, you misunderstand. F=AρV2D2 is merely an example of Fρ/ (μ^2) = f(ρvD / μ) . Specifically, it is if the function f is f(x)=Ax2.
oh , no ... i am getting more confused now , can you explain in another way round ?

Orodruin
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oh , no ... i am getting more confused now , can you explain in another way round ?
The theorem only states that the relation must be in the form of a function of dimensionless variables. It does not tell you what this function is and depending on your choice of repeating variables, the function describing your relationship will be different (because you will have different dimensionless variables). What haruspex gave you was an example.

Yes.
for repeating factor ( v, μ , D ) , i can form pi group with ( v, μ , D , F ) and ( v, μ , D , ρ) , i got (F/ v μ D) = f(v D ρ / μ) , after rearranging , i got F = ρ (D^2)(v^2) , is my working correct ?

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Orodruin
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