Sur le théorème de Vaschy-Buckingham 1 Introduction J.F. Debongnie 3 juin 2016

XIXe
Π
Π
Π

x
[x] ˆx
x= ˆx[x]
v= 25 m/s [v] =
m/s
ˆv=v
[v]=v m/s= 25
L T
L2
L/T
[u1],...,[um] [x1],...,[xn]
x1, . . . , xnn≥m
[x1] = k1[u1]p11 [u2]p21 . . . [um]pm1
· · ·
[xn] = kn[u1]p1n[u2]p2n. . . [um]pmn
ki[x1],...,[xn]
[u1],...,[um]
m/s
km/h = 1000 m
3600 s =1 3,6m/s
n x1, . . . , xn[x1],...,[xn]
m[u1],...,[um]
Π = xq1
1. . . xqn
n
[x1]q1. . . [xn]qn= 1

Π = ˆxq1
1. . . ˆxqn
n·([x1]q1. . . [xn]qn) = ˆxq1
1. . . ˆxqn
n
p11 p12 . . . p1n
. . .
pm1pm2. . . pmn 
q1
qn
= 0
Pq =0
r≤mP
x1, . . . , xn[u1],...,[um]
PAA PAB
PBA PBB  qA qB= 0
PAA
Pr×r r P
PAB r×(n−r)PBA (m−r)×rPBB
(n−r)×(m−r)
(m−r)Pr
PBA =LPAA,PBB =LPAB
L(m−r)×r
qA=−P−1
AAPABqB
PBAqA+PBBqB=−LPAAP−1
AAPABqB+LPABqB=0
(n−r)
qBN(Π)
n
rP
N(Π) = n−r
P
N(Π) = n−m
(r < m)

qB=IqB
In−r
q=qA
qB=−P−1
AAPAB
IqB=QqB
Q=−P−1
AAPAB
I
keQΠk(n−r)
Πk=xQ1k
1· · · xQnk
n
Πk=xQ1k
1· · · xQrk
r·xr+k
[uj], j = 1, . . . , m
[uj]∗=λj[uj]
[xi]∗[xi]∗=γi[xi]
γi=λp1i
1. . . λpmi
mi= 1, . . . , n
ln γi=p1iln λ1+. . . +pmi ln λmi= 1, . . . , n
ln γ1
ln γn
=PT
ln λ1
ln λm
ˆx∗
i
xi= ˆxi[xi] = ˆx∗
i[xi]∗= ˆx∗ iγi[xi]
ˆx∗
i=ˆxi
γi

x1, . . . , xn
mPr
r
ˆx1,...,ˆxr
[u1],...,[um]
[x1],...,[xn]x1, . . . , xn
ˆx1,...,ˆxn[uj]∗=λj[uj]
[xi]∗=γi[xi] ˆx∗
i=1
γiˆxi
γi= ˆxii= 1, . . . , r
r=m
ln ˆx1
ln ˆxr
=PT
AA 
ln λ1
ln λr
PAA
r < m
ln ˆx1
ln ˆxr
=PT
AA PT BA 
ln λ1
ln λr
ln λr+1
ln λm
λr+1 = 1, . . . , λm= 1
τ
`
τ= 2πs`
g
ˆτ= 2πsˆ
`
ˆg