66
6 Herz’s Task
where the operator (x, y) is defined by formulas (5.23)–(5.24).
The obtained equation is identical to (6.24). Therefore, we will find the continuous solution of Eq. (6.41) using formulas (6.32), (6.33) at the values M, N, and β 1
determined using formulas (6.38)–(6.40).
6.6 Elongated Contact Area
Assume that M N. With this assumption, the eccentricity of the elliptical contact
area
e =
1 −
a
b
2
will be close to one, and the value of the elliptical integral K(e) unrestrictedly
increases at e → 1.
In these conditions, the ellipsis eccentricity e can be found by transforming
formula (6.29). We notice that
(1 − e
2 )K(e) =
π/2
0
1 − e
1 − e 2 sin
2 ϕ
dϕ,
(6.42)
and taking into account that
lim
e→1
1 − e 2
1 − e 2 sin
2 ϕ
= 0,
(6.43)
from the first formula (6.29) we obtain
M
N
≈ (1 − e
2 )K(e), (e ≈ 1).
(6.44)
It is known [3] that at e ≈ 1 can be represented as
K(e) ≈ ln
4
1 − e 2 = ln
4a 2
b 2 ,
(6.45)
it is accounted that due to formula (6.43)
1 − e
2
=
b 2
a 2 .
Taking into account these comments, formula (6.44) may look as follows: