# Difference between revisions of "Profile of a large drop"

Line 9: | Line 9: | ||

and <math>\kappa=\sqrt{\frac{\rho g}{\sigma_{lv}}}</math> was the inverse capillary length. | and <math>\kappa=\sqrt{\frac{\rho g}{\sigma_{lv}}}</math> was the inverse capillary length. | ||

− | To visualize what the edge of a large drop then looks like, you can use the following two MATLAB programs, which will solve and plot the above differential equation. Figure 1 also shows what some of these plots look like. | + | To visualize what the edge of a large drop then looks like, you can use the following two MATLAB programs (see below), which will solve and plot the above differential equation. Figure 1 also shows what some of these plots look like. |

[[Image:Dropshape1.jpg|300px|thumb|right|Figure 1: Drop profiles for various contact angles and using the physical constants given in class.]] | [[Image:Dropshape1.jpg|300px|thumb|right|Figure 1: Drop profiles for various contact angles and using the physical constants given in class.]] | ||

Line 19: | Line 19: | ||

Nevertheless, the bottom left corner of the 179 degree profile does resemble a quarter-circle. | Nevertheless, the bottom left corner of the 179 degree profile does resemble a quarter-circle. | ||

+ | dropshapeODE.m | ||

<pre> | <pre> | ||

− | function df = | + | function df = dropshapeODE(dt,y) |

global angle; | global angle; | ||

Line 32: | Line 33: | ||

X = cos(theta) + kappa^2/2*(2*E*y-y^2); | X = cos(theta) + kappa^2/2*(2*E*y-y^2); | ||

df = sqrt(1/X^2-1); | df = sqrt(1/X^2-1); | ||

+ | </pre> | ||

+ | |||

+ | dropshape.m (the parameter to change here is "angle") | ||

+ | <pre> | ||

+ | global angle; | ||

+ | |||

+ | angle = 100; | ||

+ | theta = pi/180 * angle; | ||

+ | sigma = 30*10^(-3); | ||

+ | rho = 10^3; | ||

+ | g = 9.8; | ||

+ | kappa = sqrt(rho*g/sigma); | ||

+ | E = 2/kappa*sin(theta/2); | ||

+ | |||

+ | y0 = [0]; | ||

+ | [Xout,Yout] = ode45('dropshapeODE',[0 5*kappa^(-1)],y0); | ||

+ | |||

+ | dydx = zeros(size(Xout)-1); | ||

+ | if theta > pi/180 * 90 | ||

+ | dydx = (Yout(2:end)-Yout(1:end-1))./(Xout(2:end)-Xout(1:end-1)); | ||

+ | [p q] = max(dydx); | ||

+ | for i = 1:q | ||

+ | Xout(i) = 2*Xout(q)-Xout(i); | ||

+ | end | ||

+ | Xout = Xout-Xout(1); | ||

+ | end | ||

+ | |||

+ | figure(1); | ||

+ | clf; | ||

+ | hold on; | ||

+ | plot(Xout,Yout); | ||

+ | axis equal; | ||

+ | A = axis; | ||

+ | axis([A(1) A(2) 0 A(4)]); | ||

+ | box on; | ||

</pre> | </pre> |

## Revision as of 18:17, 18 February 2009

In class we derived the profile <math>z(x)</math> of a large drop as being:

where we had defined <math>e</math> as being the maximum height of the drop:

and <math>\kappa=\sqrt{\frac{\rho g}{\sigma_{lv}}}</math> was the inverse capillary length.

To visualize what the edge of a large drop then looks like, you can use the following two MATLAB programs (see below), which will solve and plot the above differential equation. Figure 1 also shows what some of these plots look like.

In class we had a brief discussion as to the nature of the drop's periphery on superhydrophobic surfaces. In the superhydrophobic limit (i.e., as <math>\theta_e</math> approaches 180 degrees), will the edge of the drop have a semicircular shape? The shape is apparently not semicircular at 179 degrees, since the horizontal range of the drop is only 1 mm, while it vertically extends to a height of

Nevertheless, the bottom left corner of the 179 degree profile does resemble a quarter-circle.

dropshapeODE.m

function df = dropshapeODE(dt,y) global angle; theta = pi/180 * angle; sigma = 30*10^(-3); rho = 10^3; g = 9.8; kappa = sqrt(rho*g/sigma); E = 2/kappa*sin(theta/2); X = cos(theta) + kappa^2/2*(2*E*y-y^2); df = sqrt(1/X^2-1);

dropshape.m (the parameter to change here is "angle")

global angle; angle = 100; theta = pi/180 * angle; sigma = 30*10^(-3); rho = 10^3; g = 9.8; kappa = sqrt(rho*g/sigma); E = 2/kappa*sin(theta/2); y0 = [0]; [Xout,Yout] = ode45('dropshapeODE',[0 5*kappa^(-1)],y0); dydx = zeros(size(Xout)-1); if theta > pi/180 * 90 dydx = (Yout(2:end)-Yout(1:end-1))./(Xout(2:end)-Xout(1:end-1)); [p q] = max(dydx); for i = 1:q Xout(i) = 2*Xout(q)-Xout(i); end Xout = Xout-Xout(1); end figure(1); clf; hold on; plot(Xout,Yout); axis equal; A = axis; axis([A(1) A(2) 0 A(4)]); box on;