Recent content by thepopasmurf

Thank you for your responses and help,

Hmm, maybe I should demonstrate what I mean for clarification: Cylindrical tube: fc = γ cos θ dS/dx, Surface area in cylinder, S = 2πrx Cross-sectional area, A = πr2 So Pc = fc/A = 2πrγ cosθ / πr2 Pc = 2γ cosθ / r Yeah! Correct answer Rectangular tube, w*h*x (x is direction of fluid...

Yes, that's right

Does this help?

When figuring out the capillary pressure on a liquid in a tube of a certain cross-section, the typical approach is to consider the Young-Laplace pressure and the curvature etc. I was looking through some of my old notes and I had an equation for the capillary force: fc = γ cosθ dS/dx where γ...

@Asymptotic, this effect seems close to what I am looking for. The wikipedia entry doesn't give much to build on, I guess 'stretching' and 'fingering' are the best things to look at.

There are some plastics I am using that when melted, pull apart in a stringy way. My best comparison is like melted cheese in a sandwich. I want to know more about this, but other than looking up 'polymer rheology', I'm not sure what I should call this type of flow. I want to know what causes...

Homework Statement The problem is what are the odds of an incident object of radius r1 colliding with any of a collection of target objects of radius r2, where the r2 objects have a number density N / m^3 = n and the incident object travels a distance L. Incident object is moving much faster...

Ok, so here's my reworking of the problem: In lab frame four-momentum is: p_p + p_{\gamma} The square of this is invariant (p_p + p_{\gamma})\cdot(p_p + p_{\gamma}) = \frac{E_p^2}{c^2} - p_p^2 + \frac{E_{\gamma}^2}{c^2} - p_{\gamma}^2 + \frac{2E_pE_{\gamma}}{c^2} - 2p_p\cdot...

Homework Statement A particle with known rest mass energy, m_{p} c^{2} pass through a cloud of monoenergetic photons with energy E_{\gamma}. The particle collides with a photon and a particle A, with mass m_A is created. Show that the minimum energy of the particle required for the interaction...

I'm sorry that this is frustrating for you, but I'm not yet convinced that the area of the triangles is exactly what I want. I'll try to explain my approach in more detail. I have a mesh of an object made up of triangular faces. I'm trying to get the spherical harmonics components of the mesh...

So should I change my integrand from \int Y R d\Omega to \int \frac{Y}{R} dA to keep the units correct?

I think it's fine that it's on a unit sphere because the distance part is accounted for in the integrand. And I'm not sure why I would have to add an extra R just because the function is the radius rather than any other function. Anyway, I think we're getting distracted by this point. I'm...

As far as I'm aware that is the correct element to integrate over. http://shtools.ipgp.fr/www/conventions.html Look for f_lm around a quarter way down the page http://en.wikipedia.org/wiki/Spherical_harmonics#Spherical_harmonics_expansion See the integrals in this section.

The purpose of the integral is not to find the area, but to find the harmonic coefficients of R.