psychlone

Amr Morsi, I'm just curious, If you were to rewrite your above equation, it would look like be the following, Or not; I'm just pondering your post. [math]\theta_{n}= S_{n} - \frac{f^{(n)}(0)}{n!}S^{n}[/math] where [math]\theta = theta [/math] [math]a = \theta[/math] [math]x = S[/math] Taking the formula notation from the following Wiki page: http://en.wikipedia.org/wiki/Taylor_series.

The answer to your question; how many polygon(s) to theoretically construct a 2D sphere/ sphere cross section it’s limitless. But check out this link, it might be of some interest to you. Cheers. http://en.wikipedia.org/wiki/Method_of_exhaustion

I didn't make an error, I stand by what I wrote. I'm not talking down to you in any way but! a very important word of advice; if you want answers to questions you seek, then you must make the effort to write questions that make sense! (my apologies to the administrators, and to the Science Forums community)

So, when you say the “optimum shape” you’re actually referring to the geometrical shape that will generate the greatest velocity up and down the slope, assuming that the greater the height achieved up the other side of the sphere (“sphere like shape”) is dependent upon maximising velocity? But from memory the equal frictional forces acting on both the bowl’s and marble’s surfaces are dependant not so much on the frictional coefficient of the surface selected, but dependant more on the mass of the object exerting its self on the stationary surface. The small changes in the magnitude of the frictional coefficients has only a marginal impact on the actual friction force applied.

Yes, there is an easy mathematical equation. Here are 3 hints! 1. First, though your shape is 3D, the template to create the 3D shape is still only a 2D template. 2. Think more in terms of geometry. 3. Also, try and think about using sine or cosine functions to determine the 2D template’s shape.

Thanks insane_alien, I don't know why I didn't realize before; I'm using Internet Explorer and found the RSS Feed Icon on the command bar like you said. Psychlone.

Hi to all, I was wondering if any of the scienceforum.net sub forums have any URL addresses written specifically to access RSS Feeds? psychlone.

Your equation dcowboys107 from the substitution of equ 1 into equ 2 is correct. [math]\ h = [/math] [math]\frac{370}{cot(25^{\circ})-cot(38^{\circ})}[/math] What you've done is forget to take the reciprocal of tan to equal cot, and then subtracting the two cot’s and divide into 370.

Just like to correct a simple error on the above post. [imath]\frac{d}{dx}y(u(x))=\frac{dy}{du}\frac{du}{dx}[/imath] & rewrite the expression of the last post in a more familiar notation. [imath]\frac{d}{dx}f(g(x))={f'}{(g(x)) ^ . }{g'(x)}[/imath]

To recap on earlier post. [math] f(r,M,N)=\frac{(1+r)^N M r}{(1+r)^N -1} [/math] where [math]f(r,M,N) [/math] is equal to monthly repayments. Take these numbers for example. $100,000.00 Principle. @ 0.00604166% monthly interest (7.25% year). over 300 months (25 years). Yeilds. $722.80 Monthly repayments. ** ONLY AN EXAMPLE** This equation can calculate the entire amount of interest incurred of the full life of the loan (Instantly). What you must do is when plotting the above function; reduce the amount of time remaining on the loan (from the first repayment payment of 300 monthy units @ t=0 month to calculate principle & interest to 299 monthly units @ t = 1 month for example). So reduce the time steps in each entry (spreadsheet cell) by either monthly or weekly interval and so on. As the loan time frame becomes shorter the slope of the amount owing on the principle becomes steeper. This is due to your ratio of the amount owing due to interest in comparison to principle is reducing. .

**I'm not sure if I' too late for this discussion, but I'd like to add**. I concur with Cap't Refsmmat, the washer and shell isn't a shape of a volume but are names of two methods of determining a volume. Although, I'd like to revise the statement of; " The shell method can easily find the area of a volume rotated around the y axis when it's in terms of x. " Both the washer and shell methods use rotation around the same axis regardless of taking the rotation with respect to x or y axes. Yes, in your case, the two methods, integrating with respect to x along the x-axis with give a washer volume integration, while integrating along the y-axis (but bear in mind that rotation is still taken around the x-axis, an integration of an infinite amount vertical lines making up the bounded region multiplied by a circumferential distance [math]\bar{y}[/math] will give you the shell method of volume integration. Both methods yield the same answer if you algebraically implement correctly and integrated with correct limits, you’ll be fine. Click on link, I did a quick search on Google also. This will explain more clearly. http://learning.mgccc.cc.ms.us/math/mathdocs/calc/revol2.pdf

my apologies.

I don’t mean to be rude, and I’m sorry it come across like that & I greatly appreciate your time and effort and knowledge, but as much of a help you are your beginning to frustrating me. My original question is; *But if [math] \phi [/math] & [math] \eta [/math] are variables but are still dimensionless, taking a partial differential with respect to those variables has to be valid. Because you said that taking partial derivative to these variable are similar to taking the derivative of a constant. I meant to back to you sooner with my qustion but I had to go do something. Don't take this the wrong way but let's just forget about it.

you haven't answered my question

I've looked at my problem again, and the coordinates are in Cartesian form. The reason for this is the problem is; The derivation is uses a Cartesian coordinate system for fluid flowing around the circumference of a cylinder. Where x, is not flow though a pipe but x is flow of flow of fluid parallel to the surface of the cylinder and y is always perpendicular to the surface of the cylinder, which is the thickness of the film. 2 dimensional only, looking only at a cross section of the cylinder. My apologies once again I should have mentioned all this earlier. The answer to the derivation; [math]\frac{\partial U}{\partial \phi} - \frac{\eta}{\delta_{a}} \frac{d \delta_{a}} {d \phi} U \frac{\partial U}{\partial \eta} + \frac{1}{\delta_{a}} \frac{\partial V}{\partial \eta} [/math] Where: [math] u_{r} = \frac{K Re}{\rho R} [/math] Where: [math] K = [/math] Fluid consistency index (Pascals sec) (SI uits) [math] V = \frac{v Re}{u_{r}}[/math] [math] \eta = \frac{y}{\delta (x)} = \frac{y Re}{R \delta_{a}}[/math] *But if [math]\phi[/math] & [math]\eta[/math] are variables but are still dimensionless, taking a partial differential with respect to those variables has to be valid. This derivation is from; Laminar flow and heat transfer or non-Newtonian falling liquid on a horizontal tube with variable surface heat flux (D. Oudhadda & A. II Idrissi) Int. Comm. Heat Transfer. Vol. 28. No 8, pp 1125- 1135, 2001. What I meant with the comment "substituting scalars with vectors" was scalars are dimensionless but are taken as individual components in the direction of the vector. Can a single characteristic velocity [math] u_{r} [/math] be used for both [math u [/math] & and [math] v [/math] velocity to convert them. **Bearing in mind in the above Reynolds number term uses the [math] u [/math] velocity, which is the velocity in the direction parallel to the surface of the cylinder.

I'm using the velocity in the x - direction [math]\left(u\right)[/math]. As I’m solving for the first term in the continuity equation [math]\frac{\partial u}{\partial x}[/math]. Thus, your right, the velocity in Reynolds number with cancels with the [math] u [/math] variable term. I'm going about the problem the wrong way, doing strictly substitution across different coordinate systems, which can be done by focusing on first is using the Polar form of the continuity equation and then the second step is to the substitute the vectors with scalars. Thanks for your time, it is very much appreciated.

Reply to 1). my apologies I had to edit my text to correct the LaTex errors, But my post should be reading math symbols now. Reply to 2). I didn't include the characteristic velocity variable when multiplied by dimensional velocity yields a non-dimensional velocity. where: [math] u = \frac{U}{u_{r}}[/math] [math]u_{r}=\frac{K Re}{\rho R}[/math] And I agree, based on the Cartesian coordinate system the dimensional and non dimensional continuity should be the same. And Your right, I've over looked building the equation with respect to the correct the coordinate system, which should be cylindrical or polar. This something I need to focus. This has probably the stumbling block that has prevented me from going any further. Thanks very much for your help it's greatly appreciate. And where did you study Fluid Mechanics I'm impressed with your insight. You can ignore this question if you like.

Thanks for the tip on entering equations and for your reply, it is much appreciated. To the dimensionless variables; Then; x=[math]\phi[/math]R [math]\phi=const=\frac{dx}{dR}[/math] That makes sense. the solution I have to the continuity uses a partial differentiation with respect to dimensionless variables. Could I do this; [math]\frac{\partial u}{\partial x} = \frac{\partial}{\partial\left(\frac{x}{R}\right)}\left(\frac{UKRe}{\rho R}\right) = \frac{\partial}{\partial\phi}\left(\frac{UKRe}{\rho R}\right)[/math]. Where I take the derivative with respect to [math]x[/math]. and divide it through by [math]R[/math]. Thanks.

Hi, I start with the continuity equation, for a CFD application. ∂u/∂x+∂v/∂y=0 (1) I’m interested in substituting the following (2)(3)(4)(5) into the above equation (1) to produce a non-dimensional continuity equation. u = (UKRe/ρR) (2) v = ((UK〖Re〗^(1⁄2))/ρR) (3) x=ϕR (4) y=ηδ (5) which produces; ∂/∂(ϕR) (UKRe/ρR)+∂/∂(ηδ) ((UK〖Re〗^(1⁄2))/ρR)=0 (6) ∂/∂ϕ (UKRe/ρR).∂/∂R (UKRe/ρR)+∂/∂η ((UK〖Re〗^(1⁄2))/ρR).∂/∂δ ((UK〖Re〗^(1⁄2))/ρR)=0 (7) My question is; when I do the substitution for x & y, and manipulating equation (6) would I yield equation (7)? For completeness I’ve listed all the variables and constants used. Where: u = velocity (x-direction, vector) v = velocity (y-direction, vector) U = Velocity (x direction Dimensionless) R = Radius K = Constant Re = Reynolds Number Φ= Angle (Radians/ non-dimensional) δ = film thickness (vector) η = Dimensionless coordinate normal to the surface ρ=density Thanks.

Hi CalleighMay, I can solve you problem assuming that z= z(bar) = 0 (the bar goes above the z) which is the centroid or the centre of mass about z = 0. Then I can solve your problem. y(bar) = Mxz/V --------- Where V = Volume and Mxz is the moment in the xz plane. dMxz = y dV dMxz = y Pi (r^2 - y^2) dy - Where r = the radius - the geometry of the equation a circle. Therefore 4 in your original equation is the radius squared which is 2. Mxz = Int(dMxz) - Where int = integral. the integral of dMxz = Pi r^4 /4 - WHere Pi = 3.141592654....... V = 2/3 Pi r^3 y(bar) = Mxz/V = 3/8 r Now you have to do the same for x(bar) and build an equation for Myz in the yz plane. Bearing in mind that the 3 dimensional shape your solving for is a hemishpere, with the y axis travelling up through the top of the dome. Good luck.