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visits member for 4 years, 2 months
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Started programming on a ZX spectrum in the 80's and have moved through Assembly, Turbo Pascal, C++, C#, Fortran. My main area of focus is engineering and scientific computing like numerical methods and 3D graphics.


2d
comment Sense of rotation. How would the rotation matrix look like for this “arbitrary” axis?
Yes. See the arrow in the picture showing the direction of the rotation axis.
2d
answered Sense of rotation. How would the rotation matrix look like for this “arbitrary” axis?
Jan
20
answered What are the degrees of freedom of $F=ma$ and $F =mdv/dt$?
Jan
13
comment Factors of the RSA modulus
See herongyang.com/Cryptography/…
Jan
12
answered Using the chain rule backwards
Jan
4
comment Arc length of a 3D Curve
If you want higher precision, you create a cubic spline from the points and integrate the spline.
Jan
4
answered RK4 method applied to $\frac{dy}{dt}=-\frac{y-t}{2}$ with $y(0)=1$
Jan
4
comment RK4 method applied to $\frac{dy}{dt}=-\frac{y-t}{2}$ with $y(0)=1$
Why didn't you try it with RK4? Juts Google it if you are lost.
Jan
1
comment solving systems of equations for m and b when you know they are both positive?
If you expected a positive result and didn't get it, then there is something wrong in your model (the equations or the solution). Maybe a coefficient or a sign somewhere. I get $m=1125$ and $b=88700$. I suggest showing how you get your values.
Dec
15
awarded  Caucus
Nov
24
comment Midpoint of the shortest distance between 2 rays in 3D
The vector from $a$ to $p$ projected onto the line is $b\left(\frac{b\cdot (p-a)}{b\cdot b}\right)$ and the above is another way of doing projections using the vector triple cross product.
Nov
24
revised Midpoint of the shortest distance between 2 rays in 3D
added 491 characters in body
Nov
24
answered Midpoint of the shortest distance between 2 rays in 3D
Nov
19
comment Ratio of Areas of Similar Triangles
The image link is broken for me.
Nov
10
comment Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$
The difficulty is the for a thin shell $\int dx = 0$, but $\int \rho dx >0$. Real mass density is infinite since the material volume of a hollow sphere is zero.
Nov
10
comment Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$
If real mass density $\rho$ is used then $w = \int \rho {\rm d}x$ Now the limits of this integration depend on $y$. To do this right your slicing has to be done with a polar angle ${\rm d}\theta$ instead of ${\rm d}y$. Then $dm = \rho \pi x^2 R d \theta d R$ instead of $dm=\rho \pi x^2 dx dy$
Nov
10
comment Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$
Dimensionally you integral is incorrect. The result should be of the mass times distance squared units.
Nov
10
answered Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$
Nov
10
comment Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$
I think your $dm$ is suspeect. If you do $\int dm = \int \frac{M \sqrt{R^2-y^2}}{2 R^2}\,{\rm d}y = \frac{\pi}{8} M R \ne M$. Shouldn't by definition $M=\int dm$?
Nov
10
comment Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$
Moment of inertia calculation should not have a fractional exponent. Can you show more of how you arrived at this.