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Fundamental constants

 

c    = 2.9979E8;             %[m/s]          Speed of light
mu0  = 4*pi*1E-7;            %[V*s/(A*m)]    Permeability
eps0 = 8.8542E-12;           %[A*s/(V*m)]    Permittivity
G    = 6.672E11;             %[N*m2/kg2]     Gravitation constant
me   = 9.109E-31;            %[kg]           Electron rest mass
mp   = 1.673E-27;            %[kg]           Proton rest mass
e    = 1.602E-19;            %[C]            Elementary charge
k    = 1.381E-23;            %[J/K]          Boltzmann constant
h    = 6.626E-34;            %[J*s]          Planck’s constant
Me   = 5.98E24;              %[kg]           Earth’s mass
Re   = 6.37E6;               %[m]            Earth’s radius
Ms   = 1.989E30;             %[kg]           Solar mass
Rs   = 6.966E8;              %[m]            Solar radius
Md   = 8E15;                 %[T*m3]         Earth’s magnetic dipole moment
R    = 1.985E-3;             %[kcal/(mole*K)]Gas constant

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Spin axis, precession

Example 10.2, Orbital Mechanics for Engineering Students, 2nd Edition.

A cylindrical shell is rotating in torque-free motion about its longitudinal axis.

r = 1;      % [m]
l = 3;      % [m]
m = 100;    % [kg]
theta = 20; % [deg ]nutation angle
% How long does it take the cylinder to precess through 180 ° if the
% spin rate is 2*pi/radians per minute
ws = 2*pi;                      % [rad/min]
C = m*r^2                       % [kg*m^2]
A = 1/2*m*r^2 + 1/12*m*l^2
C =

   100

A =

  125.0000

A > C, Direct (prograde) precession

wp = C/(A - C)*ws/cosd(theta)   % [rad/min]
wp =

   26.7457

The time for the spin axis to precess through an angle of 180 ° is

t = pi/wp*60                    % [sec]
t =

    7.0477

Published with MATLAB® 7.10

 

Best Fit Solution,Straight Line Fit

%https://smallsat.wordpress.com/

clear all
clc
T = [0,20,100,100,200,400,400,800,1000,1200,1400,1600];
R = [10.96,10.72,14.1,14.85,17.9,25.4,26,40.3,47,52.7,58,63];
plot(T,R,'r*');
title('Straight Line Fit');
xlabel('Temperature (deg C)');
ylabel('Resistivity (Ohm cm)');
sy = 0;
sx = 0;
sxy = 0;
sx2 = 0;
n = 12;
for i=1:n
sy = sy +R(i);
sx = sx + T(i);
sxy = sxy + T(i)*R(i);
sx2 = sx2 + T(i)*T(i);
end
a = (sy*sx2-sx*sxy)/(n*sx2-sx^2);
b = (n*sxy-sx*sy)/(n*sx2-sx^2);
fprintf('\n\n Best Fit Coefficients')
fprintf('\n x = %6.4f T + %6.4f \n',b,a)
cm_fit =a+b*T;
hold on;
plot(T,cm_fit,'r');
legend('Data Points','Best Fit Solution','location','northwest');
 Best Fit Coefficients
 x = 0.0337 T + 11.4695

 

Published with MATLAB® 7.10

Type K Thermocouple, Voltage vs Temperature

Source https://smallsat.wordpress.com/

% Polynomial Coefficients 0-500 °C[4]
% Type K thermocouple
c= [0,25.08355,7.860106e-2,-2.503131e-1,8.315270e-2,-1.228034e-2,...
    9.804036e-4,-4.413030e-5,1.057734e-6,-1.052755e-8];
E= 0;
% Measured Voltage difference in mv
V = [1,E,E^2,E^3,E^4,E^5,E^6,E^7,E^8,E^9];
T_delta = c*V';
fprintf('Cold Bath Temp.  = %0.3f C\n' ,T_delta );
fprintf('Cold Bath Temp.  = %0.3f F\n\n' ,T_delta*1.8+32 );
E= 0.74;
% Measured Voltage difference in mv
V = [1,E,E^2,E^3,E^4,E^5,E^6,E^7,E^8,E^9];
T_delta = c*V';
fprintf('Room Bath Temp.  = %0.3f C\n' ,T_delta );
fprintf('Room Bath Temp.  = %0.3f F\n\n' ,T_delta*1.8+32 );

E= 3;
% Measured Voltage difference in mv
V = [1,E,E^2,E^3,E^4,E^5,E^6,E^7,E^8,E^9];
T_delta = c*V';
fprintf('Hot Bath Temp.  = %0.3f C\n' ,T_delta );
fprintf('Hot Bath Temp.  = %0.3f F\n\n' ,T_delta*1.8+32 );
Cold Bath Temp.  = 0.000 C
Cold Bath Temp.  = 32.000 F

Room Bath Temp.  = 18.526 C
Room Bath Temp.  = 65.346 F

Hot Bath Temp.  = 73.576 C
Hot Bath Temp.  = 164.436 F

Published with MATLAB® 7.10

 

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