This is a little idle reasoning. it is coherent. but the data may be false. so please help me. all the calculations are doable with a TI30S calculator. I put 'B' into A and 'C' into B and 'c' into C and 'M' into D
then B^(1/4) * M^(1/2) == Surface temperature B *M^2 == Operating temperature -- inner temperature. B*C*M^4 == #Core# temperature M for our sun is == 15.04664737 surface of sun is == 5778 OPT of sun is == 5778^4 #Core# of sun 0s == B*C *m^4
plsease let me know if I am wrong anywhere.
stars should collapse in a predictable manner.. Let the maximum Radius be 17 light days and let N be such that the motion from N to 1 increases the pressure and the Temperature. N/(N-1)! Is the expression for a ring at N to fall to one light day position. This is a geometrical exploration of star formation NOTE: #CORE# is the center of the star. It is ~~ *E23 in our sun.
The calculated temperature F(N) * MASS2 is the ignition temperature.
Just one F(N) to define the collapse of rings of gas. F(17)
Solve for T >1/2 G*T^2 = C*D YR = (365.25* 86400) T= (2*C*D/G)^(1/2) /YR = 27922 years to fall one light day. *18 = 500000 years
ASSERT a laminar homogeneous gas cloud. ASSERT F(N) = (N-1)!/N *4K = Temperature in KELVIN. N= 17 B = 4.932009385 E12 K ASSERT without proof that all falling masses will rotate. ASSERT no other stars are made just this one. M = mass in solar masses.
If the gas falls in a regular sense with no other stars around. It should fall N/(N-1)! down to one light day. The T vs. P formula is P^(-1( *4K = temperature Kelvin. This implies that (N-1)!/N *4 K is a general formula for star formation. (still a constant)
F(17) = 4.923009385E12 -A Constant B = F(17) My reasoning involves the SUN - 1.0 solar mass. Surface operating temp core temp 5778 1.114577188E15 6.179363346E23 Formulas: Surface^4 ~~ OPT OPT* 12!/13 ~~ #CORE# C= 12!/13 == 3.684627692E 7 K a minor constant - just a long step D= 15.04665 Sun surface ~~ 5778K
s^4~~ 1.114577188E15 OPT* C~~ 36846275,92 *D= #CORE# Note: B^(1/4)*D^(1/2) = 5778.00000 B^1/4 B B*C 5778 1.114577188E15 4.106801971E22 SUN Temperature 1489.6 4.023009385E12 1.813945671E20 New solar mass temperatures D^1/2 D2 D3
B*M^2 = 5778^4 5778^4*M^(1/2) M star surface temperature 57784*M^2 M star operating temperature 5778^4*C*D*M^3 B*C*M^3 both calculate core temperature for an M class star.
D= 15.04665 =SOLAR MASS ( ??) from my mass calculations. where M may run from 1 to 350
this is the formula to find how much gas is in a cubic meter.
The general period for a star to fall and ignite is 18*27922 = 502596 years. 5 E5
Sun Kg/(4/3*(C*D*18))^3 gives Kg/M^3
1.989044385E30/(4/3*C^3*D^3*18^3 ) = 4.685273976E-15 Kg/M^3 * MASS
Kb/m^3 call this value K.*(245) -> massive star
Use the above formula to calculate the gas in Kg/M3 * mass of target in solar masses. Of gas The correct with my step function. F(N) = B is the constant that allows me to create T for various stars. I am using the temperature model of our sun to extrapolate to the surface, ignition sequence, and core. M^(1/2) M^2 M^3
C = 12!/13 the heat step for the sun modified by M^3 and D