solo Posted December 12, 2006 Posted December 12, 2006 The long standing formula for the volume of a sphere is: [math]\left(r^3\right)*\left(\frac{4\pi}{3}\right)[/math] = volume of sphere I have calculated a new and much simpler formula: [math]\left(\frac{dia^3 \pi}{6}\right)[/math] = volume of sphere Anyone have something to add?..............solo
Cap'n Refsmmat Posted December 12, 2006 Posted December 12, 2006 That's the same formula, simplified a tad.
solo Posted December 12, 2006 Author Posted December 12, 2006 That's the same formula, simplified a tad.Absolutely Cap'n........., simplified for the sake of convenience.....................solo
Klaynos Posted December 12, 2006 Posted December 12, 2006 It's not much simpler, and nothing new in any way... 2Pir == Pid ya know...
TriggerGrinn Posted December 12, 2006 Posted December 12, 2006 Thats cool. I like expressing formulas in different ways. Formulas are more than just math. They are a descrpition of objects and the relationship between things. Kinetic energy: [math] KE = \frac {1}{2}mv^2[/math] > [math] KE = \frac{(M*V) (M'*V)}{M + M'}[/math] Kinetic energy that will be spent between two frames of mass that collide in 1d collisions. M = Mass, and must be used as the lesser mass of the interaction. That is M and M' are equal values of mass in the equation but represent the lesser mass of two objects colliding, and when masses are equall the value can be used from either of course. M = Moving object M' = The object that the moving object collides with, (at rest) V = velocity of object in motion relative to frame of reference at rest. Equall masses convserve velocity and momentum in (perfect collisions) and experience the exact same force. value of force of -acceleration and value of foce for +acceleration F+F. If two objects have different mass and collide, the force involved will only be as great as what the lesser mass can cause, with the considered velocity. This equation : [math] KE = \frac{(M*V) (M'*V)}{M + M'}[/math] describes the energy spent between a simple 1d interaction between two objects of mass. For no object has energy without a partner frame to measure its velocity. and of course this all can be expressed right into quantum behavior. [math]\frac {1}{2}mv^2 = \frac {(m1 v) (m1' v)}{m1 + m1'} = \frac {p^2}{2m} = \frac {n^2 h^2}{8mL^2} = En [/math] 1. The energies are quantized and can be characterized by a quantum number n 2. The energy cannot be exactly zero. 3. The smaller the confinement, the larger the energy required.
TriggerGrinn Posted December 12, 2006 Posted December 12, 2006 see the application of this equation at this link http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html#c2
solo Posted December 17, 2006 Author Posted December 17, 2006 It's not much simpler, and nothing new in any way... 2Pir == Pid ya know... Actually, there is something new here. Examine the following formula: [math]re=\frac{\left(\sqrt[4]{\frac{6}{\pi}}\right)\left({\sqrt[3]10}\right)^{25}}{c^2}[/math]............in cgs units and; [math]re=\frac{\left(\sqrt[4]{\frac{6}{\pi}}\right)\left({\sqrt[3]10}\right)^{7}}{c^2}[/math].............in SI units Notice the involvement of [math]\left(\frac{6}{\pi}\right)[/math] within these formulae.
Ragib Posted December 18, 2006 Posted December 18, 2006 What the hell is a? and what is i? If you mean the imaginary unit..how the hell does that come in...and its different, but not simpler or more convenient...you still have to cube a quantity..divide by a bigger number...i just dont get it. Edit: O crap, this is in applied math section..my bad i thought it was pure math...
uncool Posted December 18, 2006 Posted December 18, 2006 see the application of this equation at this link http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html#c2 I looked, no sphere there at all. Please stop making plugs for your site.
swansont Posted December 18, 2006 Posted December 18, 2006 I looked, no sphere there at all. Please stop making plugs for your site. I think he was referring to the equations in his post. And I doubt hyperphysics is his site.
Chainsaw Jim Posted December 28, 2006 Posted December 28, 2006 Since we mostly work in Radius rather than diameter I suggest 'four thirds pi r cubed' is the easier. it's the same number of components.
Ragib Posted December 29, 2006 Posted December 29, 2006 I agree, just easier to stick to what's simpler.
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