Intersecting spheres of equal energy density and unequal volume: I’m an accountant and good at math when it comes to a checkbook or a balance sheet. But can someone help me put some thoughts together mathematically to deal with the intersection of two spheres? Refer to these diagrams: Spherical Cap, Sphere-sphere intersection, and circle-circle intersection. The volume V of a Spherical cap = Vcap = 1/3 pi h^2 (3 R – h) and where two spheres intersect they form an overlap in the shape of a 3-D lens. The math for the volume of the lens uses the above formula for each cap and adds the volumes together. My question adds an assumption that the spheres are spheres of energy density and that the overlap is a combination of the energy density of the two spheres. Using that premise, if the two spheres were equal in size and energy density, then the 3-D lens would have twice the energy density as the individual spheres. But if the spheres were different sizes and have equal total energy but different energy density I wanted a formula that would work. Since there are two spherical caps, one from each sphere, and each cap contains overlapping or combined energy from each sphere I thought that by using the volume of the cap as a percentage of the sphere and adding the percentages together would work. Since I am assuming different size spheres and different energy density but equal total energy in each sphere and since the lens is made up of energy from each sphere, I came up with this for the calculation of the combined percentage of energy in the 3-D lens shaped intersection: \(\frac{V_{cap1}}{V_1} + \frac{V_{cap2}}{V_2} + \frac{V_{cap1}}{V_2} + \frac{V_{cap2}}{V_1} = \frac{1/3 \pi h^2 (3 R – h)}{4/3 \pi R^3} + \frac{1/3 \pi h^2 (3 r – h)}{4/3 \pi r^3}+ \frac{1/3 \pi h^2 (3 R – h)}{4/3 \pi r^3}+ \frac{1/3 \pi h^2 (3 r – h)}{4/3 \pi R^3}\) R and r are the radii. If I read the examples correctly, h is the same for both spheres, x is the distance from the center to line h, and d is the distance between centers. The resulting formula adds up the four energy percentages. The percentage of energy in cap 1 from sphere 1, the percentage of energy in cap 2 from sphere 2, the percentage of energy in cap 1 from sphere 2 and the percentage of energy in cap 2 from sphere 1 that occupy the lens. But the formula is ugly. I think it can it be simplified because the term (1/3 pi h^2) is in each piece but I can’t make it work when I try to simplify it.
Consider this a public service Please Register or Log in to view the hidden image! PS---LaTeX makes everybody's life easier. \(\frac{V_{cap1}}{V_1} + \frac{V_{cap2}}{V_2} + \frac{V_{cap1}}{V_2} + \frac{V_{cap2}}{V_1} = \frac{1/3 \pi h^2 (3 R – h)}{4/3 \pi R^3} + \frac{1/3 \pi h^2 (3 r – h)}{4/3 \pi r^3}+ \frac{1/3 \pi h^2 (3 R – h)}{4/3 \pi r^3}+ \frac{1/3 \pi h^2 (3 r – h)}{4/3 \pi R^3}\)
Thanks for the public service. It looks pretty. I guess it can't be simplified? Testing tex \(12*12=143+7*8-(11*5)\) \(pi\) \(pi r^2\) \(3.14159265\) \(square root of 2\) \(rhubarb pie\) Just testing my new skill (lack thereof). I suppose there is a character set or font that has the appropriate symbols? Oh wait, \(\frac{1/3 \pi h^2 (3 R – h)}{4/3 \pi R^3}\) \(\pi\) \(\pi r^2\) \(\3.14159265\) \(/square root of 2\) \(\rhubarb \pi\) Please Register or Log in to view the hidden image!
Referring to the Sphere-Sphere Intersection page at mathworld.wolfram: Last night I realized that \(h\) is not the same for both spherical caps if the spheres have different radii. That means that my equation for the percentage of volume included in the lens needs to use \(h\) & \(h’\). To get the height of the caps you need to calculate \(d\) and \(d’\), which requires the radius of the circle formed at the intersection of the two spheres. If you know \(R\) and \(r\), then using \(x^2+y^2+z^2=R^2\) you can calculate \(x\). Then using \(x\), the radius of the circle at the intersection can be calculated. According to the wolfram example it looks like the height of each cap can be calculated once you know the radius of the intersection circle. If I amend my formula using \(h\) and \(h’\) it should then be right, I think. \(\frac{V_{cap1}}{V_1} + \frac{V_{cap2}}{V_2} + \frac{V_{cap1}}{V_2} + \frac{V_{cap2}}{V_1} = \frac{1/3 \pi h^2 (3 R – h)}{4/3 \pi R^3} + \frac{1/3 \pi h’^2 (3 r – h’)}{4/3 \pi r^3}+ \frac{1/3 \pi h^2 (3 R – h)}{4/3 \pi r^3}+ \frac{1/3 \pi h’^2 (3 r – h’)}{4/3 \pi R^3}\) I'm making an assumption that the energy density in the overlap (volume of the combined caps) is twice the average energy density of the two original spheres.
I'm having a little trouble with LaTex and my simple formula for gravity. I need a little help with the use of tex? Gravity \(g={-e}+{(e*(1-c))}\), where \(g\) is gravity per quanta of mass, \({-e}\) is the negative energy of the pull wave per quanta, \(e\) is the positive energy of the push wave per quanta, and \(c\) is the containment ratio in percent. Gravity of mass \({g} sub {m}\); I don't know how to phrase that in tex. A little help please. is \(q*((-e)+{(e*(1-c)))}\) where \(q\) is the number of quanta in the mass. So the whole thing is \({g sub m}={q}*(({-e})+{(e*(1-c))})\) I need someone who will help me phrase some equations in LaTex by private message or maybe a link to a simple quick reference guide to help me do it myself.
\(g_m\) = g_m I fear, however, that this is the least of your worries. Could you try and answer the following: what is the answer to the following: 1kg + 12m = ? Please Register or Log in to view the hidden image!
Thanks, let's see if I have it : \(g_m=q*((-e)+(e*(1-c)))\) Ah, thank you. And the units of measure worry is yours not mine. In my personal cosmology (QWC) where this formula was developed the unit of measure is the quantum of energy. Therefore the \(g_m\) is in terms of energy quanta. You can gleefully attach QWC on my thread if you want.
Ok. Please Register or Log in to view the hidden image! You became very defensive when I last asked questions.
Defensive is always the way in mathematics. I think a great deal of stems will revolve around the contemplation of impervious math, which may lead to problems of predicting itself.
Really? I spend most of my days doing mathematics of some form or another, and I'm certainly not defensive about it. In fact, I actively welcome discussions about the particular area I happen to be working on. The same would be said of my colleagues. Please Register or Log in to view the hidden image!
Oh, don't get me wrong. Some people are born to take the obsolete course, where a line never equals a curve. Point is, is that some people, do proclaim a math, which is never ever really realized by those who work in an area of math. This can be interesting, and de-pregressive at best.
Einsteins work was ridiculed for years. His work is now the maintream. Here is a perfect example. Mathematicians cannot do physics work, unless they learn about physics themselves.
How about, that... '' a mathematician cannot determine the works of physics alone...'' Does this make it clearer, friend?
But who said they could? And what does this have to do with "Defensive is always the way in mathematics". You're all over the place this evening! Please Register or Log in to view the hidden image!
Oh, you know... most peeps around here actually believe that mathematics alone can reveal trhe secrets of the universe. Maybe as a math teacher, you can explain that is far from the truth as it being, than the subordinates being being all over the place.
Do people really think that? I'd have thought that people think mathematics is an essential tool to describe the theories of physics. Do you have something against me being a mathematics teacher? You keep referring to it!
If it helps, it is mearly dogma itself. I have nothing against you. But i do when people claim that physics is based on mathematics alone. There is a multitude of aspects that requires the proper function of physics itself. Mathematics is obsolete without physics. And vice versa. But things are never one way.
