The proof of mass vector.
in [Math]
|
Prev: infinity ...
Next: The set of All sets
From: Eric Gisse on 30 Dec 2005 22:30 [snip] > > In the above, the "area vector division by length vector" is suggested. Vector division is not a defined operation for vectors. [snip]
From: Ka-In Yen on 1 Jan 2006 21:50 Ka-In Yen wrote: > Ka-In Yen wrote: > > The proof of mass vector. > > > > 1. The unit of vector. > > In physics, The unit of three-dimensional cartesian coordinate > > systems is meter. In this paper, a point of 3-D coordinate > > system is written as > > (p1,p2,p3) m, or (p:3) m > > and a vector is written as > > <a,b,c> m, or <a:3> m > > or > > l m<i,j,k> = <a,b,c> m > > In the above, the "area vector division by length vector" is suggested. > We can divide a length vector by a velocity in a different way. > Assuming > a length vector is l m<i:3>, and a velocity is v (m/s) <j:3>. <i:3> and > <j:3> are unit vectors. > > l m<i:3> / [ v (m/s) <j:3> ] > =l <i:3>o<j:3> / v s > <j:3> is moved to numerator. o is dot product. > =l cos(theta) / v s ---------(1) > theta is the angle between two vectors. > > OR > > v (m/s)<j:3> / [ l m<i:3> ] > =v cos(theta) / l s^(-1) --------(2) Length vector division by velocity gives two solutions: l cos(theta) / v s or l / (v cos(theta)) s. Depending on your application, you choose one operation. > > Both length vector and area vector have two directions; we can choose > one of their directions to keep cos(theta)>0. Correction: It should be cos(theta)>=0. If negative time is your application, then cos(theta)>=0 is unnecessary. Ka-In Yen Magnetic force: Drag and Bernoulli force of ether dynamics. http://www.geocities.com/redlorikee
From: Ka-In Yen on 2 Jan 2006 19:11 Eric Gisse wrote: > [snip] > > > > > In the above, the "area vector division by length vector" is suggested. > > Vector division is not a defined operation for vectors. > Dear Eric, Thank you for your comment. Then I am the first one. Ka-In Yen Magnetic force: Drag and Bernoulli force of ether dynamics. http://www.geocities.com/redlorikee
From: Ka-In Yen on 9 Jan 2006 20:18 Ka-In Yen wrote: > Eric Gisse wrote: > > > In the above, the "area vector division by length vector" is suggested. > > > > Vector division is not a defined operation for vectors. > > Dear Eric, > Thank you for your comment. Then I am the first one. Mathematically I prove that Einstein was ill-trained on three dimensional vector algebra. ^_^ Ka-In Yen Magnetic force: Drag and Bernoulli force of ether dynamics. http://www.geocities.com/redlorikee
From: Pmb on 12 Jan 2006 07:12
>> l m<i,j,k> = <a,b,c> m >> >> where l=abs(sqrt(a^2+b^2+c^2)) is the magnitude of the vector, >> and <i,j,k> is a unit vector which gives the direction of >> the vector. <i,j,k> is not a unit vector. It has a magnitude of sqrt(3) >> 2. Linear mass density is a vector. >> >> The mass of a string is M kg, and the length of the string >> is l m<i:3>. Where l m is the magnitude of the length, and >> <i:3> is a 3-D unit vector which gives the direction of the >> string. Then the linear mass density of the string is: >> >> M/(l<i:3>)=(M/l) (kg/m)<i:3> That doesn't mean that you can call mass a vector. Mass is a scalar which has a value at every point on the string. As you have it there are two vectors each of which are tangent to the string and point in opposite directions. Pete |