« Last post by butchmurray on Today at 03:27:02 »
The formulation of gamma.
This formulation is from the critically acclaimed Cal Tech lecture series “The Mechanical Universe and Beyond” Lecture No. 42 “The Lorentz Transformation” starting about 12 minutes into the video. Dr. Feynman and Dr. Einstein held professorships at Cal Tech University. Here is a link to Lecture No. 42.http://www.learner.org/resources/series42.html#
One of the methods to formulate the Lorentz factor gamma uses a right triangle and the Pythagorean theorem.
Inertial frame K’ is in motion relative to inertial frame K at speed v. The observer is at rest in frame K.
In the right triangle the vertical leg, Ct’, is the length of a light path perpendicular to the direction of motion in frame K’.
The length of horizontal leg, vt, is the distance frame K’ advanced relative to frame K at speed v in the time t.
The hypotenuse, Ct, is the length of the path of the light in light path Ct’ seen in frame K by the observer.
Subtract (vt)² from both sides
Simplify right side
C²*t’²= t²*(C²- v²)
Divide both sides by C²
Square root all
Divide both sides by sqrt(1-v²/C²)
Substitute gamma for 1/sqrt(1-v²/C²)
Divide both sides by gamma
The seminal equation:
At speed v=0, (vt)²=0 substitute 0 for (vt)²:
Square root all.
Ct’=Ct At speed v=0
C is the constant speed of light.
t is time in frame K. Time t in frame K is constant with respect to speed v.
The length of Ct is then constant with respect to speed v.
The length of Ct’ is perpendicular to the direction of motion. In accordance with the y’=y equation of the Galilean transformation and the Lorentz transformation its length is constant with respect to speed v.
At speed v=0, Ct’=Ct. As the lengths of Ct and Ct’ are constant with respect to speed v for the reasons stated above, v>0 Ct’=Ct.
For v=0, Ct’=Ct. For v>0, Ct’=Ct. Then Ct’=Ct.
Substitute Ct’ with its equivalent Ct in the equation:
(Ct)²+(vt)²=(Ct)² The equation is invalid for any non-zero value of speed v
Therefore, the equation from which gamma is derived is invalid.
Thorntone E Murray, Houston, Texas USA April 22, 2014
« Last post by demografx on Today at 02:05:18 »
Today, April 23, 2014 is:
English Language Day
I wish I knew English sign language, it's pretty handy.
« Last post by CliffordK on Today at 01:11:56 »
I think you're off by about a factor of 10.
40 NiMH batteries (2.5 AH, 1.2V) arranged in parallel is equivalent to 1 battery, 1.2V, 100AH.
Or, you can arrange 10 banks of 4 in series/parallel to make a 12V, 10AH battery.
So, to make them equivalent to your 100AH, 12V deep cycle car battery, you'll need 10 of these blocks of batteries, or 400 of your AA batteries in total.
Going with D cells, you would need about 1/4 that number, or 100 batteries or so, which at least will make it easier to wire up. Perhaps even consider larger batteries such as F size.
I believe the original Teslas used blocks of small Lithium batteries, although I think they also had fairly complex electronic control circuitry. But, the Lithium batteries may be more touchy with over charging as well as low voltage cycling than the NiMH batteries.
You may also be able to find used Prius battery cells, 7.2V, 6.5AH. Your trolling motor should run just fine wired to 14.4V.
I'd certainly go for some kind of a smart charger. Trickle Charging might be ok, but most of the 14.4V NiMH chargers I'm seeing are for less than 5AH batteries, and may struggle with a 100+AH battery pack. This one should work with your pack, if you go with 12V. Absopulse Electronics BCP 130-PEL, but I'm not seeing retail pricing on it.
I'm not sure what it would take to convert a standard auto charger for use with NiMH batteries.
Personally I never just "run down" batteries, although it may not be bad to wait to recharge until they are mostly depleted, assuming you can wait that long with ordinary use. Lead Acid batteries should probably be stored full though.
Oh, one note about running 12V car batteries in parallel. Kill one cell, and it will quickly kill the whole set. I don't know if NiMH batteries suffer with the same low cell problems that lead acid batteries have.
0.02/c = 6.6712819E-11 and is a baseline value for G which does not take into account the density of a mass. The factor for varying density needs to be worked out from mathematical manipulation (differentiation of point mass sources over relative distances).
« Last post by jeffreyH on 22/04/2014 22:21:10 »
Some data here on gas interacting with Sagittarius A*.
« Last post by thedoc on 22/04/2014 20:30:02 »
Peter Steadman asked the Naked Scientists:
I found out the other day that when a rock is compressed almost to breaking, it generates an electric current that seems to be constant and, unlike nuclear energy, does not deplete over time.
This being the case, could we use the planet's gravity and the mass of rocks crushing down on themselves in the earth's crust to give us a free source of electricity?
Whilst writing this I just remembered where I saw it.... It was a BBC programme about 'earth quakes', there was a scientist trying to figure out why there is sometimes glowing skies before an earthquake.
Peace Peter Steadman.
What do you think?
« Last post by jeffreyH on 22/04/2014 20:03:49 »
Data on the distance in Schwarzschild radii against velocity would be very interesting. How would this match against expected time dilation? Is there any data on an accretion disk? This looks promising.
« Last post by thedoc on 22/04/2014 19:30:01 »
Can I donate my living body to research?
Asked by Em Green
Visit the webpage for the podcast in which this question is answered.
...or Listen to the Answer or [download as MP3]
« Last post by David Cooper on 22/04/2014 19:22:27 »
Can 40 x 2500mAh AA batteries (cost £100) replace one 100Ah lead acid battery (cost £60)? I'm not suggesting this for use in a car, but for small boats (specifically sailing dinghies) where weight is important. There are electric outboard motors available (sometimes called trolling motors - popular with anglers when noise isn't wanted), but they're usually powered by heavy lead batteries. I would like to be able to use such a motor with a much lighter power source, mainly so that it becomes practical to carry 4 or 5 hundred Ah of power without the weight becoming prohibitive.
Lithium batteries of the kind used in electric cars might be best in terms of grunt per kg, but the problem with them is that you practically have to keep them on life support all the time - they don't like being left flat for long and they don't like being kept full charged either. NiMH looks as if it could be a better alternative, just so long as you always run batteries close to flat before recharging them fully.
My idea would be to use several blocks of 40 AA batteries, putting 8 sets of 5 batteries in series (to add up to 12 volts), with each set of 5 connected in parallel. (By using them in parallel, it should stop them getting hot and lengthen their lifespan.) I know that there can be problems when using multiple batteries if some of them run down before others (they can be reverse charged, and that damages them) but it would be possible to identify the individual batteries that run down most quickly or slowly and move them into other sets more like themselves so that they are better matched.
So, has anyone ever tried doing anything of this kind? I don't want to repeat an expensive experiment that may be doomed to failure from the start if it turns out that combining too many of these batteries together results in too high a battery death rate. Even if running them down in such a block isn't a problem, it may be that trying to charge them as a block will be the killer instead. The purpose behind all this is to make long coastal and river journeys safer and less gruelling, but without needing to carry around a tank of explosive liquid and a noisy oil-spewing machine attached to the transom. It should also be possible to do some of the charging with a solar panel, and ultimately it may be possible to use a sail which serves as a solar panel.
« Last post by thedoc on 22/04/2014 18:30:02 »
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