Hi All
I am working on a device to control both intensity and direction of gravity, note. My test vehicle weighs 1,035 pounds, and adding my weight makes 1200 pounds total (544.32 kg). Any thoughts on the gravity power I will need to convert in order to achieve 5,000 feet altitude (starting from 803.76324 feet above sea level)? Also... the power I will need after redirecting a portion of gravity 90 degrees in order to reach mach 2. I do want to go fairly fast, note.
JC the elder
sorry, the secret planetary governing body has ruled that unsanctioned manipulation of gravitational forces will not be permitted.
My test vehicle weighs 1,035
that should say 'masses' not weighs.
why limit yourself to 5,000 ft?
Also... the power I will need after
sorry mach 2 lateral will violate noise ordnances in most planetary airspace jurisdictions. you are limited to about mach .98
I do want to go fairly fast, note.
several power sources are available. a few are:
the spindizzy the Bergenholm (inertialess) drive the improbability field drive. the TARDIS matter/antimatter converter to transwarp drive. one early method used grand pinot wine (but that was simple reaction mass)
Sorry, your technical jargon is amuck. Gravity results in a force, not power. And moving to a different altitude takes energy--power is merely a measure of how fast the energy is applied. And the same for velocity , it takes energy, boy, and that's spelled e.n.e.r.g.y.
If your dream is to be a SF writer, your really need to get the technical jargon right, or you'll be laughted out of your local Trek lodge.
Exactly. Basic. Point is, I don't want to apply the energy to fast and I need to measure the rate of flow. Please try to understand the problems involved here.
Again, I want to go fast, but not too fast, and the rate of flow is key to this. I figure mass is the key, since the energy of gravity is pretty much constant for a given altitude. Perhaps this will help you to understand the problem:
velocity = sqrt (KEmax/mass)
JC the elder
Yes, quite astute of you. Without gravity I shall have to contend only with the masses. Recalculations are in order.
JC the elder
No faith in my device, eh? I'm marking you off my list for contributors to the project.
JC the elder
Are you for real?
...
Newtons 2nd law plus a little calculus will let you play thru various scenarios.
No, but I'm working on it. :-)
JC the elder
You do realize that Gravity is not a force. Its more like Terrain, and matter that congregates together is a result of the matter falling into a pot-hole if it were. Since you cant reshape the terrain you cant control it.
Sorry I failed to notice your question earlier. The reason is that I'm afraid of heights.
JC the elder
Yes, when matter falls in a pot-hole it just levels up the terrain. The equation is
pot-hole level = Fill-in/pot-hole volume
JC the elder
---------- You know your height above sea level to 8 significant figures? Wow! How do you get that sort of accuracy? Just because your calculator gives you that many digits doesn't mean that it is meaningful.
As to power, it may be a milliwatt or a gigawatt- it all depends on how soon you want to get to altitude. I suggest that you look at "energy" not "power".
I hope that you are trolling
- --
Don Kelly @shawcross.ca remove the X to answer
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snipped-for-privacy@noaspamadelphia.net
perhaps no too afraid, after all you head does seem to be in the clouds :)
Actually to 12 significant figures, but I rounded, note.
KEmax=Power*mass/Resistance. I would like to get to 5,000 feet in about 1 minute. Just a leisurely 60 MPH average, note. Suggestions?
JC the elder
Yes, I am attempting to find it a more lofty and *dignified* location... :-)
JC the elder
snipped-for-privacy@noaspamadelphia.net
Ha ha, are you suggesting it is better "in the clouds" than "up your ..."
Goggle Dialitic Crystals
MG
To get to height h with a mass m the "energy" required is mgh Consideration of KE is not required. This is the same with any source of energy input including the ways we do it now. The same story is true for getting to mach 2. However, one must consider the efficiency of the device so that the actual energy and input is larger.
As to KE:- it appears that you have mixed up some factors here.
KE =1/2(mass) *( velocity^2)
which is definitely NOT power*mass/Resistance. Have you mixed up v for velocity for v for voltage?
I was politely giving you the benefit of the doubt when I suggested that you are trolling. By the way, how can you know your base elevation (rounded off) to 1/10000 of an inch ? This is in line with the rest of the nonsense- which has, by the way, nothing to do with alt.engineering.electrical.
Bye,
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