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Thread: "Your way of stopping an asteroid the size of Texas from hitting Earth."
Posted by: AvengerTam1010
launch texas at the asteroid
Posted by: Mutoid Log
Your maths and physics are complete nonsense and there's no reason for me to look further into it. The very fact that you wanted to figure out a force in newtons invalidates everything you tried to do. Sorry, but your post sucks, hard. Nothing personal.
You did get the numbers we need, though. We have the mass, we have the velocity. E = 11/2 * 0.0097kg * (905 m/s)^2 = 43694.98375 J =~ 44 kJ of energy.
I've tried to decipher what you did, and this is what your post consists of:
You assume that the bullet starts at velocity 0, accelerates during the entire flytime, with the average velocity being the 905 m/s. Based on this ridiculous assumption (in real life, halo canon and halo gameplay, the entire acceleration of the bullet happens inside the weapon), you calculate the acceleration of the bullet correctly. Then you use this acceleration to calculate a force, again correctly. However, this force is the force that must be given to the bullet to make it accelerate, and has nothing to do with the final impact.
So, it appears that you did manage to calculate something proper: the required propulsion force output of a rocket-powered AR bullet. Too bad that has nothing to do with spartan shields.
Sorry, but your solution isn't really much use. You could either a) Work out the applied energy required to take out a Spartan's shields or b) Work out the applied force required to take out a Spartan's shields.
I chose the latter. If you were approximating using plasma bolts (don't know why, using bullet values is easier?), you'd obviously chose to approximate using energy.
Here is an example (albeit, a silly one): Imagine some kind of uber-robot who can take out a Spartan's shields in a prod. It's going to need a certain balance of mass and acceleration to generate the force needed, i.e. APPROXIMATELY 3,520 Newtons of applied force. I know what you're thinking: "BUT WAIT! THAT'S THE FORCE REQUIRED!!!!1". Well, at a mere fifty metre range or a prod, factors such as drag aren't really going to do much to an approximated value, so the force in NEARLY equals force out. For the approximated value I achieved, it's negligible.
So, back onto the robot example thingbob. That robot's going to need one of the factors - either mass or acceleration - for his prodding finger to eliminate the Spartan's shields. It'll either have to have a damned heavy hand or a hugely fast prodding hand, assuming he's going to fully eliminate the shields with a considerable 3,520 Newtons of force. Technically, the robot could use a 1,000,000,000 Kg prodding finger moving at just 0.1mph and still inflict the same force - so long as the force was applied through a finger with a surface area that entirely made contact with the shields. (Would this mean one would have to take into account the surface area of the point of a bullet? Head is swimming.)
ANOTHER example to help what I'm explaining there ^ : You brush past someone and your shoulders barge at 5mph. Not much force applied, really. Then, at 5mph, you walk straight into someone. Bam, full force applied, captain.
I hope this clears things up if the OP was a bit rushed.