swansont Posted May 10, 2014 Posted May 10, 2014 if you agree with how laws of physics are connected and you can make use of them for good; then you must explain the following since s( displacement) does not change. and work is non conserved then momentum must be non-conserved. correct me if I am wrong with a good illustration if possible. don't just stick to conservation of momentum I know that law. The fundamental law to be breached if this concept has to work is: "center of of gravity of a system of objects will not change its initial velocity or position(if stationary) if external force is not applied" a part of conservation of momentum Work is path dependent, and so is s. You're canceling the path dependence out of the problem. It's like saying 10^49/10^49 must be a huge number because look, there's 10^49! You have a TIME integral, not a PATH integral. At least you now seem to agree you have to violate conservation of momentum for your idea to work. It's not going to happen. There is no software that will tell you it can happen. The only people who will tell you it can happen are crackpots, who don't understand physics (i.e. they will be wrong). We know the below relation between work and momentum p=(W/s)t There are 3 stages of momentum one is initial momentum (p1) which is zero. Second is just after initial path independent force (p2) and third is after path dependent force (p3). p1 - p2 - p3 = 0 if momentum is conserved Assume path independent force is applied in both stages then from the line integral and regular work example done above; we know that work is 4 and -4 for each case respectively. And s is 2 and assume time is equal to 1 then inserting values into (W/s)*t. we get 0 - 4/2 - (-4/2)=0-2+2=0 So momentum is conserved if both forces are path independent. Now take the case with one path independent and one path dependent. From the path integral example done above we know that work of path independent force is 4 and work of path dependent force through arc is -6.28. Inserting these values in equation, 0 - 4/2 - (-6.28/2) = 0 – 2 + 3.14 = 1.14 So momentum is not conserved. So there could be thrust. You can get any answer you want if you just make up the numbers. But that's not physics.
dijinj Posted May 11, 2014 Author Posted May 11, 2014 You can get any answer you want if you just make up the numbers. But that's not physics. these are not just made up numbers these are from calculations i made earlier I am quoting both calculations for your reference if you can not view the image of calculation go to bottom of link https://sites.google.com/site/nonmassexpulsiontypethruster/contact correct me if I am wrong these results are put in following calculations We know the below relation between work and momentum p=(W/s)t There are 3 stages of momentum one is initial momentum (p1) which is zero. Second is just after initial path independent force (p2) and third is after path dependent force (p3). p1 - p2 - p3 = 0 if momentum is conserved Assume path independent force is applied in both stages then from the line integral and regular work example done above; we know that work is 4 and -4 for each case respectively. And s is 2 and assume time is equal to 1 then inserting values into (W/s)*t. we get 0 - 4/2 - (-4/2)=0-2+2=0 So momentum is conserved if both forces are path independent. Now take the case with one path independent and one path dependent. From the path integral example done above we know that work of path independent force is 4 and work of path dependent force through arc is -6.28. Inserting these values in equation, 0 - 4/2 - (-6.28/2) = 0 – 2 + 3.14 = 1.14 So momentum is not conserved. So there could be thrust.
swansont Posted May 11, 2014 Posted May 11, 2014 these are not just made up numbers these are from calculations i made earlier I am quoting both calculations for your reference Then you did the calculation wrong.
CaptainPanic Posted May 12, 2014 Posted May 12, 2014 ! Moderator Note Multiple physics experts have explained all the ins and outs of the topic. This thread has gone around in circles since the very beginning, and now that it has lost all momentum, it is time to close it. Thread closed.
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