ninus maximus

he used the wrong equation and if he knows physics , and thats a strong if , he knows he used the wrong equation. he has lost trust in his ability to perform. he can gain trust only by posting the correct equation , and by posting correct results obtained by a correct equation that pertains to the issue. from what I have seen so far he doesnt know diddly , squat. sorry but thats just the way he presents his knowlege of physics. perhaps there is someone in this forum who knows physics that would like to discuss this concept , as it stands no one that knows physics and how to apply physics has stepped forward.

the below equation is the only equation you have provided. is the above equation that uses two seperate masses where you gave a equation? because it is the only equation you have gave! V1f = M1-M2/M1+M2 * V1i WE ARE NOT CHANGING THE MASS WE ARE ONLY CHANGING THE VELOCITY OF THE MASS you said my math violates the laws of physics , then you use an impossibility such as a infinite mass to give a reason why the concept wouldnt work. since mass does not change durring acceleration , it would be more factual to say that your math violates the laws of physics. as I said earlier a simple nay sayer will go to great extents just to prove a point. tie a string to a mass , then drop the mass from a height , does the mass stop as it gets closest to the ground. or does the mass continue to swing until it reaches its apx dropped from height? the speed is zero before you drop the mass. the speed is zero after it swings to the other side and stops , before it swings again. it did not loose a great amount of KE just as the mass in my concept does not loose a great amount of KE passing through the turnarounds. in the above example of a mass tied to a string the mass is propelled by gravity and it is met by the same exact force as it swings upward again. the mass in my concept is propelled by a force and is not met with a opposing force as the mass and string example. no resistance = no loss in KE. little resistance = little loss in KE. perhaps there is someone in this forum who knows physics that would like to discuss this concept , as it stands no one that knows physics and how to apply physics has stepped forward.

jjjjj consider a radio controlled truck , a toy like the ones you can buy at radio shack. you place the truck inside a sphere , the sphere can be seperated like one of the small plastic easter eggs that people put candy in. you turn the truck on so the the wheels of the truck are turning. you place the truck inside the sphere and completely seal it up. then place the sphere on the ground. matter does not cross the boundary of the closed system. and in a closed system only matter is denied crossing a boundary in physics. so the truck wheels turn pressing against the inside of the sphere , the sphere then turns pressing against the ground. the sphere moves , this is called reactionless propulsion because there is no force presented to the ground due to a force that is pressed against the outside of the sphere. A reactionless drive or inertial propulsion engine (also reactionless thruster, reactionless engine, bootstrap drive, and inertia drive) is any form of propulsion not based around expulsion of fuel or reaction mass. there is no mass or fuel being expelled from the sphere. The name comes from Newton's Third Law of Motion, usually expressed as: "For every action, there is an equal and opposite reaction." Such a drive would use a hypothetical form of thrust that does not require any outside force or net momentum exchange to produce linear motion, and therefore necessarily violates the conservation of momentum, a fundamental principle of all current understandings of physics. In addition it can be shown that conservation of energy is violated. In spite of their physical impossibility, such devices have been often proposed in recent history. my my sorry jjjjj I guess it wouldnt work afterall. better luck next time. I guess we understand now LOL

point to where it violates the conservation of energy then big nose "Maths Expert". show the readers how much of an expert you are. just saying it violates something , isnt saying much. and having "Maths Expert" underneath your name only means you have posted alot here in this forum. something anyone can accomplish , by posting meaningless replies such as you do. were not talking about acceleration as the mass enters the first turnaround. momentum is given as kgm/s 1 newton = 1 kgm/s/s so I should have used 7071.1 kgm/s/0d that unit discrepancy can in no way prove the overall concept is flawed. however I immediately followed with this force of momentum of 7071.1 N/0d will be presented to the first turnaround as the mass passes through the first turnaround and changes its momentum from P=7071.1 N/0d to P=7071.1 N/180d you cannot present a momentum to a object. momentum is not a force. the force that an moving object presents to another object is quantified by its mass * its acceleration. so when I used P=mv I was quantifying the masses force due to its momentum. Momentum measures the 'motion content' of an object, and is based on the product of an object's mass and velocity. Momentum doubles, for example, when velocity doubles. Similarly, if two objects are moving with the same velocity, one with twice the mass of the other also has twice the momentum. Force, on the other hand, is the push or pull that is applied to an object to CHANGE its momentum. Newton's second law of motion defines force as the product of mass times ACCELERATION (vs. velocity). Since acceleration is the change in velocity divided by time, you can connect the two concepts with the following relationship: force = mass x (velocity / time) = (mass x velocity) / time = momentum / time Multiplying both sides of this equation by time: force x time = momentum if he was refering to impulse measure in N*s then he was wrong in assumming that the mass would be experiencing an impulse or change in momentum as it enters the turnaround. after it enters the turnaround it will experience a change in momentum however and I clearly showed that change. would you mind telling the readers how much the speed reduces? obviously you have compared my numbers to your numbers , wouldnt you think that the readers would like to know the speed reduction also. and with your proof using your numbers then you can have a basis for your assumption that the concept would not work. Im not sure why exactly you dont think it would work. suppose you and big nose were both in separate boats that weigh the same. and you both weigh the same , one of you two pushes the other ones boat with a force of 100 N one boat will go one way because of the 50 N force that is applied to it and the other will go the other way because of the 50 N force applied to it. if just 1 of you were in a boat and push off from the pier with a 100 N force you would travel further because the pier is not elastic.

just what I thought , you are a simple nay sayer that will go to great extents just to prove a point , normaly these nay sayers will choose to accuse a concept as violating a physical law as you have , yet nay sayers offer nothing viable as a reason. they will type hundreds or even thousands of words so that the readers will think that they are correct , but they only offer lip service as a reason. when a single calculation can prove them right if they are right. thats why they dont use calculations , they arent right , they are simply nay saying. they dont even show where a physical law is suspected as being violated. notice how he took plenty of time telling me that my math was wrong , that I used wrong units. but did he exchange the units for the correct units and then work the calculation , no. he cant prove it wont work using the math side of physics , and its not like it would take a rocket scientist or a "expert nay sayer " to do that. but maybe a expert physicist could use math. they simply say they have been violated. until you can offer something other than lip service as in a mathematical reason my concept stands. and your lip service stands.

I have often wondered why a 90 % efficient gas turbine is not used to charge the battery set vs using a 10-15 % efficient ice engine. a gas turbine is usually around 90% efficient , and running the turbine at a set rpm would be perfect for a generator head designed to run at that set rpm. good point there swansont

will exchange momentum with the railgun+track. yes , and the railgun and track are attached to the spacecraft ! and that exchange of momentum is the equal and opposite reaction of the spacecrafts momentum due to the force applied in accelerating the mass. I am treating this as a collision , in fact 2 collisions. the mass turning around in the turnaround will push the spacecraft , just as if the mass were compressing a spring. although the spring completely stops the mass and then accelerates the mass in the opposite direction , the spring also applies the compression force to the spacecraft. the only real difference is the mass does not stop as it passes through the turnaround in fact its speed does not greatly reduce. thats taking it to an extreme wouldnt you think , I sure hope that you are not simply a nay sayer that will go to any expence just to try and prove a point. I have found no viable reason why this would not work , and I have looked , yet you pick something infinite from thin air and use it. if you have found a reason why it would not work would you mind posting the reason , also I am looking for a viable reason so if you dont mind , please avoid posting violations to physics laws , as simply posting a physical law without the inclusion of your reason the physical law appears to be violated will never prove or disprove the viability of any concept. ie ... F=ma 0=1000kg*1000m/s/s the above shows a physical law violation... vs that wont work because it violates newtons laws of motion. the above does not show a physical law violation. I personaly cannot find a violation of any law of physics in this concept.

Hello swansont proposing that a force of 500 N is used to accelerate the 100 kg mass in the rail gun for the 500 meter distance. we can assume that the direction that the rail gun accelerates the 100 kg mass is 0 degrees in a 360 degree range. since the mass will be changing direction it is best to include direction when describing the motion of the mass using velocity , so I will include direction in the calculations to avoid confussion. the velocity of the 100 kg mass will increase from 0 m/s/0d to 70.711 m/s/0d in 14.142 seconds. Durring acceleration the mass has a average velocity of 35.3555 m/s/0d so the mass travels a distance of 35.3555 m/s/0d * 14.1422 seconds = 500.0045521 meters/0d in 14.142 seconds time. since the masses force of momentum is the product of its mass * its velocity P=mv the momentum of the mass as it enters the first 180 degree turnaround will be P=mv 100 kg * 70.711 m/s/0d = 7071.1 N/0d this force of momentum of 7071.1 N/0d will be presented to the first turnaround as the mass passes through the first turnaround and changes its momentum from P=7071.1 N/0d to P=7071.1 N/180d now the 100 kg mass free floats to the second turnaround requiring 0 additional force as it is not being accelerated and it still has a momentum that is the product of its mass * its new velocity , and it is met by 0 resistance as the mass free floats in zero g to the second turnaround. the mass then enters the second turnaround and presents a force to the turnaround of 100 kg * its new velocity = slightly less than 7071.1 N / 180d I have not included any resistance to movement as the mass passes through the (2) 180 degree turnarounds , however using rolling friction such as .0001 which is common in low friction bearings available on todays market should deliver a useable calculation for speculation purposes. so we can say that there will be a .0001 N loss of the momentum that was the result of the 500 N force * 14.142 seconds used to accelerate the mass initially. for instance the loss of momentum of the mass as it travels through the first turnaround would be 7071.1 N * .0001 N = 0.70711 N Knowing that the .70711 N momentum will be lost as the mass passes through the first 180 degree turnaround , and that there will also be a loss of momentum as the mass passes through the second 180 degree turnaround we can establish that the mass will loose momentum passing through both turnarounds. 0.70711 N * 2 = 1.41422 N and we can establish that the new momentum of the mass after it has passed through both turnarounds will be the product of its mass * its new velocity. to provide for a easier to follow topic we can say that we have established that the mass looses a momentum of 2 N through the entire cycle. we have not established a mass for the space craft yet , so we can assign a mass to the space craft. since we do not yet know the new velocity that the mass attains while passing through the 2 turnarounds we can only give a estimation of the space crafts movement. suppose the space craft has a mass of 5000 kg the force applied to the space craft is 500 N its acceleration is 0.1 m/s/s its initial velocity is 0 m/s/180d its average acceleration is 0.70711 m/s/s/180d its final velocity is 1.41422 m/s/180d the space craft moves a distance and direction of apx 10.000091042 meters/180d in 14.1422 seconds we can establish that the space craft has gained a momentum of apx 7071.1 N/180d therefore your question is certainly one that needs addressing as this will have a grave impact on determining the velocity and distance that the space craft will travel.

hello everyone , I'm new to the forum. and I have a theory of a new type of propulsion for space travel. consider a rail gun that fires a 100 kg mass along a 500 meter rail. then the mass enters a turnaround at the end of the 500 meter rail. having accelerated the 500 meters , it will present a force to the turnaround. the mass turns 180 degrees through the turnaround and begins traveling the opposite direction and free floats in zero g for 500 meters in the opposite direction. it then enters a second 180 degree turnaround and presents a force to the second turnaround. the mass turns 180 degrees again in the second turnaround and now enters the rail gun a second time and it is accelerated even faster the second time. the two forces that are presented to the turnarounds by the 100 kg mass will cancel each other out and the force that was used to accelerate the 100 kg mass will propel the spacecraft. so you end up with 2 positive forces for propulsion ; 1) the force used to accelerate the mass 2) the force presented to the second turnaround by the mass and 1 negative force for propulsion ; 1) the force presented to the first turnaround by the mass this process repeats giving the space ship the ability to accelerate as long as power is used to accelerate the mass. as the mass inside is following a path inside the space ship the momentum of the mass in reference to the space ship is conserved. and since force is not momentum the conservation of momentum is not violated. it is a closed system but matter is not crossing the boundary. the forces that are applied to the space ship from the inside , push the space craft the same way a mass ejection propulsion system works by ejecting or throwing mass out of the spaceship. the force that accelerates the mass inside the space ship is transfered to the space ship which gives the space ship acceleration. nothing needs to be ejected or thrown from the ship for propulsion.