ninus maximus

yep , I was thinking of something that would fit in the bathtub. of course a boat would be next I suppose. I was thinking about just mounting a pipe on a wooden board. then mounting a rail gun inside the pipe , let the mass move up in a 180 electrical conduit fitting then float through the air into another 180 electrical conduit , then back into the rail gun. I suppose I would need to build the rail gun first , then the 1st turn. then find out where the other 180 electrical conduit would need to be placed by firring a mass through the 1st 180. I should be able to just give the rail gun and horizontal pipe a little angle to use 2 identical 180's as the mass will want to drop between 180's . thats pretty much the same thing. put it in the tub and fire up the rail gun. if the concept is viable as Im sure it is , then it would also work in space.

I already have , even swansont says it will move , even if it only went around in a circle, its reactionless propulsion because theres no friction in space. I dont think that any TRUE laws of physics are broken however. newton never said what is said today. other people added to what newton said. because they thought their interpretation of what he said was correct. and we have shown here that this can happen in space , and engineers knowing this can happen as a result of mass being forced away from a spaceship both internally and externally can avoid tragic accidents.

it probably would rotate , but it would have linear motion as the mass accelerates down the accelerator. it would then cause the rear of the spaceship to move slightly to the right if you were observing it from the rear of the spaceship , then it would stop the spaceship , then it would slightly move the rear of the spaceship to the left. but if opposing sets of accelerators and masses are used then it would not move the spaceship lateraly or rotate , it would have linear motion then it would stop. then it would just sit there as the masses free float. then as the masses pass through the second turn the spaceship would again move in the (+) direction. so you cant just say something will not work and expect people to believe you , just as you cant just say something will work and expect people to believe you. but showing how something might or might not work will allow people to make there own decissions.

so your saying that the equation you used shows that the mass reduces in velocity? to -32 m/s vs my -40 m/s given that how would that velocity reduction of the mass show that the scenario is impossible? the mass can easily float to the other turn. there it will push the spaceship in the (+) direction again as it passes through the 2nd turn. then the mass will have a velocity of < +32 m/s as it enters the accelerator again. I used your equation again and I get +26 m/s through the 2nd turn. so the mass has another velocity reduction from -32 m/s to +26 m/s. I can see that it is a correct scenario even if the mass does slow slightly through the 2 turns , Im not sure why you say the scenario is impossible however. I would like to point out that these days friction can be almost completely negligible if using a almost frictionless surface such as an air table like the one used at MIT. of course there are more cost effective methods to reduce friction , and in my original example I had noted that I had not included resistance.

100kg * 40 m/s = 4000N this is not how you calculate force. what you calculated there was momentum, not force. the force comes from whatever its colliding with. again you have the forces all wrong. it didn't. when something changes directon it does not stop. it changes direction , and that is all. the mass does not stop it turns. 4000N resistive force to stop it in 1 second = 4000N-s whats the difference? are you saying that the mass does not go through the turn?

so you are saying that the mass m1 stops? and that the mass m1 then accelerates in the opposite direction? 100kg * 40 m/s = 4000N it could be stopped but it would require a resistive force to do so , if there were enought resistance , and it would require 4000N resistive force to stop it in 1 second , but where does it get the needed force to reverse? ie...100kg * 32 m/s = 3200N plus there is the resistive force that the mass met entering the turn so we need to include that also , there is a resistive force of 4000N that the mass meets as it accelerates backwards , so now the mass needs a force of 7200N in order to achieve 32 m/s in the opposite direction as it backtracks. it only had 4000N when it entered the turn , how did it gain 7200N by entering the turn?

Because it shows that the 100kg mass stops and then backtracks , there is no obstruction causing the mass to stop , and there is no additional force causing the mass to accelerate in the opposite direction. if it looks fine to you , then perhaps you can explain what the result tells you. what does v1f describe. m1 or m2 or both? is it the final velocity of m1 or m2? if it does describe a mases velocity , how did you determine which masses velocity it describes. only the velocity of m1 is given in the equation , so just exactly how does the equation know what the velocity of m2 is? one other point I would like you to touch on , there is no resistance force of any kind mentioned in the equation , so in that case the mass would not even slow down. but the results I get show the mass stops then goes backwards it backtracks the way it went into the turn , Im not saying its momentum changed Im saying that it never completed the turn , it reversed its momentum and headed backwards. thats not possible. could you elaborate a little on how you believe the answers to be correct?

I used your description of the equation and gave it a try. m1 = 100kg m2 = 1000kg m1-m2 / m1+m2 *v1i = v1f 100-1000/100+1000 * 40 m/s = v1f -900/1100 * 40 m/s = v1f = -32.727272727272727272727272727273 m/s am I to believe that the mass accelerates from 40 m/s to -32.72 m/s? because that is the results from your equation. for that to happen a force would need to first stop the mass , then accelerate the mass in the opposite direction. how exactly do you use that formula if the above is not correct? why dont you work the equation for me , because I cant see any possibility of that answer being correct.

a 80 kg mass traveling at a velocity of 18 m/s collides with a 80kg mass traveling at a velocity of 18 m/s KE=.5 (m1v)^2=12960 j KE=.5(m2v)^2= 12960 j non elastic collision and boom energy destroyed. the two masses merge but they dont have velocity so no kinetic energy.

yes , the wiki deffinition is vague but it is all that we have to work with. perhaps they need to clarify the deffinition to include a stipulation that stipulates that any reactionless drive cannot react against matter because there is matter everywhere even in deep space. this way there would never be a possiblity of ever achieving a 100% reactionless drive. there isnt anything to misunderstand , its just basic physics. the example I used was a force of 500N applied to a 100kg mass and a 500N force applied to a 1000kg mass one force pushing a mass and one force pushing another mass. theres nothing "technically, demonstrably inaccurate " about that. then the 100kg mass passes through a turnaround theres nothing "technically, demonstrably inaccurate " about that. the 100kg mass goes through the turn and applies a force to the 1000kg mass causing the 1000kg mass to stop. theres nothing "technically, demonstrably inaccurate " about that. assuming no friction , and friction can actually be negligible if engineered correctly , the 100kg mass would exit the turn at the same velocity it entered the turn only in the opposite direction. theres nothing "technically, demonstrably inaccurate " about that. in fact all of the above are technically, demonstrably accurate.

I personaly dont think that the sphere is a reactionless drive , it is the deffinition that causes me to say that it is. I had a perfect example of a inertial drive in another thread but it was closed because the concept was not understood by those in this group and name calling and insults resulted from the lack of understanding. that missunderstanding resulted from swantsonts use of the below equation that requires the weight of the mass changing , but the weight of the mass does not change. notice the M1-M2 in the above , this calcuates a change in velocity due to a change in mass , not a change in the masses velocity becauses it encounters a turnaround. this equation can only be used to describe the change in velocity a mass will undergo if its mass changes while it is moving. I would like to discuss that concept but it seems that theres no point in trying. this forum seems to restrict thought if the thought doesnt fall within the moderators realm of reality be it reality or not.

I believe that the deffinition on the internet does not stipulate friction. it only stipulates that it is "not based around expulsion of fuel or reaction mass." nothing comes out of the sphere , so the sphere qualifies as a reactionless drive. there was a time when the sphere being a closed system would violate newtons laws because the sphere could move without a external force being applied to it. this time has passed , but I remember it well. back then , nothing could cross the boundary , work , energy , were bound inside a closed system. now work and energy can pass the boundary , only matter is denied. this too will change.

I suppose that the above shows that you are abiding by the rules by giving me an example of how I should conduct myself. Bollocks is a word of Anglo-Saxon origin, meaning "testicles". The word is often used figuratively in British English, as a noun to mean "nonsense", an expletive following a minor accident or misfortune, or an adjective to mean "poor quality" or "useless". Similarly, the common phrases "Bollocks to this!" or "That's a load of old bollocks " generally indicate contempt for a certain task, subject or opinion. Conversely, the word also figures in idiomatic phrases such as "the dog's bollocks" and "top bollock(s)", which usually refer to something which is admired, approved of or well-respected. Bollocks I was suspended for a week for doing what you just did , now will a moderator please perform their duty and warn swantsont and begin to count his violations to the forum rules , that is if the rules also apply to the moderators. I think that what is good for the goose is also good for the gander. now to continue. the earth is external to the sphere , but the earth does not apply a force to the sphere if the truck inside the sphere is not turning its wheels. the sphere will not simply move without the trucks wheels turning. so the force is comming from within the sphere. not from the earth. the force aplied by the trucks wheels are the action and the inside of the sphere provides the reaction force , and this force is transfered to the sphere as a action force and the earth provides the reaction force. it is a closed system and according to the deffinition of a reactionless propulsion drive , it is a reactionless propulsion drive. reactionless drive but it may be that you have a link to a better deffinition of a reactionless drive that you could share.

I dont want to argue , only dissagree if that is allowed. In thermodynamics, a closed system can exchange heat and work (for example, energy), but not matter, with its surroundings. the sphere is sealed , matter cannot be exchanged. closed system nothing comes out of the sphere and reacts with the spheres surroundings so the sphere is an example of reactionless propulsion.

I'm dissagreeing with you swansont so I suppose I will be banned or suspended again , but anyway for the sake of Physics here goes. The Sphere does not have a external force acting on it. the sphere is a closed system. and the friction is not a external force it is the result of the internal force.

I was curious as to why so many seemed so interested in commenting off topic about my concept. so I checked swansonts friends list to see if there might be some background traffic occurring. low and behold , it appears that I was suddenly looking at a list of moderators that were defending swansont. which tells me that what we have here is more of a social network than a physics forum. and if the moderators are the only ones responding to post and if someone doesnt agree with a moderator because they dont agree with the moderators logic , and then all the moderators friends and fellow moderaterators suddenly jump in and defend the moderator which eventually leads to the person being banned as I have also checked out in the banned member list , or should I say the moderators trophy room as that is more what it resembles. I suppose you definitely have physics JUST THE WAY you want it. none of the intruding posters added any content to the topic , that is considered trolling isnt it. there main intent was to defend swansont because I questioned his refusal to supply a correct equation that would fit into the discussion. I have noticed one thing in common that they dont have is a failure to communicate with each other. listed in order of apearance. swansont---Sorry swansont is a moderator/admin and you are not allowed to ignore him or her. BigNose----Sorry Bignose is a moderator/admin and you are not allowed to ignore him or her. insane_alien ---- on my ignore list mooeypoo---Sorry mooeypoo is a moderator/admin and you are not allowed to ignore him or her. Moontanman ---- on my ignore list Charles Langley - not friends with any of the above , so not ignored. -----------------swansont------FRIENDS--LIST--------------- mooeypoo BigNose insane_alien Moontanman -----------------mooeypoo------FRIENDS--LIST--------------- ydoaPs insane_alien Moontanman swansont -----------------insane_alien--FRIENDS--LIST--------------- mooeypoo -----------------Moontanman----FRIENDS--LIST--------------- mooeypoo -----------------BigNose-------FRIENDS--LIST--------------- ydoaPs -----------------ydoaPs-------FRIENDS--IN--COMMON--LIST--------------- BigNose insane_alien mooeypoo Moontanman swansont

Done , its a waste of intellect conversing with you guys , you have none to share. I suppose there may have been intellectuals in this forum in the past , but like me they realized that this forum is a waste of their time. you guys drove the smart smart people away . leaving only stupid smart people.

I highly doubt that any one I have conversed with on this forum has a PHD in math or physics. I surmised this by their lack of the USE of math , and their heavy reliance on philosophy , things others have said. if they were PHD's they would know how to use math to explain their side of the discussion. thus my scientific summation of there abilities.

a picture is worth a thousand words , now this forums intellectual properties has been defined. a bunch of school kids on summer vacation , mingled with a few elderly people who just happen to be here for some reason. Merged post follows: Consecutive posts merged did you look at the MIT video , the momentum of the steel ball did not change until gravity accelerated , slowed and stopped it. still it presented a force that is the product of its mass * its velocity to the table. If any one in this thread lacks comprehension of physics , it is not me. if the turnaround drains so much momentum then why does the steel ball accelerate to exactly the same speed when the ball is just dropped?

and you guys call this a science forum? Moontanman , falling off of a mountain is not similar to my concept , a more similar example would be a race car accelerating from zero m/s to 70 m/s then before he enters the first turn he takes his foot off the gas pedal. his car then coast through the first turn. his car then coast to the next turn. his car then coast through the next turn. his car still has momentum. he then accelerates his car again. and because his car has a initial velocity his car will gain more acceleration before his car enters the first turn. so when his car enters the first turn his cars velocity is greater than 70 m/s its really extremely simple to understand , I am about to give up on you guys as you dont appear to be very inteligent at all. and perhaps you should rename this forum to the pseudo-scienceforums.net I dont see any science in this thread comming from you guys , only brainwashing results. I see myself being banned from this forum , because you guys cant comprehend physics. if thats the case , then so be it , this way you guys can maintain your pseudo-science image. and I can find a forum where physics is comprehended.

well then insane allen why dont you put some numbers into the equation and give it a try , let me help you. V1f=M1-M2/M1+M2 * V1i V1i = 70.711 m/s M1 = 100 kg there , now solve for V1f I want to see what you will use for M2 and when your through explain the following. The law of conservation of energy is an empirical law of physics. It states that the total amount of energy in an isolated system remains constant over time (is said to be conserved over time). I think it is said that our universe is an isolated system. but there are none within our universe. perhaps you can show me a isolated system.

absolutely not !!! Im showing conservation of momentum. point to where Im not! the item he questioned shows conservation of momentum. he said , she said , why not just stay out of it. and let swansont mind his buisness , and you can mind yours.

yes , it is. thats why I question his ability to apply physics equations to find solutions to problems. L2mdAvdPhT4 Collisions involve forces (there is a change in velocity). Collisions can be elastic, meaning they conserve energy and momentum. in my concept the mass passing through a turnaround is a elastic collision. the mass conserves its energy and its momentum because it is not met by an opposing force. other than a slight almost negligible resistance force as it presses against the turnaround while passing through a turnaround. you cant?

having a job as a physicist does not prove you have a working knowlege of physics , if you have a job as a physicisist and dont comprehend physics you can kill people. having a job as a physicist and not knowing physics and how to apply physics only sudgest you used cheat sheets in college.

wow , i was right you dont have a clue. none of the mass in the entire system changes. only the velocity of the masses changes. the equation you used deals in changing masses , not changes in velocity. ie............... the changes that would occur in velocity if the mass changes in magnatude durring acceleration. the only way to get V1f = -V1i momentum ---- your talking about momentum ------ momentum ----- is not a force. if I drive my car south at 50 mph then turn the car around and drive it north at 50 mph I get V1f=-V1i if you will take the time to read the post and comprehend them before you reply to them with your expert opinion you might save a little face. I stated that I had not included resistance yet , as the mass passes through the two turnarounds as follows