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The force generated upon the release of this air is: F=[m*Ve]+[(PtPa)*Ae]
Because we don't have a cone, m*Ve is zero. Therefore, the amount of thrust at a full tank is: F = (PtPa)*Ae F = (10132kPa101kPa)*0.00064516m^2 F = 6471N one load after another... theres something VERY important when calculating force  YOU NEED MASS  Try looking up the unites within a pascal, moron. A pascal is a N/m^2. A newton is a kg*m/s^2. kg, BTW, is the mass... Funny, you're the one who insists on using pressure as a proxy for force, and are not even aware of what the units of pressure represent...LOL F=MA
not
F=whatever you want it to be.
And, if you plug the units for pressure and area into a formula, you get...exactly what F=mA gives you... it looks like your using the initial pressure in the tank (pt)then subtracting the initial pressure in the pipe (pa?)then multiplying that number by the nozzle area (Ae?).
so where are you going with this ?
That's how you calculate the force produced by a fluid moving from a region of higher pressure, to lower pressure. I though that would be selfevident... Because we don't have a cone, m*Ve is zero so since the mass comes out of a tube as you said , the mass has no velocity? Nope. The air that comes out of the tank is still at pressure when it leaves the tube connecting the pipe to the outside world. If you have a coneshaped nozzle you can harness that pressure, and use the pressure to accelerate the gas within the nozzle. As the air expands through the nozzle accelerates it, and the degree of acceleration determines the force generated. Ve is measured relative to the speed of the air before it enters the nozzle, and if that air doesn't accelerate, Ve = 0. Any mass * 0 = 0. it just covertly sneeks through the nozzle or perhaps it is being transported as they do on star trek !! Nope, it obeys the well understood laws of fluid dynamics  the exact same laws used for designing everything from aromatizers to rockets. you know the mass of air. you know the force , its 100 psi. (believe it or not) My god, you are dumb. Pressure is not a measure of the amount of force when the air is released  it is a measure of the force exerted on the tank by the air. The two are not the same  a 1000L tank at 100PSI will generate far more force than a 1L tank at 100PSI. Strange, isn't it, how these simple concepts elude you! all you need now is the acceleration.
a=f/m acceleration = force / mass
SP ) standard Pressure dry air has a density of 1.2754 kg/m3.
this will give you the mass and the initial velocity for your formula. What the hell do you think F=[PtPa]Ae calculates? EXACTLY WHAT YOU JUST WROTE! What you are missing is the area  pressure is FORCE PER UNIT AREA NOT FORCE. now you have the rest of your formula..
No, you've just shown that you cannot read mathematical formulas, and that you don't understand the difference between pressure and force. Pressure is not force, and more than kinetic energy is momentum. but you now have the full formula that you obviously already knew , you just didnt want to allow others to see the outcome of actual numbers , you only want them to believe your twisted math and the endless jargon that flows from your knowlege of what other people said.
I presented the numbers, for every one to see. You, on the other hand, simply displayed your inability to read a mathematical formula. LOL Bryan
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reactionless propulsion is even being experimented with today. heres a good example , the box can be considered a closed system there is no external forces that apply agaist the box , but the box moves. click here to watch video did you even watch the video  the tracks the box travels on are clearly attached to the switch. Having the box on an electrified track is hardly a closed system. Nor, for that matter, is having it on the ground. Ground = friction. Bryan
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your formula will not work without the (((((( MASS ))))) of the air. and your formula will not work without the velocity of the ((((((( MASS )))))))  F=[m*Ve]+[(PtPa)*Ae] Because we don't have a cone, m*Ve is zero. Therefore, the amount of thrust at a full tank is: F = (PtPa)*Ae F = (10132kPa101kPa)*0.00064516m^2 F = 6471N  suppose the mass = 1.29 and the velocity is 5000 m/s 1.29 * 5000 = 6450 we begin with a much larger number than zero ... 6450 + 6471 = 12921 Ntwice the amount that you first posted. however even your math below is incorrect. F = (10132kPa101kPa)*0.00064516m^2 F = 6471N the answer to the above is 6.471 not 6471 and your calling me a moron? LOL and later you say that I cannot read a formula , when it is you that seems to lack that ability , not in just comprehending them but also in performing calculations with them. they dont just put elements in formulas because they dont have anything else to do. they put them there because they are a critical part of the formula , otherwise they would have left them out in the first place. without mass you cannot have a force. and by assumming that the air has both no mass or velocity will never render a close solution to the formula. and how does the formula know how heavy air is? unless you tell it.... theres no sence in calling people names , it just shows your inability to deliver a inteligent responce therefore you revert to name calling. did you even watch the video  the tracks the box travels on are clearly attached to the switch.
yes I did watch the video , and if the wires are attached to the tracks it only shows that he didnt want people thinking that the box was being pulled by the wires. it obviously has an electric motor in it , we cant transmit electricity through the air anymore like Tesla did , and maybe the electric motor is too big and power consumming to have a battery in the box, after all he is just demostrating an reactionless effect. meaning there is no direct force applied to the box that leaves the box as thrust or a direct push with a force.
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your formula will not work without the (((((( MASS ))))) of the air. and your formula will not work without the velocity of the ((((((( MASS ))))))) and your calling me a moron? LOL Yes, I am. And you just proved me right. Mass is an intrinsic value needed for the calculation of pressure. You cannot calculate pressure without first knowing the amount of gas molecules present. pV=nRT, assuming an ideal gas. Ergo, if you know the pressure, you can calculate force without knowing the mass of the gas itself. And since it bears repeating: Pressure is measured as force/area. In Si units that is newtons and meterssquared respectively. A newton is a kg*m/s^2 You'll notice mass  kilograms  is an intrinsic part of force. And since force is an intrinsic part of pressure, mass is also an intrinsic part of pressure. without mass you cannot have a force. and by assumming that the air has both no mass or velocity will never render a close solution to the formula. Since I assumed no such thing, your point is? The formula is so simple, I'm amazed you still don't get it. NASA gets it, high schoolers making bottle rockets get it. Heck, even revlon (makers of lipstick, amoung other things) gets it. Apparently, you don't. When talking about a compressible fluid emerging from a pipe or other closed structure, the force it produces is equal to the the force generated due to the pressure difference pushing the fluid out of the pipe, PLUS the energy harnessed from the expansion of the gas (if you have a nozzle to take advantage of that expansion). F=[m*Ve]+[PtPa]Ae [m*Ve] calculates the energy derived from the nozzle, and is determined by how effectively the nozzle uses the pressure to accelerate the gas. No nozzle and this value is zero. [PtPa]Ae calculates the force generated due to the pressure difference, taking into account the size of the opening between the highpressure and lowpressure regions. No mass is necessary, since mass is already accounted for by pressure. The above IS rocket science, but its still pretty simple science. theres no sence in calling people names , it just shows your inability to deliver a inteligent responce therefore you revert to name calling. LOL, pot meet kettle... Bryan
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F=[m*Ve]+[(PtPa)*Ae] do you see the (+) sign in the above formula? lets do this one number of oranges = 0 + 0 lets put some numbers in there number of oranges = 8 + 8 = 16 then lets do it your way. number of oranges = 0 + 8 = 8 all you are calculating is pressure times the nozzle area. what if the fluid is methane do you think that you will get the same result in you formula if worked the way you say it can be worked? methane .717 kg/m^3 liquid methane 415 kg/m3 using 5000 m/s methane .717 * 5000 = 3585 liquid methane 415 * 5000 = 2075000 you cant leave this out of this or all you have is this suppose each of the above also have 1000 psi pressure we get your exact result for each of the above fluids no matter what fluid we use. and where does the fluid velocity come from? if you dont include it. this is rocket science F = q × Ve + (Pe  Pa) × Ae where F = Thrust q = Propellant mass flow rate Ve = Velocity of exhaust gases Pe = Pressure at nozzle exit Pa = Ambient pressure Ae = Area of nozzle exit not what you put up. The product of q*Ve is called the momentum, or velocity, thrust and the product (PePa)Ae is called the pressure thrust. you are only including the pressure not the momentum of the mass , because you didnt include mass. q = the amount of mass that passes through the nozzle. Ve= the velocity of the mass that exits the nozzle. you left out the most important part of the formula.
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OK, you want maths?
Referring to my previous picture
p1 = momentum of plane p2 = momentum of air pushed back by plane's propeller p3 = momentum of pipe
total momentum of the whole system ptotal = p1 + p2 + p3
Initially everything's stationary and the plane's in the middle: p1=0 p2=0 p3=0 ptotal=0+0+0 = 0
Then the plane starts flying: p1=10 p2=10 p3=0 ptotal=1010+0 = 0
Then the air from the propeller hits the pipe p1=10 p2=0 p3=10 (notice that the pipe is moving!!!) ptotal = 10+010 = 0
Then the plane hits the end of the pipe p1=0 p2=0 p3=10 + 10 = 0 (now the pipe stopped moving) ptotal = 0+0+0 = 0
TOTAL MOMENTUM REMAINS ZERO AT ALL TIMES!!! You might be confused by the small movement of the pipe, and think that's propulsion, but turn down think street and work out why it's no use for a spaceship.
Last edited by kallog; 06/04/10 04:30 AM.




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F=[m*Ve]+[(PtPa)*Ae]
do you see the (+) sign in the above formula? Yes, and had you read what I wrote above you'd understand why its there  the force generated as a compressed gas moves from a region of higher pressure to lower pressure is equal to the sum of the force generated by the expansion of the pressurized gas against a nozzle (if there is one) PLUS the force exerted due to the pressure differential. Strangely enough, when describing sums mathematically, we use '+' signs. I know, its shocking  using the appropriate mathematical symbol to represent its concordant mathematical procedure. all you are calculating is pressure times the nozzle area. Which described the force produced given the pressure differential and the area connecting the highpressure region to the lowpressure one. If you think that is wrong, certainly you can provide the mathematical evidence it is wrong. what if the fluid is methane do you think that you will get the same result in you formula if worked the way you say it can be worked? Assuming it is under the same pressure and doesn't go through a phase transition  yes. The force generated due to a pressure differential is always the same for a given pressure difference and crosssectional area of the opening. you cant leave this out of this or all you have is this But I can leave it out m*Ve specifically deals with the force generated as the gas expands and pushes against the nozzle. Since we don't have a conical nozzle to harness the energy, Ve is zero. m*Ve will always be zero if Ve itself is zero. Basic math, the fact you are arguing about it says wonders about your understanding of what is quite simple math. suppose each of the above also have 1000 psi pressure we get your exact result for each of the above fluids no matter what fluid we use. Exactly  although the amount of gas required to achieve 1000psi will vary gastogas. this is rocket science F = q × Ve + (Pe  Pa) × Ae
where F = Thrust q = Propellant mass flow rate Ve = Velocity of exhaust gases Pe = Pressure at nozzle exit Pa = Ambient pressure Ae = Area of nozzle exit
not what you put up. LOL, and once again paul shows his total ignorance of math. The formula you just wrote is THE EXACT SAME ONE I HAVE BEEN WRITING ALL ALONG. Substituting one symbol for another  in your case, q for m  doesn't magically make it another formula. If anything, your good for a morning laugh. Oh, and by the way F is force, as in newtons (N). You need to integrate to get thrust (which is in newtonseconds [N*s]). The product of q*Ve is called the momentum, or velocity, thrust and the product (PePa)Ae is called the pressure thrust. you are only including the pressure not the momentum of the mass , because you didn't include mass. Nope. q*Ve (m*Ve in the way I wrote it) is the force produced by the nozzle, due to the expansion (and thus acceleration) of the gas by the conical shape of the nozzle. The actual pressuredriven force is accounted for by the [PtPa]*Ae. Ve is calculated using: The only real elements of that you need to understand is the ratio: 1[Pe/Po] [PePo] is the pressure difference at the entrance verses exit of the nozzle; essentially you are calculating the force generated by the change in pressure as the air expands in the nozzle. In our case there is no expansion, as the nozzle is just a hole. Thus, Pe = Po, ergo: 1[Pe/Po] = 1[1] = 0 Since the remainder of the formula is multiplied by that value, the Ve will also end up being zero. you left out the most important part of the formula. Nope, I understand the formula and thus calculated it correctly. Since Ve is zero, m*Ve (or q*Ve in your case, although its the same bloody thing) is also zero, ergo that part of the math can be ignored. Its funny though, you lecturing me on math, when you seem to have completely missed the fact that you "corrected" my math by providing the same formula and changing one letter. LOL, stupid is funny! Bryan
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I used to think that you were just refusing to admit that the pipe would move , by not putting mass and velocity into the formula. now I just think your a complete bone head .... LOL a force caused by a object that has a mass the same as a car with a velocity of 5000 m/s would cause a greater force than a object that has the mass of your brain , slightly greater if the brain were someone elses that had something useable in it. you really are a college "gadget". the only reason that the pipe could move is the momentum of the mass moving inside the pipe. ie.. in clasical mechanics momentum newtons second law F = (d/dt)*(m*v) using your boneheaded method by removing the (m*v) in the above formula will result in force = distance / distance*time YOUR WAY PRODUCES NO FORCE , BECAUSE YOU USE NO MASS if it moves slowly , unconstricted then not much force is produced. if it moves faster by constriction , then much more force is produced. Im right and your wrong , your just too boneheaded to admit it. Its funny though, you lecturing me on math, when you seem to have completely missed the fact that you "corrected" my math by providing the same formula and changing one letter.
LOL, stupid is funny!
no I didnt I included the correct formula , and I included all the elements of the formula, if you gadgets would have learned this stuff or at least that you cant remove portions of formulas to get a correct answer we wouldnt have so many buidings and bridges collapsing. thankfully your just a programmer that only deals in imagery , so you cant really make things fail. I did correct your math several times though. and you know it.
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the only reason that the pipe could move is the momentum of the mass moving inside the pipe. Exactly  and [PtPa]Ae calculate the portion of that derived from the differential pressure. if it moves slowly , unconstricted then not much force is produced. if it moves faster by constriction , then much more force is produced.
Im right and your wrong , your just too boneheaded to admit it. No, you're just too stupid to understand what that formula is calculating. There are two sources of force that move the pipe  the pressure differential and the expansion of the gas once it leaves the pipe (captured by the nozzle). To calculate the former you need only the differential pressure and crosssectional area connecting the two regions. To calculate the later you need the mass and the acceleration provided by the nozzle. Its funny though, you lecturing me on math, when you seem to have completely missed the fact that you "corrected" my math by providing the same formula and changing one letter.
LOL, stupid is funny!
no I didnt I included the correct formula In that case you can certainly show where your formula varieties from mine. Of course, we both know you won't, as your formula is the exact same one I have been using since day 1. Swapping symbols doesn't change the formula itself. Bryan
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I already have shown you , bonehead. the q*ve = a sum then you add that sum to the (pepa)*Ae LOL F = q × Ve + (Pe  Pa) × Ae you are leaving out one of the sums. There are two sources of force that move the pipe true , ..... it just so happens that you are leaving one of them out !!! so when you add them together your answer is incorrect because one of them is zero. the pressure differential and the expansion of the gas once it leaves the pipe (captured by the nozzle).
Wrong ... the pressure differencial and the expansion are the (PePa)*Ae side of the formula... the two forces are the mass flow rate * velocity (q*Ve) + the expanding gasses (PePa)*Ae F = (q*Ve)+(PePa)*Ae "you're just too stupid to understand " you have already stated these previously. Momentum, BTW, is a specific physical quantity equal to mass * velocity.
so why did you leave out Momentum , mass * velocity Sorry, you don't. Missing half the equation does not equal a valid answer.
and why did you leave out half the equation. perhaps to try and twist around a result? one that would prove you wrong... There are two sources of force that move the pipe There are two sources of force that move the pipe There are two sources of force that move the pipe There are two sources of force that move the pipe There are two sources of force that move the pipe
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I already have shown you , bonehead.
the q*ve = a sum then you add that sum to the (pepa)*Ae
LOL And that is different from my formula how? The plus is in there  are you blind, or just stupid? F = q × Ve + (Pe  Pa) × Ae
you are leaving out one of the sums. No, I am not. m*Ve is zero when you do not have a nozzle which allows for expansion. Ergo m*Ve will always be zero in the case of your perpetual motion machine, as you specified the air was coming out of a tube  as in a nondivergent nozzle. 0 + anything is the same as anything by itself... There are two sources of force that move the pipe true , ..... it just so happens that you are leaving one of them out !!! so when you add them together your answer is incorrect because one of them is zero. I did not leave one of them out  I provided you with the formula for Ve, and in the case you have described it's value will be zero. Don't take my word for it  do the math for yourself. Without a divergence cone Po = Pe. No matter how much you try and deny it, having a Po and Pe which are the same means Ve is zero. Basic math. the pressure differential and the expansion of the gas once it leaves the pipe (captured by the nozzle).
Wrong ... the pressure differencial and the expansion are the (PePa)*Ae side of the formula... Nope. (PePa)*Ae refers to the pressure differential between the pressurized tank and the outside environment. Ve is determined by the pressure differential in the stream of gas as it expands once it leaves the tank and enters the nozzle. A primer on these calcs: http://www.nakkarocketry.net/th_thrst.htmlMore specifically: " If the pressure ratio (and thus expansion ratio) is 1, then F = 0. The only thrust produced by such a nozzle is the pressure thrust, or Ftotal = (PePa)Ae. Such a nozzle, of course, would have no divergent portion, since A*/Ae=1, and would be a badly designed rocket nozzle!" Momentum, BTW, is a specific physical quantity equal to mass * velocity.
so why did you leave out Momentum , mass * velocity Because that's not what m*Ve is. m (q in your case) is the mass flow rate  as in kg/s, not the mass (as in kg alone). Hence, that portion of the formula calculates the force  not momentum  produced. That should be selfevident to an "expert" such as yourself  you cannot summate a momentum and force, as the units (and what they describe) are different. The fact you seem to think you can summate different units of measure doesn't exactly help your credibility  not that you have any left that is. Momentum = mass * velocity = kg*m/s Force (in simple terms) = mass * acceleration = kg*m/s^2 Force (in m*Ve) = kg/sec * m/sec = kg*m/s^2 Bryan EDIT: I'm off to the cottage for three days. Maybe use that time to read through Richards pages and learn a little about how rockets generate thrust. Then maybe we can have a conservation that extends beyond me trying to explain simple formulas to you.
Last edited by ImagingGeek; 06/04/10 06:17 PM.
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No, I am not. m*Ve is zero when you do not have a nozzle which allows for expansion. if its ZERO then NOTHING will come out. even if there was nothing but a hole in the tank , there would still be air comming out = m*Ve through the hole , but that might hinder your result. bonehead. I dont think theres anything further to discuss as you dont seem capable of performing math calculations. but have a nice trip.
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Let's knock off with the insults, please.
If you don't care for reality, just wait a while; another will be along shortly. A Rose




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Hey, not that I know what that formula's all about but have you two both totally overlooked this really simple miscommunication?
 Paul says Ve is velocity  Imaginggeek says Ve is zero
Just saying it back and forth to each other doesn't change the fact that you're obviously talking about different quantities!!!!




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F = (10132kPa101kPa)*0.00064516m^2
1 kPa = .1450377 psi 10132kPa = 1472.959764 psi we start with 100 psi... not 1472 psi 100 psi / .1450377 = 689.47 kPa 14.7 psi / .1450377 = 101 F = (689.47kPa101kPa)*0.00064516m^2 F = 588.47kPa * 0.00064516m^2 F = 0.3796573052 Nnot your result of But the above is only the pressure thrust. you still need to add the momentum or velocity thrust. the part you left out (m*Ve) or (q*Ve) F = (m × Ve) + (Pe  Pa) × Ae air has a mass of 0.0807 lb / cu ft 0.0807 / 1728 = 0.000046701388 lbs / cu in the area is 1 sq inch so air at 6.828 atm has a mass of 6.828 X 0.000046701388 = 0.000318877077264 pound mass 0.000318877077264 pounds = 0.000144640209 kilograms m = 0.000144640209 kilograms 100 pounds = 45.359237 kilograms a = F/m 45.359237 kg / 0.000144640209 kg = 313600.466382069 m/s^2 F=(0.000144640209 * 313600.466382069 m/s^2)+ (Pe  Pa) × Ae F=(45.359)+ (Pe  Pa) × Ae F=(45.359)+ (689.47kPa101kPa)*0.00064516m^2 F=(45.359)+ 588.47kPa * 0.00064516m^2 F= 45.359 + 0.3796573052 F = 45.738894305199934012421 N45 kilograms = 99.208018 pounds the momentum of the air leaving the tank through the nozzle has transfered to the pipe.
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you still need to add the momentum or velocity thrust.
the part you left out (m*Ve) or (q*Ve)
Havn't you noticed that ImagingGeek is saying Ve=0 ? Obviously he's not talking about velocity! Did he ever say Ve means velocity? Maybe he's got the wrong formula, why don't you bother to understand what he's thinking instead of doing trivially obvious calculations?




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Havn't you noticed that ImagingGeek is saying Ve=0 ? yes I have. and your point? NASA Rocket Thrust Equasionsif you will notice in the above formula the mass flow rate and velocity = the part he left out. the part that supplies the most thrust. he insisted that he could leave it out. that it wouldnt make a difference. he was wrong. why don't you bother to understand what he's thinking instead of doing trivially obvious calculations? because what hes thinking is wrong , thats why. you can believe him if you choose , but I wont because I think. and from what I have seen , I think his thinking is wrong.
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Yea certainly looks nonzero to me.
He gave another equation for Ve in message 34741, and allowed Po=Pe, but I'm not sure you can do that. For any nozzle shape there'll be more pressure on the inside than the outside  and thus some gas expansion. Without a pressure difference there wouldn't be any flow.
But we'll see better when he returns.
Anyway, how does this show that momentum isn't conserved? Or how does it show that you can get sustained propulsion of a closed cylinder?




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yea , he put up a image of a formula to calculate Ve.
I didnt show that momentum isnt conserved , in his way of thinking momentum isnt conserved.
he seems to think that the pressures inside the pipe will cause the pipe to not move and he is right , but he is claiming that no force will be felt.
he claims that because a force is comming out of the nozzle that force will be canceled out by the other end of the pipe that is 500 ft away and that is why the pipe wouldnt move.
I say that none of the pressures inside the pipe will cause the pipe to move , only the momentum of the mass inside will cause the pipe to move.
the important part here is the mass inside the pipe that is moving , and the momentum that it transfers to the pipe.
like attaching two gas cylinders together between a valve and 1 is full of 1500 psi gas and the other is empty.
if you place them in a float , then in a swimming pool then open the valve the mass inside will move until the pressure equalizes in the two cylinders , they will transfer momentum to the float and the float will move.
he is so poisoned by things he has heard that he cannot think.
suppose there is a large spinning space station that is supplying 1g to allow people inside the space station to walk around.
and suppose there is a ladder that would allow people to climb up to the center of rotation.
as people climb up the ladder more and more acceleration is given to the space station that increases its angular velocity.
the peoples angular velocity decreases because their radius from the center decreases , therefore they transfer their angular velocity to the space station.
the result is increased artificial gravity.
those still at the peripery will feel a higher gravity. the space station has not been pushed from the outside but its spinning faster.
this is reactionless propulsion.
and the more people that climb up to the center , the faster the space station rotates.
internal momentum changes affect the momentum of the overall system.
I think this is great because this means that we dont have to pollute the rest of our solar system and beyond by using gasses that burn and pollute.
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Yea you're mostly right. You can cause movement of a closed pipe, or change the speed of a spinning space station by moving things around inside. This is exactly what physicists already know.
BUT! It isn't sustainable, so you can't actaully travel and useful distance. In the spinning space station example, Those people who climbed to the center are now in the center! If they ever go back to the edge the thing will slow down again.
For linear motion, after you've transferred some momentum internally and made the outer pipe move, the mass you moved has to stop, and when it does, it cancels out any momentum it gave the pipe to begin with. So the pipe moves a short distance then stops. You can only repeat the process as long as you have mass at one end of the pipe. If you try to recirculate it in any way, you'll only move the pipe back to where it started from.
This is all clearly obvious by considering that the _total_ momentum of the closed system remains constant. You're just transferring it between the pipe and the internal moving parts.
Science education can't "poision" people because it allows them to think more rigorously, so they have the skills to challenge anything authorities try to tell them. In fact trusting an authority is considered to be nonscientific.
Science isn't just a random collection of facts, it's mostly a tightly interwoven system of many compatible ideas. If one 'fact' was inconsistent with the others, then it'd be easily detected. That's not something that people can fake, especially not when the whole world has access to experiments they can do themselves to discover the truth.
Although I do see your point that when you're immersed in one common set of knowledge it's hard to come up with new, different ideas. Most good new ideas come from young people who don't have such a long lifetime of the same old thing. But equally when you have hardly any knowledge, it's easy to think of millions of ideas that are wrong.
Last edited by kallog; 06/07/10 05:18 AM.




