Why does reverse thrust work?

[Absolute]

To elaborate further - the fact that gas molecules can “communicate” with each other in this way is why physicists often ignore what is happening on the level of the individual molecules, and simply refer to what is happening to the entire volume of gas.

When physicists say “a rocket produces thrust by accelerating gas out the nozzle”, they simply mean that the geometry of the rocket is designed to extract energy from gas molecules that are moving in the forward direction, leaving the gas molecules with an average velocity that is biased in the backward direction, and leaving momentum conserved.

However, again, because the gas molecules are colliding and changing directions billions of times every billionth of a second, the rocket ends up extracting some energy from every gas molecule, since every gas molecule will spend some fraction of it’s time randomly bounced in the forward direction, regardless of where it is in the combustion chamber, and what the average motion of the surrounding molecules is.

When you consider all these effects together, you find that the thrust produced by the rocket is equal to the product of the velocity of the exhaust gas, and the mass of the exhaust gas. That is the total mass of all the exhaust gas, without regard to which averaged direction the gas was originally moving inside the combustion chamber (again, because in reality, all the molecules are moving in random directions anyway, with just a slight bias towards whatever the macroscopic flow direction is).

Now, the math for that statement I cannot explain in layman’s terms. You will just have to take my word for it. But I hope you understand, in terms of molecular motion, what I am saying. A rocket/jet engine/etc. does not get thrust from the initial expansion wave moving in the forward direction hitting the wall and bouncing off. It gets thrust by extracting the forward motion from the totally random motion of the molecules.

Note that these effects only occur in a gas which is sufficiently dense, meaning there are enough molecules that collisions occur sufficiently frequently. A rocket driven by a very expanded (not dense) gas at very high temperature would begin to approach the behavior you are imagining - many of the molecules would free stream straight out the nozzle without colliding and transferring energy at all. But rockets/jet engines/etc. are designed so that the propellant is dense enough for there to be enough molecular collisions for it to extract energy from all the propellant molecules, and this free streaming effect occurs only at the very outside edge of the gas, and is so small as to be negligible.

[Absolute]

I will make on final post before going to bed.

When people refer to the “temperature” of a gas, they are simply talking about the average speed of the random motions of the molecules within it. These random motions are referred to as “thermal motion” for that reason. If a region of gas is not moving as a whole, all of the particles within it are still moving around in random directions, at an average speed defined by the temperature of the gas, and constantly colliding and changing direction. However, when you average the direction and speed of all these molecules, you find that they all cancel out and average to 0.

In contrast, when a region of gas is moving in a certain direction at a certain speed, we refer to the “macroscopic velocity”, “bulk velocity”, or “flow velocity”, as a parameter describing the behavior of the entire region of gas. Inside a region of flowing gas, the particles are still moving around randomly in different directions at a defined temperature - but the distribution of velocities is simply shifted by a bit towards the macroscopic flow direction. So the particles are still moving randomly and colliding randomly, it is just that their overall, averaged movement is towards one direction. Think of it as people on a cruise ship - the entire cruise ship, and all the people on it, may be moving south at 30 mph, but all the people inside are still walking about in (essentially) random directions.

The reason people have been harping on the difference between explosions and controlled combustion is that they both produce different types of particle motion. An explosion releases a lot of energy in macroscopic / bulk flow - jets of gases emitted in certain directions, shock waves, debris flying in straight lines, etc. In contrast, controlled combustion releases energy almost exclusively as thermal energy - the particles are still moving about randomly in random directions, just at a higher average speed than before the combustion. The particles heated by controlled combustion expand “gently”, simply as a result of their random motions carrying them outwards, as opposed to being propelled outwards in bulk flows as from an explosion.

A rocket/jet engine/etc. is designed to capture the thermal energy released by the propellant, and transform it into forward motion. The nature of a gas and the billions upon billions of collisions that occur between gas molecules and cause them to transfer energy make this possible, and allow the engine to extract energy from the entire volume of heated gas by preferentially causing the gas to expand in a given direction, and control the direction of the bulk velocity the expanding gas ends up with (consequently, controlling the direction of the velocity imparted to the rocket by the gas molecules).

This would not be possible using the results of an explosion, because a significant fraction of the energy from an explosion goes directly into the bulk velocity of the gas, and as you note, energy in the bulk velocity of the gas going straight out the nozzle cannot be recovered, and energy in the bulk velocity of the gas going straight into the sides of the chamber is also wasted, etc.

The difference is, the thermal energy of the gas that is going straight out the nozzle still can be recovered and transferred to the rocket. Again, because the particles in the gas collide so frequently, energy can be transferred to the combustion chamber walls through long sequences of particle collisions, even if the energy originates from particles that are themselves far away from the chamber walls, and might even (on average) be moving away from it.

[virtually_yours]

@Absolute

Okay, let’s back up. The reason we ended up in this discussion is your claim that, in a jet engine, there is “a combustion emanating from a central point”. You seem to have a view that half of the “thrust” goes towards the front of the engine, where it collides with the wall and pushes the engine forward, and half of the thrust is “wasted” because it goes the other way, straight out the back.

Thanks for coming down earth. But if you read my posts you would see where I elaborate and said a perceived central point.. The thrust wasted issue I changed to not used and if I remember correctly stated something to the effect that if is could all be used the engine would be 100% efficient. Which is debatable yes because the molecules don’t know direction and the heat/expansion is necessary for the process to happen. I said before that there is force acting on ALL sides of the chamber, but i will use the front back vector because that pertains to the point.

@Absolute

You are correct in that some of the thrust of a rocket comes from the pressure of the gas molecules hitting the forward combustion chamber wall. You are incorrect in your claim that anything about the gas that initially expands out the back of the nozzle is “wasted”.

I clarified that in other posts I called it unused energy in the context that it didn’t actually act on the plane, but I know of course it is necessary otherwise all the other actions would not happen. so in this context it is not wasted. I never meant wasted in the context of waste of time I meant if somehow it could all be used if would be more efficient which again is debatable I guess because of the necessary factor, the nature of the combustion so to speak.
[/QUOTE]

@Absolute

That was how this whole discussion got started, and you’re right, I was too hasty. We can have a reasonable discussion about a jet engine in the context of your tube.

I really appreciate this, maybe I ran a bit hot also.

@Absolute

Your understanding would be correct if we were talking about an explosion of macroscopic particles (i.e. chunks of debris), which all move outwards in pretty much a straight line from the source of the explosion. If you were trying to drive your tube through space by somehow creating explosions that released only a spray of small metal BB pellets, instead of a gas, your intuitive understanding would be correct. Half those chunks of metal would have some component in the forward direction and collide with the front of the tube, driving it forward, and the other half would have a component in the backward direction, and go straight out the back without doing jack shit to the velocity of the tube.

However, a gas is not like that, unless it is a very, very low density gas. Gas molecules are so small, and there are so many of them, that they are constantly colliding with each other and changing direction. Only a very few gas molecules at the outer edge of the expanding cloud are moving in a straight line - the others are bouncing around, colliding with each other, and only moving outwards on average, when you consider the averaged motion of all the quadrillions of gas molecules together.

Here’s my statement from a few posts up" “I put it to you the rocket engine works on the same principle, heat producing agitated particles moving at extremely high speeds in all directions” . I guess I should give more detail because you can’t read my mind, but I thought you would have assumed from this that I understood all directions to mean colliding ,bouncing off the side ect.
My point is that eventually some hit the plane and that is the actionable force. For anything to move, something has to act ON it. The other molecules hitting the sides of the chamber are a necessary part of the reaction so they do their part as that but banging on the side of the can doesn’t help directly with the forward motion of the plane..But yes if they weren’t there the other molecules wouldn’t be there to do the actual work so yes they are all in it together.
Eventually of course they all make it out the back because n more fuel and air is coming in. And that was my claim to the where’s the thrust part of this post.

@Absolute

There is a term for this in physics, it is called the mean free path - the average distance that a gas molecule will travel before colliding with another gas molecule and changing direction. At 1 atmosphere pressure, this is approximately 0.00006 mm. At higher than atmospheric pressure, such as inside the combustion chamber of a jet engine or rocket, it is yet smaller. Even at 1 atmosphere, a 1-foot-wide nozzle has approximately 500,000 mean free paths across it - meaning a gas molecule would on average collide and (essentially randomly) change direction 500,000 times before traveling across that 1-foot distance.

As a result, gas molecules can transfer energy to and from each other in a way that an expanding field of macroscopic objects can not. Gas molecules are constantly colliding with each other and distributing energy and pressure throughout the gas - an expanding field of metal BBs is not.

I never thought is was like bbs, I was just using the pertinent force vectors forward and back., and explosion for illustrative purposes; now we are down to the nitty gritty these terms not needed at this time.

@Absolute

So, in a rocket driven by an explosion in the center, yes, you will have two clouds of gas moving in opposite directions - on average. But inside, the molecules are constantly colliding and bouncing and moving every which way.

I agree. but the pertinent issue is that there is no movement until one finds a way to bounce or transfer its energy to the front part of the chamber.
If somehow all found their way out the back, the rocket would never move, if they could all find their way forward all the energy would be used.(just a side point)

@Absolute

So, when the gas hits the forward wall of the combustion chamber, yes, it exerts a pressure and drives the rocket forward. But then what happens? The gas molecules bounce off the forward wall, and are now heading in the backward direction. The reason the rocket/jet engine/whatever continues to experience a forward thrust is that those molecules move only 0.00006 mm (on average) backward, before colliding with some other molecule, and being reflected back into the forward wall of the chamber, like a billiard ball. And then the molecule that that molecule colliding with gets reflected in the backward direction, and collides with another molecule, and so forth. This process repeats billions and billions of times (actually more like a billion^2 times), and over time, it transfers energy from the heated gas molecules to the forward wall of the rocket. On average, you end up with a situation in which the rocket is moving in the forward direction, and the average velocity of the gas molecules has a bias in the backward direction (this means that, even though the velocity of any single molecule could be pointing in any direction, if you pick one at random, you are slightly more likely to find a particle that is going backward than forwards).

yes but remember when the molecules transfer heat they cool hence less movement, that’s why more fuel is needed to keep things hot.
I agree with all this, I see nothing different.

@Absolute

even the gas which is moving away from the central point on average can transfer energy to the forward wall of the combustion chamber!* And so the motion of the gas moving in the “backward” direction is not wasted, but in fact, pretty much contributes just as much thrust as the gas initially moving in the forward direction.

Excellent! This I didn’t perceive! And great laymans terms!
But, a question, when the heated molecules hit the molecules of the chamber metal they are cooled so their lose their energy so to speak, wouldn’t this happen to the molecules further away away from the chamber so to speak, and would there be any energy left by the time the last molecule bounced hits the ac? Does the energy decrease in any way exponentially or not as one molecule of the same mass bounces off another? At rest I assume not, but we are talking of molecules out of their element (pardon the pun if it is one).
But once again excellent, concise,clear explanation from you. I couldn’t ask for anymore
Thank you.. now, on to read your next post..
vy VY

[virtually_yours]

@Absolute

To elaborate further - the fact that gas molecules can “communicate” with each other in this way is why physicists often ignore what is happening on the level of the individual molecules, and simply refer to what is happening to the entire volume of gas.

When physicists say “a rocket produces thrust by accelerating gas out the nozzle”, they simply mean that the geometry of the rocket is designed to extract energy from gas molecules that are moving in the forward direction, leaving the gas molecules with an average velocity that is biased in the backward direction, and leaving momentum conserved.

However, again, because the gas molecules are colliding and changing directions billions of times every billionth of a second, the rocket ends up extracting some energy from every gas molecule, since every gas molecule will spend some fraction of it’s time randomly bounced in the forward direction, regardless of where it is in the combustion chamber, and what the average motion of the surrounding molecules is.

When you consider all these effects together, you find that the thrust produced by the rocket is equal to the product of the velocity of the exhaust gas, and the mass of the exhaust gas. That is the total mass of all the exhaust gas, without regard to which averaged direction the gas was originally moving inside the combustion chamber (again, because in reality, all the molecules are moving in random directions anyway, .

Clear again, I am sure we are in agreement but maybe terminology is different. This is how I am seeing it, correct me if I am wrong. For my layman understanding to come from the dumb-oops- ground up; I don’t say the rocket is producing thrust, I envision the rocket as being thrusted Thrusted by the action of the molecules.

@Absolute

with just a slight bias towards whatever the macroscopic flow direction is).

Question: 1) For clarity:When you say bias towards you mean relative to what? I assume you mean that inside the gas itself the molecules themselves have no bias towards a particular direction but because the gas as a whole is moving out the back (of course the gas is the molecules) and taking the molecules with it; relative to the rocket the molecules all moving eventually away out the back.Or because of the firewall, the bouncing off it creates a bias in the opposite direction?

@Absolute

A rocket/jet engine/etc. does not get thrust from the initial expansion wave moving in the forward direction hitting the wall and bouncing off. It gets thrust by extracting the forward motion from the totally random motion of the molecules.

As stated before I am aware of the movement in all directions and bouncing, I was narrowing it down to specific force directions that were pertinent, that was dealing with the force issue. but as I said also I didn’t realize that the molecules on the way out actually could transfer their movement back. That is a gem.That’s cool, thanks again.

@Absolute

"When you consider all these effects together, you find that the thrust produced by the rocket is equal to the product of the velocity of the exhaust gas, and the mass of the exhaust gas. That is the total mass of all the exhaust gas, without regard to which averaged direction the gas was originally moving inside the combustion chamber:

Now, the math for that statement I cannot explain in layman’s terms. You will just have to take my word for it. But I hope you understand, in terms of molecular motion, what I am saying.

Note that these effects only occur in a gas which is sufficiently dense, meaning there are enough molecules that collisions occur sufficiently frequently. A rocket driven by a very expanded (not dense) gas at very high temperature would begin to approach the behavior you are imagining - many of the molecules would free stream straight out the nozzle without colliding and transferring energy at all. But rockets/jet engines/etc. are designed so that the propellant is dense enough for there to be enough molecular collisions for it to extract energy from all the propellant molecules, and this free streaming effect occurs only at the very outside edge of the gas, and is so small as to be negligible.

Once again my imagining was based on what was ACTING on the actual metal to push it. I always said it was the force in the direction of the aircraft that pushed it, not the force coming out the back. Yes from what you showed me on the molecular level; the molecule even though it was going backwards still did work by bouncing back and forth, but only when that work ends up at the metal of the plane can it be used to push the plane. It the molecule going backwards relative to the plane bounced molecules all over the place sides etc. and the none of the molecules in the chain hits the forward part of the plane, no work to that respect is done.. you say this is minimal though to the overall reactions taking place, so until I see something different that could explain otherwise this I agree

@Absolute

Now, the math for that statement I cannot explain in layman’s terms. You will just have to take my word for it. But I hope you understand, in terms of molecular motion, what I am saying

You sad person hehe you’re so clouded by formulas you don’t even know when you are being clear..You made the statement in layman terms (as best you could)! that means you don’t need complex math formula. Good work.
and thanks again \vyvY

[Absolute]

@virtually_yours

Excellent! This I didn’t perceive! And great laymans terms!
But, a question, when the heated molecules hit the molecules of the chamber metal they are cooled so their lose their energy so to speak, wouldn’t this happen to the molecules further away away from the chamber so to speak, and would there be any energy left by the time the last molecule bounced hits the ac? Does the energy decrease in any way exponentially or not as one molecule of the same mass bounces off another? At rest I assume not, but we are talking of molecules out of their element (pardon the pun if it is one).
But once again excellent, concise,clear explanation from you. I couldn’t ask for anymore
Thank you.. now, on to read your next post..
vy VY

Well, the next posts are simply expounding on this basic point, so if we are in agreement already from the first post you may probably just skim or skip them.

I’m glad I could be helpful!

To answer this question: collisions between molecules in a gas, on average, may be considered to be perfectly elastic, meaning no energy is lost. I say on average because any given molecule may end up with a substantial amount of energy in what are called “internal degrees of freedom” (which you can think of as the molecule simply being “twisted”), but this energy will be released in short order when the molecule relaxes (“un-twists”), and turned back into thermal motion.

There are still plenty of ways for the gas to lose energy. It emits black-body (thermal) radiation, which is absorbed by the engine wall and heats the engine. It loses energy by thermal conduction into the walls as well. And some small fraction of the internal degrees of freedom do not relax into thermal motion, but instead release a photon (light) of a specific frequency. All of these contribute to efficiency (or lack thereof) of the entire engine, in addition to various theoretical limits on how much energy it is feasible to extract from a heated gas (see Carnot cycle - Wikipedia). A rocket nozzle, if carefully design, can actually be something like 70% efficient.

I will take a look at your next post after breakfast.

[virtually_yours]

@Absolute

I will make on final post before going to bed.

When people refer to the “temperature” of a gas, they are simply talking about the average speed of the random motions of the molecules within it. These random motions are referred to as “thermal motion” for that reason. If a region of gas is not moving as a whole, all of the particles within it are still moving around in random directions, at an average speed defined by the temperature of the gas, and constantly colliding and changing direction. However, when you average the direction and speed of all these molecules, you find that they all cancel out and average to 0.

I agree. I assume that’s why we feel heat? friction of the fast moving molecules bouncing off us?

@Absolute

In contrast, when a region of gas is moving in a certain direction at a certain speed, we refer to the “macroscopic velocity”, “bulk velocity”, or “flow velocity”, as a parameter describing the behavior of the entire region of gas. Inside a region of flowing gas, the particles are still moving around randomly in different directions at a defined temperature - but the distribution of velocities is simply shifted by a bit towards the macroscopic flow direction. So the particles are still moving randomly and colliding randomly, it is just that their overall, averaged movement is towards one direction. Think of it as people on a cruise ship - the entire cruise ship, and all the people on it, may be moving south at 30 mph, but all the people inside are still walking about in (essentially) random directions.

" it is just that their overall, averaged movement is towards one direction".
I asked before but since it is here also: the bias towards " you mean in respect to an observer or some kind of coriolis force thing?
I don’t think you mean this but asking for clarity: you don’t mean the people will all end up at the front of the boat because their randomness is biased towards the front?

@Absolute

The reason people have been harping on the difference between explosions and controlled combustion is that they both produce different types of particle motion. An explosion releases a lot of energy in macroscopic / bulk flow - jets of gases emitted in certain directions, shock waves, debris flying in straight lines, etc. In contrast, controlled combustion releases energy almost exclusively as thermal energy - the particles are still moving about randomly in random directions, just at a higher average speed than before the combustion. The particles heated by controlled combustion expand “gently”, simply as a result of their random motions carrying them outwards, as opposed to being propelled outwards in bulk flows as from an explosion.

yup that’s why I abandoned using explosion but explained why I used it. I pictured the shrapnel/debris etc. as the aircraft to try and illustrate my rapid expansion point. A bit extreme ..I agree.

@Absolute

A rocket/jet engine/etc. is designed to capture the thermal energy released by the propellant, and transform it into forward motion. The nature of a gas and the billions upon billions of collisions that occur between gas molecules and cause them to transfer energy make this possible, and allow the engine to extract energy from the entire volume of heated gas by preferentially causing the gas to expand in a given direction, and control the direction of the bulk velocity the expanding gas ends up with (consequently, controlling the direction of the velocity imparted to the rocket by the gas molecules).

This would not be possible using the results of an explosion, because a significant fraction of the energy from an explosion goes directly into the bulk velocity of the gas, and as you note, energy in the bulk velocity of the gas going straight out the nozzle cannot be recovered, and energy in the bulk velocity of the gas going straight into the sides of the chamber is also wasted, etc.

The difference is, the thermal energy of the gas that is going straight out the nozzle still can be recovered and transferred to the rocket. Again, because the particles in the gas collide so frequently, energy can be transferred to the combustion chamber walls through long sequences of particle collisions, even if the energy originates from particles that are themselves far away from the chamber walls, and might even (on average) be moving away from it.

No prob, like I said I was just trying to show what the thrust actually was using that example. through the discussion I realized that people were having a problem with it, so i dropped it in favour of expansion of gases/combustion. In my mind an explosion is that expansion..I’ve made a note to self..vyVY

[virtually_yours]

@Absolute

Well, the next posts are simply expounding on this basic point, so if we are in agreement already from the first post you may probably just skim or skip them.

I’m glad I could be helpful!

To answer this question: collisions between molecules in a gas, on average, may be considered to be perfectly elastic, meaning no energy is lost. I say on average because any given molecule may end up with a substantial amount of energy in what are called “internal degrees of freedom” (which you can think of as the molecule simply being “twisted”), but this energy will be released in short order when the molecule relaxes (“un-twists”), and turned back into thermal motion.

There are still plenty of ways for the gas to lose energy. It emits black-body (thermal) radiation, which is absorbed by the engine wall and heats the engine. It loses energy by thermal conduction into the walls as well. And some small fraction of the internal degrees of freedom do not relax into thermal motion, but instead release a photon (light) of a specific frequency. All of these contribute to efficiency (or lack thereof) of the entire engine, in addition to various theoretical limits on how much energy it is feasible to extract from a heated gas (see Carnot cycle - Wikipedia). A rocket nozzle, if carefully design, can actually be something like 70% efficient.

I will take a look at your next post after breakfast.

cool. thanks
vyVY

[Absolute]

@virtually_yours

Clear again, I am sure we are in agreement but maybe terminology is different. This is how I am seeing it, correct me if I am wrong. For my layman understanding to come from the dumb-oops- ground up; I don’t say the rocket is producing thrust, I envision the rocket as being thrusted Thrusted by the action of the molecules.

These are both just different ways of looking at the same process.

Question: 1) For clarity:When you say bias towards you mean relative to what? I assume you mean that inside the gas itself the molecules themselves have no bias towards a particular direction but because the gas as a whole is moving out the back (of course the gas is the molecules) and taking the molecules with it; relative to the rocket the molecules all moving eventually away out the back.Or because of the firewall, the bouncing off it creates a bias in the opposite direction?

All there is in a gas is the molecules, so it is not correct to say the gas “as a whole is moving and and taking the molecules with it”.

In a gas that is macroscopically at rest, if you were to add up the random motions of all the molecules, you would find it all averages to 0, meaning that even though molecules are all moving around, the gas as a whole (meaning, the billions of molecules when considered together) are not going anywhere.

In a gas that is moving, if you add up the random motions of all the molecules, you find that the motion of the molecules in one direction does not average to zero. They are still moving in random directions, but statistically, any given molecule is more likely to be found moving in that direction. Macroscopically, this means the gas is moving. To calculate the temperature of such a gas, you subtract the average velocity component from each molecule’s motion, and consider only the motion of the molecules relative to the average movement.

Returning to the cruise ship analogy - all the people inside the ship are randomly moving around. The cruise ship itself, and thus all the people in it, is nevertheless definitely moving in one direction, when averaged over time. From the perspective of an outside observer who is stationary outside the ship, the people are moving around randomly, but every one one of them is undoubtedly moving in the same direction the ship is - when averaged over time.

Let’s say the average person walks at perhaps 4 mph, and the ship is traveling north at 3 mph (a slow ship). Obviously, if a person is walking towards the front of the ship, they will appear to an outside observer be traveling north at 7 mph. If someone is walking towards the back of the ship, they will appear to be traveling south at 1 mph. If someone is walking across the ship to the east, they will appear to be traveling northeast at sqrt(4^2 + 3^2) = 5 mph. Yet, over time, all of these people still travel with the ship, and their average direction is wherever the ship is going, even if their instantaneous direction at any given time might be different.

It the molecule going backwards relative to the plane bounced molecules all over the place sides etc. and the none of the molecules in the chain hits the forward part of the plane, no work to that respect is done.. you say this is minimal though to the overall reactions taking place, so until I see something different that could explain otherwise this I agree

Yes, and in a very low-density gas, this can happen. As the gas density increases, the collisions become so frequent that it is statistically impossible that any given molecule will not transfer some energy to the forward motion of the engine through this process. However, the fact that there is a maximum theoretical efficiency for this type of process (see the link I posted earlier) is related to the fact that it is impossible to extract all of the energy out of the molecules through these purely random collisions.

You sad person hehe you’re so clouded by formulas you don’t even know when you are being clear..You made the statement in layman terms (as best you could)! that means you don’t need complex math formula. Good work.
and thanks again \vyvY

Heh, well, the problem is, if you want to actually analyze or design one of these things quantitatively, you need to work through the math. And often, when you do go through the math, you discover properties of the system that you would not have been able to discern through intuition alone, or which may in fact be counter-intuitive.

[Absolute]

@virtually_yours

I agree. I assume that’s why we feel heat? friction of the fast moving molecules bouncing off us?

Friction is probably not the right word, but yes, close enough.

" it is just that their overall, averaged movement is towards one direction".
I asked before but since it is here also: the bias towards " you mean in respect to an observer or some kind of coriolis force thing?
I don’t think you mean this but asking for clarity: you don’t mean the people will all end up at the front of the boat because their randomness is biased towards the front?

No, I mean relative to an outside observer. The people will end up wherever the boat is going, even though they are walking around “randomly” inside the boat. They will presumably stay relatively evenly distributed around the inside of the boat.

[Absolute]

By the way, the general term for this entire field of physics (studying the individual motions of particles and how they give rise to macroscopic behavior) is called Statistical Mechanics.

Statistical mechanics

That Wikipedia article is not all that great, but it has links to many others, and terms you can Google for.

[virtually_yours]

@Absolute

By the way, the general term for this entire field of physics (studying the individual motions of particles and how they give rise to macroscopic behavior) is called Statistical Mechanics.

Statistical mechanics - Wikipedia

That Wikipedia article is not all that great, but it has links to many others, and terms you can Google for.

Good Thanks .. I think were done with this issue of this particular topic. all the others basically fall in line now… if you want to address the other clarification questions I raised, fine… take a break, I need one. and let me know when you want to take on this continuous acceleration in spacetopic..
vyVY

[virtually_yours]

hehe you folks think this is bad. you should have been in my Flight school instrument class, every time I raised my hand there where awwws and arghhs. But I got the highest grades in the exam. He asked just about everyone in the class their grade but me…(of course he knew already I think) So i said my grade… and he joked- “we’re still trying to find out how you cheat”… they couldn’t believe this seemingly not able to understand person got top marks…looking back I was just trying to be 100% sure as possible that all other possibilities were ruled out to get a clear picture of what was going on…
vyVY

[virtually_yours]

@Absolute

By the way, the general term for this entire field of physics (studying the individual motions of particles and how they give rise to macroscopic behavior) is called Statistical Mechanics.

Statistical mechanics - Wikipedia

That Wikipedia article is not all that great, but it has links to many others, and terms you can Google for.

eureka? I think I got the acceleration pictured in my mind in layman terms..If I am standing on an object in space and the direction of motion is down from my feet, and I crouch and stand up rapidly the object would move.. let’s say at ten mph away from me. If I come back to the object and stand on it me and the object are moving ten mph to someone observing us. But to me the object is stationary. If I crouch and push again the object goes away from me at ten mph but the speed to the observer is now 20 mph, even though I used the same force. I come back to the object and stand on it, The object to me is stationary, but to the observer we are moving 20 mph. I push again.. 30 mph and so on and so on.. so that little bit of force from my legs would eventually push that mass up to the speed of light..YES!!!
So put an engine or pulse engine on the mass instead of me and it is on their merry way to light speed..
I hope this is the F=MA in layman terms what you talk about. The formula I can’t grasp as it is right now.. maybe later, but that scenario I described I understand clearly..
If that is it.
woohoooooo
vyVY

[Absolute]

Yes, that is pretty much it. Of course, if you were standing on an object in space, and you pushed on it, the two of you would move away from one another - just as if you were both standing on an icy pond and tried the same thing. So you would need some way of catching up to it to push on it again. But yes, you have the basic idea. The object in space will continue to move, neither speeding up nor slowing down, until another force is applied to it. And if you could continue to apply a force to it, it would eventually reach very close to the speed of light.

The formula F = ma is probably easier to understand if written as a = F/m, where a is the acceleration an object of mass m will experience when subject to a force of F.

Acceleration has units of distance / time^2, for instance, meters per second per second. So, an object with acceleration 5 m/s^2 will accelerate by 5 m/s every second. If it starts at rest and accelerates at 5 m/s^2 for 4 seconds, it will be traveling at 20 m/s after 4 seconds.

So, let’s say you are standing on an object in space with mass 50 kilograms (such an object will weight about 110 pounds on Earth). You push on it with your legs with a force of 900 Newtons (about 200 pounds) for 0.75 seconds. According to a = F/m, it experiences an acceleration of 18 m/s^2, and so its final velocity is 18 * .75 = 13.5 m/s, or about 30 mph.

As you are pushing it, the object’s inertia resists your pushing with a force of 900 Newtons as well (remember, for every action, an equal and opposite reaction). However, assuming you weigh about 100 kilograms (220 pounds on Earth), the acceleration you experience as a result will be only half that of the object - 9 m/s^2, and your ultimate velocity ends up 6.75 m/s, in the opposite direction.

F = ma simply tells you how hard you much push (what force you must apply) to an object of mass m to accelerate it at acceleration a.

If you could constantly apply a force F to an object m, it would accelerate at rate a. Note that the object’s velocity does not enter into this equation - a constant force will continue to accelerate the object at a constant rate regardless of how fast it is already going.

[virtually_yours]

@Absolute

Yes, that is pretty much it. Of course, if you were standing on an object in space, and you pushed on it, the two of you would move away from one another - just as if you were both standing on an icy pond and tried the same thing. So you would need some way of catching up to it to push on it again.

Yes I understood this, but was just keeping my eye on the mass, so me going away an coming back wasn’t the issue because as you read I put the pulse engine on it in place of me, and that of course is fixed.

@Absolute

But yes, you have the basic idea

Cool. what a relief

@Absolute

The formula F = ma is probably easier to understand if written as a = F/m, where a is the acceleration an object of mass m will experience when subject to a force of F.

Acceleration has units of distance / time^2, for instance, meters per second per second. So, an object with acceleration 5 m/s^2 will accelerate by 5 m/s every second. If it starts at rest and accelerates at 5 m/s^2 for 4 seconds, it will be traveling at 20 m/s after 4 seconds.

So, let’s say you are standing on an object in space with mass 50 kilograms (such an object will weight about 110 pounds on Earth). You push on it with your legs with a force of 900 Newtons (about 200 pounds) for 0.75 seconds. According to a = F/m, it experiences an acceleration of 18 m/s^2, and so its final velocity is 18 * .75 = 13.5 m/s, or about 30 mph.

As you are pushing it, the object’s inertia resists your pushing with a force of 900 Newtons as well (remember, for every action, an equal and opposite reaction). However, assuming you weigh about 100 kilograms (220 pounds on Earth), the acceleration you experience as a result will be only half that of the object - 9 m/s^2, and your ultimate velocity ends up 6.75 m/s, in the opposite direction.

F = ma simply tells you how hard you much push (what force you must apply) to an object of mass m to accelerate it at acceleration a.

If you could constantly apply a force F to an object m, it would accelerate at rate a. Note that the object’s velocity does not enter into this equation - a constant force will continue to accelerate the object at a constant rate regardless of how fast it is already going.

acceleration is my prob. I understand per sec per sec clearly but when someone says acceleration doesn’t mean speeding up but change in velocity I get confused.. vyVY

[Absolute]

@virtually_yours

acceleration is my prob. I understand per sec per sec clearly but when someone says acceleration doesn’t mean speeding up but change in velocity I get confused.. vyVY

Well, informally, acceleration does mean speeding up.

Formally, “velocity” means more than just speed, it means speed and direction. Your velocity can change if you change directions while continuing to move at the same speed (e.g. a car turning a corner). So, formally, acceleration refers to all changes in velocity, be it speeding up, slowing down, or changing direction.

[virtually_yours]

@Absolute

Well, informally, acceleration does mean speeding up.

Formally, “velocity” means more than just speed, it means speed and direction. Your velocity can change if you change directions while continuing to move at the same speed (e.g. a car turning a corner). So, formally, acceleration refers to all changes in velocity, be it speeding up, slowing down, or changing direction.

yes I remember from my crude reading of introduction to mechanics book (I have to find it) there was a diagram showing angular momentum like a wheel spinning and they were indicating acceleration and said it was a change in the direction of velocity. I thought to myself where is the speeding up in this, so I just accepted that they were using the word acceleration for two different meanings one for speeding up and the other for change in velocity direction. I couldn’t understand how if I am driving along at 10 mph and take a left at 10 mph that I am somehow speeding up. So I figured that it meant two different things.

The formula F=MA still trying to picture it in lay terms but unable I think.. The closest I can get (if you can call it close) is CF on CM=CA Constant Force on Constant Mass equals Constant acceleration. (constant in asterisks because of course acceleration itself is by definition continuous)
vyVY

[AdamF]

In relation to the earlier discussion, I tried very hard to find a graph showing the speed of air as it moves through a jet/turbo-fan engine. Yet I came up empty. Even my textbook dealing with turbo-machinery simply assumes that the speed is nearly constant.

Please help?

[virtually_yours]

@AdamF

In relation to the earlier discussion, I tried very hard to find a graph showing the speed of air as it moves through a jet/turbo-fan engine. Yet I came up empty. Even my textbook dealing with turbo-machinery simply assumes that the speed is nearly constant.

Please help?

This probably doesn’t pertain to this but I want to correct something I stated earlier (remembering my old flight ground school days). I’m surprised the pilots in here didn’t pick up on it, maybe I was correct where the turbo fan was concerned unless they are cambered also… I couldn’t imagine them not being .. But The propeller does not really go forward by corkscrewing (as a screw does in wood) through the air though that is the result, it is the camber on the blade that is the same as the camber on the wing that creates a low pressure on the front of the blade and that is what pulls the aircraft forward. hope I didn’t step in it again,bracing for impact.
vyVY

[virtually_yours]

@AdamF

In relation to the earlier discussion, I tried very hard to find a graph showing the speed of air as it moves through a jet/turbo-fan engine. Yet I came up empty. Even my textbook dealing with turbo-machinery simply assumes that the speed is nearly constant.

Please help?

Adam I don’t know if you read the previous posts but it is proved that in the engine it is the combustion that provides the heat that gets the molecules in motion that act on the aircraft to move it.. The inner part of the fan provides the air needed for the exact fuel air ratio to burn, too much air,well you know what will happen, but there is a bypass for scenarios like that,That is the internal part. The outer part of the fan is used as a propeller (or lots of propellers) to pull the aircraft forward..
The air is compressed just like it is compressed in a piston engine then fuel added for the combustion.
That’s what I see basically

vyVY