Denspressure vs Reality

Quote from: JackBlack on November 25, 2017, 04:48:39 AM

Quote from: sceptimatic on November 25, 2017, 04:46:45 AM

As soon as you reach a constant speed depress the clutch pedal or turn off the ignition and see if you carry on moving forward.

Is that your problem? You just cut the ignition and have the engine act as a brake?

In bold. Try it.

I fixed the bolding to show the other option you provided and indicated was equivalent.
I have done both.
Putting in the clutch (such as to change gear, after obtaining a constant speed in one gear) the car continues moving forward, although slowing down gradually. That gives enough time to engage the next gear and release the clutch and have the car continue moving.
Meanwhile, stalling or cutting the engine turns the engine into a brake. This results in a rapid deceleration, depending upon how fast the car is moving it can seem almost instant.

Quote from: JackBlack

Quote from: sceptimatic on November 25, 2017, 04:48:03 AM

Chop the blades and cut the helicopter engine to neutral. Same scenario.

No. Completely different.

Same scenario.

No. Completely different.
In one, you put the gears into neutral (which I'm not sure is even possible on a helicopter). This results in the blades continuing to spin as they had been before, gradually slowing down.
Cutting the blades down results in sections of the blades flying out and then the remainder of the rotor speeding up due to less resistance (both as air resistance and moment of inertia) of the blades.

They are completely different scenarios.

What you need to be looking at is vertical momentum at a constant velocity.

You are yet to justify why it needs to be a constant velocity.
Until you do, we will keep looking at things with vertical momentum, regardless of if its velocity is constant.

your feet leave the deck and from then on you are decelerating into the sky.
What's the issue here?

Why are you decelerating?
Why don't you stop dead instantly?
Other than your momentum/inertia, what keeps you going?
You have already admitted (in that post) that your springboard nonsense stops when you are off the ground.

Quote from: Jonny B Smart on November 25, 2017, 07:03:25 AM

It’s just like rockets and cars. You said that rockets and cars stop when the engine is cut, but they don’t, neither do balls, and neither do you.

Vertical and constant.

So if a rocket is accelerating instead of having a constant vertical speed it can continue going? It wont magically stop dead?

Notice how yet again, by trying to manipulate your nonsense to match reality you have defeated your own objection?

Quote from: Jane on November 25, 2017, 01:32:55 PM

Not trying to get involved in this debate, looks hectic enough as it is, just trying to understand the model.

If we took, say, a platform on some kind of mechanism that moved up at a constant, fast speed, and placed a ball on top of it, then that ball would be initially trapped on the surface of the platform keeping going up at a constant speed.
Would it be correct to say that if the platform was to quickly decelerate, stop dead or reverse, the ball in question would not leave the surface of the platform but would instead match speed with it at every second?
And if instead the platform was to accelerate up before slowing/stopping/reversing then and only then would the ball leave the surface of the platform and be thrown up?

And if this is accurate then, companion question to the first situation, how could you tell the difference between the ball keeping to the surface of the platform staying at a dead stop, and the ball decelerating at the same rate as the platform?

Assuming these questions are clear this'll probably be my only post on this topic, thanks in advance!

If all ball is sitting on a flat board is clamped in place until it got up to speed, and then the ball is released, it would roll off the back due to the force of wind resistance. If it is prevented from rolling back, and the board slows rather quickly, the ball will roll off the front due to its inertia. If a board with a ball on top accelerated upward and then stops, the ball would go flying up. Gravity would immediately start to decelerate it until it reverses direction and drops to the ground.

I don't care, I'm asking for Scepti's model.

I don't care, I'm asking for Scepti's model.

From what he has said, it will depend upon how steep it is.
If the hill is over some magic critical steepness, the ball will stop dead with the board if it was moving at a constant vertical speed.
But if it was accelerating, or below this critical steepness, it keeps going.

He has no explanation of why there is this difference or critical steepness, nor has he provided what the critical steepness is.

Quote from: Jonny B Smart on November 25, 2017, 01:44:34 PM

Quote from: Jane on November 25, 2017, 01:32:55 PM

Not trying to get involved in this debate, looks hectic enough as it is, just trying to understand the model.

If we took, say, a platform on some kind of mechanism that moved up at a constant, fast speed, and placed a ball on top of it, then that ball would be initially trapped on the surface of the platform keeping going up at a constant speed.
Would it be correct to say that if the platform was to quickly decelerate, stop dead or reverse, the ball in question would not leave the surface of the platform but would instead match speed with it at every second?
And if instead the platform was to accelerate up before slowing/stopping/reversing then and only then would the ball leave the surface of the platform and be thrown up?

And if this is accurate then, companion question to the first situation, how could you tell the difference between the ball keeping to the surface of the platform staying at a dead stop, and the ball decelerating at the same rate as the platform?

Assuming these questions are clear this'll probably be my only post on this topic, thanks in advance!

If all ball is sitting on a flat board is clamped in place until it got up to speed, and then the ball is released, it would roll off the back due to the force of wind resistance. If it is prevented from rolling back, and the board slows rather quickly, the ball will roll off the front due to its inertia. If a board with a ball on top accelerated upward and then stops, the ball would go flying up. Gravity would immediately start to decelerate it until it reverses direction and drops to the ground.

I don't care, I'm asking for Scepti's model.

He hasn’t responded since he acknowledged the validity of f=ma, and then I pointed out that this equation means no acceleration means no force. I think that his model is kaput. (Apparently, if you take the logic out of an idea, it will “stop dead.”)

I don't care, I'm asking for Scepti's model.

Don't ask us, ask Sceppy.

Quote from: Jane on November 25, 2017, 02:58:26 PM

I don't care, I'm asking for Scepti's model.

Don't ask us, ask Sceppy.

Thats my line!

It doesn’t matter which direction you are moving. Moving at a constant velocity requires no force. force = mass x ACCELERATION, remember? No acceleration (change in velocity) means no force.

Moving against a force (gravity or air resistance, for example) requires a balanced force to maintain velocity.

You are massively contradicting yourself.

Except that isn't what happens.  You begin to slow down, you do not stop dead.  If you were moving at 100 kph and you instantly stopped you would likely be dead.
Your example does not match my real world experience.

You are still not grasping what I'm saying.

Not trying to get involved in this debate, looks hectic enough as it is, just trying to understand the model.

Feel free to get involved is you wish. At least I know you try to understand it even if you might not agree.

If we took, say, a platform on some kind of mechanism that moved up at a constant, fast speed, and placed a ball on top of it, then that ball would be initially trapped on the surface of the platform keeping going up at a constant speed.

Ok, let's use an iron ball just so we don't have any added extras of compression as we would on a football or tennis ball.
Just to make it more clear more than anything else.

 

Would it be correct to say that if the platform was to quickly decelerate, stop dead or reverse, the ball in question would not leave the surface of the platform but would instead match speed with it at every second?

Let's look at decelerating.
In my mind I believe decelerating can only be a thing immediately after the cut off point of acceleration and it's a massive key thought.
If something is building up a mph and the power is immediately cut, that build up will allow deceleration to happen, which means a forward movement to still happen as ever reducing mph.

In this case we would see the iron ball sit on the platform all the way up and all the way down to the ground.

And if instead the platform was to accelerate up before slowing/stopping/reversing then and only then would the ball leave the surface of the platform and be thrown up?

If the platform accelerates and then the power is cut, it simply decelerates.
If  a brake was applied to the platform then the ball would carry on decelerating, meaning it would do a quick small jump but wouldn't significantly show any large vertical gain.
A tennis ball/football would make a bigger gain and I can explain why but we'll deal with the issues first.

And if this is accurate then, companion question to the first situation, how could you tell the difference between the ball keeping to the surface of the platform staying at a dead stop, and the ball decelerating at the same rate as the platform?

It would depend on the mass of the ball being used as to what you could see clearly or marginally or not at all.


However, there is still one more thing to deal with which is the constant velocity.
This is the one where I'm basically arguing the point of, even though it seems to have jumped from the rocket topic. lol

My argument is this.

On that very same platform going up at a constant velocity, it is under a constant energy load that is keeping that platform moving at a stead mph vertically.
The iron ball on the platform is also moving at the same rate, obviously.
If I was to cut the power in a split second to leave that platform attached to nothing, then that platform and the ball stop dead before accelerating back down.
My argument is that it would not move an inch further vertically after power is cut at a constant rate of mph.

 

Assuming these questions are clear this'll probably be my only post on this topic, thanks in advance!

Hopefully they are clear but if not feel free to ask and also feel free to enter the topic.

Quote from: sceptimatic on November 25, 2017, 07:55:47 AM

Quote from: Jonny B Smart on November 25, 2017, 07:03:25 AM

It’s just like rockets and cars. You said that rockets and cars stop when the engine is cut, but they don’t, neither do balls, and neither do you.

Vertical and constant.

What do you even mean?

Project any fairly dense object vertical at 10 m/s and it will take roughly one second to stop, in which time it will have travelled up roughly 5 m.
Then it will start to fall and be back to the original level in roughly one more second travelling down at close to 10 m/s.

Video it on your camera or phone  and check the motion frame by frame - and remember, I sad roughly.

Springboard and deceleration is what you're implying.
I am NOT.
I am implying what I've just implied.


Vertical and CONSTANT

He hasn’t responded since he acknowledged the validity of f=ma, and then I pointed out that this equation means no acceleration means no force. I think that his model is kaput. (Apparently, if you take the logic out of an idea, it will “stop dead.”)

Second law: Second law: Force on an object is equal to the mass  of that object multiplied by the acceleration of the object: F = ma.

So basically you apply energy to a dense mass and accelerate that mass or move that mass.
In simple terms what you put in as energy you get out as equal energy. Nothing more and nothing less.
Simple enough as long as it's used in the reality of the physical world we actually live in.


Try and understand what it means instead of thinking I'm ducking out.

Quote from: Jane on November 25, 2017, 02:58:26 PM

I don't care, I'm asking for Scepti's model.

Don't ask us, ask Sceppy.

She is but you people just can't resist jumping in to have a pop at her for asking questions.
It's like you're all crapping yourselves in case she goes against you.

She's asking questions.
She does not accept my thoughts.
She likes to figure out each and every theory and to do so she asks pertinent questions and queries answers.

She's not your enemy.

Quote from: Jane on November 25, 2017, 01:32:55 PM

Not trying to get involved in this debate, looks hectic enough as it is, just trying to understand the model.

Feel free to get involved is you wish. At least I know you try to understand it even if you might not agree.

Quote from: Jane on November 25, 2017, 01:32:55 PM

If we took, say, a platform on some kind of mechanism that moved up at a constant, fast speed, and placed a ball on top of it, then that ball would be initially trapped on the surface of the platform keeping going up at a constant speed.

Ok, let's use an iron ball just so we don't have any added extras of compression as we would on a football or tennis ball.
Just to make it more clear more than anything else.

 

Quote from: Jane on November 25, 2017, 01:32:55 PM

Would it be correct to say that if the platform was to quickly decelerate, stop dead or reverse, the ball in question would not leave the surface of the platform but would instead match speed with it at every second?

Let's look at decelerating.
In my mind I believe decelerating can only be a thing immediately after the cut off point of acceleration and it's a massive key thought.
If something is building up a mph and the power is immediately cut, that build up will allow deceleration to happen, which means a forward movement to still happen as ever reducing mph.

In this case we would see the iron ball sit on the platform all the way up and all the way down to the ground.

Quote from: Jane on November 25, 2017, 01:32:55 PM

And if instead the platform was to accelerate up before slowing/stopping/reversing then and only then would the ball leave the surface of the platform and be thrown up?

If the platform accelerates and then the power is cut, it simply decelerates.
If  a brake was applied to the platform then the ball would carry on decelerating, meaning it would do a quick small jump but wouldn't significantly show any large vertical gain.
A tennis ball/football would make a bigger gain and I can explain why but we'll deal with the issues first.

Quote from: Jane on November 25, 2017, 01:32:55 PM

And if this is accurate then, companion question to the first situation, how could you tell the difference between the ball keeping to the surface of the platform staying at a dead stop, and the ball decelerating at the same rate as the platform?

It would depend on the mass of the ball being used as to what you could see clearly or marginally or not at all.


However, there is still one more thing to deal with which is the constant velocity.
This is the one where I'm basically arguing the point of, even though it seems to have jumped from the rocket topic. lol

My argument is this.

On that very same platform going up at a constant velocity, it is under a constant energy load that is keeping that platform moving at a stead mph vertically.
The iron ball on the platform is also moving at the same rate, obviously.
If I was to cut the power in a split second to leave that platform attached to nothing, then that platform and the ball stop dead before accelerating back down.
My argument is that it would not move an inch further vertically after power is cut at a constant rate of mph.

 

Quote from: Jane on November 25, 2017, 01:32:55 PM

Assuming these questions are clear this'll probably be my only post on this topic, thanks in advance!

Hopefully they are clear but if not feel free to ask and also feel free to enter the topic.

It’s looking from where I’m sitting that Sceptimatic May have a basic misunderstanding of how things, balls or otherwise move in space, and has a habit of not being clear and at times confused in his statements.

He states if there was a vehicle that took off and flew upward accelerating until it achieved a constant velocity and flew at this velocity for some time and then let’s say dropped something. To me this sounds like a bomber taking off and dropping a bomb.



To my understanding if a plane were to accelerate upwards and drop a bomb the bomb would continue upward for a short period until its upward velocity was zero whereupon it would accelerate downward. The bomb would however continue to move in a forward direction at a velocity initially determined by the release velocity of the aircraft.

If the same plane with a second bomb moving at a constant horizontal velocity released a bomb the bomb would fall under gravity at an initial velocity of that of the aircraft.

If a dumb object like an iron bomb is moving inside a vehicle how it eventually falls is determined by the motion of the aircraft.

The main problem is with the wording of Sceptimatic’s argument as he is invoking an impossible situation that of an object moving vertically upwards with an acceleration of zero. His argument therefore is somewhat an oxymoron.

I think he needs to clarify his argument and present it again.

My argument is this.

On that very same platform going up at a constant velocity, it is under a constant energy load that is keeping that platform moving at a stead mph vertically.
The iron ball on the platform is also moving at the same rate, obviously.
If I was to cut the power in a split second to leave that platform attached to nothing, then that platform and the ball stop dead before accelerating back down.
My argument is that it would not move an inch further vertically after power is cut at a constant rate of mph.

This simply does not match reality.

Quote from: Jonny B Smart on November 25, 2017, 08:18:55 AM

It doesn’t matter which direction you are moving. Moving at a constant velocity requires no force. force = mass x ACCELERATION, remember? No acceleration (change in velocity) means no force.

Moving against a force (gravity or air resistance, for example) requires a balanced force to maintain velocity.

You are massively contradicting yourself.

He really isn’t.

No force doesn’t mean no forces are acting at all but usually means no net force.

It’s looking from where I’m sitting that Sceptimatic May have a basic misunderstanding of how things, balls or otherwise move in space, and has a habit of not being clear and at times confused in his statements.

He states if there was a vehicle that took off and flew upward accelerating until it achieved a constant velocity and flew at this velocity for some time and then let’s say dropped something. To me this sounds like a bomber taking off and dropping a bomb.



To my understanding if a plane were to accelerate upwards and drop a bomb the bomb would continue upward for a short period until its upward velocity was zero whereupon it would accelerate downward. The bomb would however continue to move in a forward direction at a velocity initially determined by the release velocity of the aircraft.

If the same plane with a second bomb moving at a constant horizontal velocity released a bomb the bomb would fall under gravity at an initial velocity of that of the aircraft.

If a dumb object like an iron bomb is moving inside a vehicle how it eventually falls is determined by the motion of the aircraft.

The main problem is with the wording of Sceptimatic’s argument as he is invoking an impossible situation that of an object moving vertically upwards with an acceleration of zero. His argument therefore is somewhat an oxymoron.

I think he needs to clarify his argument and present it again.

I don't need to clarify anything more than I'm doing.

Quote from: sceptimatic on November 26, 2017, 01:10:55 AM

Quote from: Jonny B Smart on November 25, 2017, 08:18:55 AM

It doesn’t matter which direction you are moving. Moving at a constant velocity requires no force. force = mass x ACCELERATION, remember? No acceleration (change in velocity) means no force.

Moving against a force (gravity or air resistance, for example) requires a balanced force to maintain velocity.

You are massively contradicting yourself.

He really isn’t.

No force doesn’t means novgorces are acting at all but usually means no net force.

Give me a clear example of what you're saying which goes against what I'm saying.

Quote from: Jonny B Smart on November 25, 2017, 05:50:34 PM

He hasn’t responded since he acknowledged the validity of f=ma, and then I pointed out that this equation means no acceleration means no force. I think that his model is kaput. (Apparently, if you take the logic out of an idea, it will “stop dead.”)

Second law: Second law: Force on an object is equal to the mass  of that object multiplied by the acceleration of the object: F = ma.

So basically you apply energy to a dense mass and accelerate that mass or move that mass.
In simple terms what you put in as energy you get out as equal energy. Nothing more and nothing less.
Simple enough as long as it's used in the reality of the physical world we actually live in.


Try and understand what it means instead of thinking I'm ducking out.

No! f=ma works with simple algebra.

If you have a mass and want to accelerate it at a certain rate, you multiply to find the amount of force that you need. For example, accelerating a mass of 5 kg at a rate of 10 m/s/s requires 50 Newtons (cute how the unit is named after Him). Accelerating the same mass at 5 m/s/s requires 25 Newtons, 3 m/s/s requires 15 Newtons, 1 m/s/s requires 5 Newtons, and no acceleration requires (can be handle the math?)...5 kg X 0 m/s/s = ...0! That’s right! NO FORCE AT ALL.

But wait! What is “acceleration”? Acceleration is ANY change in velocity. But what is “velocity”? Velocity is speed and direction.

So, traveling at a constant speed in a constant direction requires NO force. The equation does not care which direction you are moving, or whether you are speeding up or slowing down. It takes the same amount of force to go from 10 m/s to 20 m/s as it does to go from 10 m/s to full stop if you make the changes in the same amount of time.

Feel free to get involved is you wish. At least I know you try to understand it even if you might not agree.

Thanks, I'll try to avoid the repeated cliches. I just always hate getting involved in ganging up.

Quote from: Jane on November 25, 2017, 01:32:55 PM

And if this is accurate then, companion question to the first situation, how could you tell the difference between the ball keeping to the surface of the platform staying at a dead stop, and the ball decelerating at the same rate as the platform?

It would depend on the mass of the ball being used as to what you could see clearly or marginally or not at all.


However, there is still one more thing to deal with which is the constant velocity.
This is the one where I'm basically arguing the point of, even though it seems to have jumped from the rocket topic. lol

My argument is this.

On that very same platform going up at a constant velocity, it is under a constant energy load that is keeping that platform moving at a stead mph vertically.
The iron ball on the platform is also moving at the same rate, obviously.
If I was to cut the power in a split second to leave that platform attached to nothing, then that platform and the ball stop dead before accelerating back down.
My argument is that it would not move an inch further vertically after power is cut at a constant rate of mph.

I guess the best set-up for this hypothetical platform would be to attach it to a rope hanging out of a fast lift, or some such thing. So the cut-off here would be to cut the rope, immediately leaving the platform 'attached to nothing' as you say.
If I'm right, I think what you're saying is that the only reason the platform in question moves is because of the lift or whatever generic force is tugging on the rope; the moment that rope is cut you've switched off that force, and an object moving at a constant speed has no way to 'store' any force to keep going with. Hence, immediate stop because you end up with zero force pushing it up, a natural resisitive force pushing it down, and the downwards force is infinitely greater than the upwards so it immediately stops its motion.
Meanwhile an object with mass that accelerates has its own force, so it can resist the resistive force. And meanwhile whatever's propelling the lift has a generator of some description so it provides its own acceleration too.

This is my best reading, at least. It's kind of a variation on f=ma, but using that more as a definition than an equation. Force only exists when an object has acceleration rather than arising at any change in velocity (as in that case coming to a dead stop would cause it, and you'd get a deceleration instead).

 

Quote from: Jane on November 25, 2017, 01:32:55 PM

Assuming these questions are clear this'll probably be my only post on this topic, thanks in advance!

Hopefully they are clear but if not feel free to ask and also feel free to enter the topic.

Thanks, I hope I wasn't too far off. I have far too much fun figuring things like this out.

Quote from: sceptimatic on November 26, 2017, 01:49:33 AM

Feel free to get involved is you wish. At least I know you try to understand it even if you might not agree.

Thanks, I'll try to avoid the repeated cliches. I just always hate getting involved in ganging up.

Quote

Quote from: Jane on November 25, 2017, 01:32:55 PM

And if this is accurate then, companion question to the first situation, how could you tell the difference between the ball keeping to the surface of the platform staying at a dead stop, and the ball decelerating at the same rate as the platform?

It would depend on the mass of the ball being used as to what you could see clearly or marginally or not at all.


However, there is still one more thing to deal with which is the constant velocity.
This is the one where I'm basically arguing the point of, even though it seems to have jumped from the rocket topic. lol

My argument is this.

On that very same platform going up at a constant velocity, it is under a constant energy load that is keeping that platform moving at a stead mph vertically.
The iron ball on the platform is also moving at the same rate, obviously.
If I was to cut the power in a split second to leave that platform attached to nothing, then that platform and the ball stop dead before accelerating back down.
My argument is that it would not move an inch further vertically after power is cut at a constant rate of mph.

I guess the best set-up for this hypothetical platform would be to attach it to a rope hanging out of a fast lift, or some such thing. So the cut-off here would be to cut the rope, immediately leaving the platform 'attached to nothing' as you say.
If I'm right, I think what you're saying is that the only reason the platform in question moves is because of the lift or whatever generic force is tugging on the rope; the moment that rope is cut you've switched off that force, and an object moving at a constant speed has no way to 'store' any force to keep going with. Hence, immediate stop because you end up with zero force pushing it up, a natural resisitive force pushing it down, and the downwards force is infinitely greater than the upwards so it immediately stops its motion.
Meanwhile an object with mass that accelerates has its own force, so it can resist the resistive force. And meanwhile whatever's propelling the lift has a generator of some description so it provides its own acceleration too.

This is my best reading, at least. It's kind of a variation on f=ma, but using that more as a definition than an equation. Force only exists when an object has acceleration rather than arising at any change in velocity (as in that case coming to a dead stop would cause it, and you'd get a deceleration instead).

 

Quote

Quote from: Jane on November 25, 2017, 01:32:55 PM

Assuming these questions are clear this'll probably be my only post on this topic, thanks in advance!

Hopefully they are clear but if not feel free to ask and also feel free to enter the topic.

Thanks, I hope I wasn't too far off. I have far too much fun figuring things like this out.

If what you are saying is correct, then no one could throw anything up. What keeps a ball moving up after you throw it? It’s called momentum.

Quote from: Badxtoss on November 25, 2017, 08:37:51 AM

Except that isn't what happens.  You begin to slow down, you do not stop dead.  If you were moving at 100 kph and you instantly stopped you would likely be dead.
Your example does not match my real world experience.

You are still not grasping what I'm saying.

You said, a bit earlier I think, that if you were driving up a very steep hill at a constant speed of 50 mph and released the gas you would stop dead.  That's not what happens, I've done it.
What am I not grasping?

Quote from: sceptimatic on November 26, 2017, 01:13:02 AM

Quote from: Badxtoss on November 25, 2017, 08:37:51 AM

Except that isn't what happens.  You begin to slow down, you do not stop dead.  If you were moving at 100 kph and you instantly stopped you would likely be dead.
Your example does not match my real world experience.

You are still not grasping what I'm saying.

You said, a bit earlier I think, that if you were driving up a very steep hill at a constant speed of 50 mph and released the gas you would stop dead.  That's not what happens, I've done it.
What am I not grasping?

I think you need to be driving at a constant speed up a perfectly vertical road...

Quote from: sceptimatic on November 26, 2017, 01:49:33 AM

Feel free to get involved is you wish. At least I know you try to understand it even if you might not agree.

Thanks, I'll try to avoid the repeated cliches. I just always hate getting involved in ganging up.

Quote

Quote from: Jane on November 25, 2017, 01:32:55 PM

And if this is accurate then, companion question to the first situation, how could you tell the difference between the ball keeping to the surface of the platform staying at a dead stop, and the ball decelerating at the same rate as the platform?

It would depend on the mass of the ball being used as to what you could see clearly or marginally or not at all.


However, there is still one more thing to deal with which is the constant velocity.
This is the one where I'm basically arguing the point of, even though it seems to have jumped from the rocket topic. lol

My argument is this.

On that very same platform going up at a constant velocity, it is under a constant energy load that is keeping that platform moving at a stead mph vertically.
The iron ball on the platform is also moving at the same rate, obviously.
If I was to cut the power in a split second to leave that platform attached to nothing, then that platform and the ball stop dead before accelerating back down.
My argument is that it would not move an inch further vertically after power is cut at a constant rate of mph.

I guess the best set-up for this hypothetical platform would be to attach it to a rope hanging out of a fast lift, or some such thing. So the cut-off here would be to cut the rope, immediately leaving the platform 'attached to nothing' as you say.
If I'm right, I think what you're saying is that the only reason the platform in question moves is because of the lift or whatever generic force is tugging on the rope; the moment that rope is cut you've switched off that force, and an object moving at a constant speed has no way to 'store' any force to keep going with. Hence, immediate stop because you end up with zero force pushing it up, a natural resisitive force pushing it down, and the downwards force is infinitely greater than the upwards so it immediately stops its motion.
Meanwhile an object with mass that accelerates has its own force, so it can resist the resistive force. And meanwhile whatever's propelling the lift has a generator of some description so it provides its own acceleration too.

This is my best reading, at least. It's kind of a variation on f=ma, but using that more as a definition than an equation. Force only exists when an object has acceleration rather than arising at any change in velocity (as in that case coming to a dead stop would cause it, and you'd get a deceleration instead).

 

Quote

Quote from: Jane on November 25, 2017, 01:32:55 PM

Assuming these questions are clear this'll probably be my only post on this topic, thanks in advance!

Hopefully they are clear but if not feel free to ask and also feel free to enter the topic.

Thanks, I hope I wasn't too far off. I have far too much fun figuring things like this out.

In a way you are sort of getting there but also to keep a constant velocity vertically you still have to have a force but that force to keep that mass in that particular atmosphere at each height always requires not only that applied energy but requires more energy to advance vertically to keep that constant velocity.

I hope I've explained that to you so you can understand it from my point.

Quote from: Jane on November 26, 2017, 06:57:27 AM

Quote from: sceptimatic on November 26, 2017, 01:49:33 AM

Feel free to get involved is you wish. At least I know you try to understand it even if you might not agree.

Thanks, I'll try to avoid the repeated cliches. I just always hate getting involved in ganging up.

Quote

Quote from: Jane on November 25, 2017, 01:32:55 PM

And if this is accurate then, companion question to the first situation, how could you tell the difference between the ball keeping to the surface of the platform staying at a dead stop, and the ball decelerating at the same rate as the platform?

It would depend on the mass of the ball being used as to what you could see clearly or marginally or not at all.


However, there is still one more thing to deal with which is the constant velocity.
This is the one where I'm basically arguing the point of, even though it seems to have jumped from the rocket topic. lol

My argument is this.

On that very same platform going up at a constant velocity, it is under a constant energy load that is keeping that platform moving at a stead mph vertically.
The iron ball on the platform is also moving at the same rate, obviously.
If I was to cut the power in a split second to leave that platform attached to nothing, then that platform and the ball stop dead before accelerating back down.
My argument is that it would not move an inch further vertically after power is cut at a constant rate of mph.

I guess the best set-up for this hypothetical platform would be to attach it to a rope hanging out of a fast lift, or some such thing. So the cut-off here would be to cut the rope, immediately leaving the platform 'attached to nothing' as you say.
If I'm right, I think what you're saying is that the only reason the platform in question moves is because of the lift or whatever generic force is tugging on the rope; the moment that rope is cut you've switched off that force, and an object moving at a constant speed has no way to 'store' any force to keep going with. Hence, immediate stop because you end up with zero force pushing it up, a natural resisitive force pushing it down, and the downwards force is infinitely greater than the upwards so it immediately stops its motion.
Meanwhile an object with mass that accelerates has its own force, so it can resist the resistive force. And meanwhile whatever's propelling the lift has a generator of some description so it provides its own acceleration too.

This is my best reading, at least. It's kind of a variation on f=ma, but using that more as a definition than an equation. Force only exists when an object has acceleration rather than arising at any change in velocity (as in that case coming to a dead stop would cause it, and you'd get a deceleration instead).

 

Quote

Quote from: Jane on November 25, 2017, 01:32:55 PM

Assuming these questions are clear this'll probably be my only post on this topic, thanks in advance!

Hopefully they are clear but if not feel free to ask and also feel free to enter the topic.

Thanks, I hope I wasn't too far off. I have far too much fun figuring things like this out.

In a way you are sort of getting there but also to keep a constant velocity vertically you still have to have a force but that force to keep that mass in that particular atmosphere at each height always requires not only that applied energy but requires more energy to advance vertically to keep that constant velocity.

I hope I've explained that to you so you can understand it from my point.

No. A constant velocity requires no force. A constant velocity opposing a force requires a balancing force so that the net forces are zero. If you were to climb high enough, gravity would decrease noticeably, and the force required to continue at a constant vertical velocity would decrease as well.

If what you are saying is correct, then no one could throw anything up. What keeps a ball moving up after you throw it? It’s called momentum.

And that's why I like to actually talk to someone rather than talking about a few choice words. You can make informed arguments rather than just convincing whoever you're talking to that you're just trolling.
What keeps the ball moving would be its acceleration. Momentum, as the gauge of how hard something is to stop, would be a factor generally speaking, but if you want to consider this as an alternative model it wouldn't be required. Plus momentum is a property, not a driving force. It's basically a measure of how kinetic energy varies with respect to velocity. It's the consequences of that Scepti's debating.

In a way you are sort of getting there but also to keep a constant velocity vertically you still have to have a force but that force to keep that mass in that particular atmosphere at each height always requires not only that applied energy but requires more energy to advance vertically to keep that constant velocity.

I hope I've explained that to you so you can understand it from my point.

Ah, yep, right.
I'm kind of thinking of this in terms of storage, seems to be a good way to visualise a lot of it.
An object accelerating up is basically getting loaded up with 'up' velocities. A few downs from resistive forces (gravity/denpressure, air resistance etc) might get into it, but they're outweighed and the net velocity is up, and increasing. When it gets cut off from acceleration the ups on board don't vanish, but there's only a finite number of them and they'd quickly be outweighed by the downs, so the upward motion slowns as more and more ups get cancelled, and then there's nothing, and then there's donwards.
Meanwhile an object moving at a constant speed gets loaded with ups at the same rate the downs cancel them out, so there's no leftover force and as soon as ups stop being added the downs dominate and hence downward velocity.

Is that an accurate interpretation?

Might just be my weird way of thinking about it, but it seems to be more about acceleration than momentum. It's less to do with energy, more to do with velocity being incremental; that's the only way a dead stop would work from what I can see, otherwise the object would be deccelerating at an infinite rate, even if just for an instant. The lack of any time to decelerate though works just fine with increments, it just goes from velocity A to velocity B.
That might be a lot of nonsense though.

Ah, yep, right.
I'm kind of thinking of this in terms of storage, seems to be a good way to visualise a lot of it.
An object accelerating up is basically getting loaded up with 'up' velocities. A few downs from resistive forces (gravity/denpressure, air resistance etc) might get into it, but they're outweighed and the net velocity is up, and increasing. When it gets cut off from acceleration the ups on board don't vanish, but there's only a finite number of them and they'd quickly be outweighed by the downs, so the upward motion slowns as more and more ups get cancelled, and then there's nothing, and then there's donwards.
Meanwhile an object moving at a constant speed gets loaded with ups at the same rate the downs cancel them out, so there's no leftover force and as soon as ups stop being added the downs dominate and hence downward velocity.

Is that an accurate interpretation?

If I'm interpreting what you are trying to interpret from my point then yes, you appear to be on the right lines.

Might just be my weird way of thinking about it, but it seems to be more about acceleration than momentum. It's less to do with energy, more to do with velocity being incremental; that's the only way a dead stop would work from what I can see, otherwise the object would be deccelerating at an infinite rate, even if just for an instant. The lack of any time to decelerate though works just fine with increments, it just goes from velocity A to velocity B.
That might be a lot of nonsense though.

You see, I class acceleration as one thing. I never use constant acceleration.
Acceleration is simply just that. If you're accelerating you are never constant, because your speed is ever changing and this is the only real way you can decelerate.

You can never decelerate from a constant velocity.

As you accelerate vertically you have to add thrust to push the mass or you have to lose the mass of the object (not the fuel) to keep the acceleration.
If you can't lose the mass of the object then you have to increase the thrust.
However, if you are already at full thrust then all you have left is to lose fuel mass to keep that thrust.

The issue with this is, at best you keep a constant velocity and you can only keep it for each portion of atmosphere you plough through and push against.
A constant velocity cannot ever result in a deceleration because you are never accelerating, so a vertical deceleration that would allow the object an ever slowing forward vertical motion would not be relevant.

It all comes back to the rope climbing and snapping scenario.
If you were to jump on a rope and yank yourself up it and then it snaps, you will still advance a little due to your initial springboard start or accelerated start.

If after your springboard accelerated start you then start the climb and hit a constant velocity, you are required to use more energy with each movement just to keep that constant velocity, because you cannot lose your mass to aid you.
If that rope snaps at any time during any of that climb, you're going to stop dead and then accelerate downwards. You will never gain another inch UP.


No matter what you're up against in whatever way vertically, you will never keep accelerating unless you lose the mass of the object and the fuel of it at the same time.
You will never do it by losing just fuel. All you can achieve is a constant velocity at best.



Obviously space rockets defy this because they kick themselves up their own arses, apparently. lol.

Quote from: Jonny B Smart on November 26, 2017, 07:13:50 AM

If what you are saying is correct, then no one could throw anything up. What keeps a ball moving up after you throw it? It’s called momentum.

And that's why I like to actually talk to someone rather than talking about a few choice words. You can make informed arguments rather than just convincing whoever you're talking to that you're just trolling.
What keeps the ball moving would be its acceleration. Momentum, as the gauge of how hard something is to stop, would be a factor generally speaking, but if you want to consider this as an alternative model it wouldn't be required. Plus momentum is a property, not a driving force. It's basically a measure of how kinetic energy varies with respect to velocity. It's the consequences of that Scepti's debating.

Quote from: sceptimatic on November 26, 2017, 07:54:10 AM

In a way you are sort of getting there but also to keep a constant velocity vertically you still have to have a force but that force to keep that mass in that particular atmosphere at each height always requires not only that applied energy but requires more energy to advance vertically to keep that constant velocity.

I hope I've explained that to you so you can understand it from my point.

Ah, yep, right.
I'm kind of thinking of this in terms of storage, seems to be a good way to visualise a lot of it.
An object accelerating up is basically getting loaded up with 'up' velocities. A few downs from resistive forces (gravity/denpressure, air resistance etc) might get into it, but they're outweighed and the net velocity is up, and increasing. When it gets cut off from acceleration the ups on board don't vanish, but there's only a finite number of them and they'd quickly be outweighed by the downs, so the upward motion slowns as more and more ups get cancelled, and then there's nothing, and then there's donwards.
Meanwhile an object moving at a constant speed gets loaded with ups at the same rate the downs cancel them out, so there's no leftover force and as soon as ups stop being added the downs dominate and hence downward velocity.

Is that an accurate interpretation?

Might just be my weird way of thinking about it, but it seems to be more about acceleration than momentum. It's less to do with energy, more to do with velocity being incremental; that's the only way a dead stop would work from what I can see, otherwise the object would be deccelerating at an infinite rate, even if just for an instant. The lack of any time to decelerate though works just fine with increments, it just goes from velocity A to velocity B.
That might be a lot of nonsense though.

Your explanation is pretty good, but common terms are helpful.

Let’s ignore changes in direction. “Acceleration” is increasing speed. Objects do not remain at a constant speed due to acceleration. If you drive your car, you accelerate until you reach cruising speed. That requires force, and you experience inertia when you feel pushed back in your seat. Once at cruising speed, you no longer feel pushed back in your seat because you are not accelerating. You still need to apply enough force to balance the resistance force of friction. If you apply the brakes, you are increasing friction to resist motion. This causes deceleration—a decrease in speed. This causes you to lean forward due to inertia.

If a rocket accelerates to 100 m/s and then cuts its engines, it will continue upward, but begin to decelerate at 9.8 m/s/s (let’s call it 10 for easy math). After one second, it will be traveling at 90 m/s, then 80, 70, 60...until it stops. After it has stopped, it will accelerate toward the ground. First 10 m/s, then 20, 30, 40...This is pretty much what you said, but a lot of research has worked out the exact math of it.

Momentum is a product of mass and velocity. By “product,” I mean “result of multiplication.”

One of the best one liners from Sceptimatic:-

"You can never decelerate from a constant velocity"

Aircraft cruising at 36,000 ft heading with a ground speed  of 560mph. According to Sceptimatic  this aircraft could never slow down!  And he things he has a firm grasp on the laws of motion!

 You see, I class acceleration as one thing. I never use constant acceleration.

Acceleration can be constant. It is a rate of increase. If I drive from a stop to 10 km/h in one second, then that’s 10 km/h/s. If after two seconds I’m going 20 km/h, and after three seconds I’m going 30 km/h, that’s a constant acceleration of 10 km/h/s.

Acceleration is simply just that. If you're accelerating you are never constant, because your speed is ever changing and this is the only real way you can decelerate.

You can never decelerate from a constant velocity.

Really? So if I’m traveling at 100 km/h for an hour, I can never slow down?

As you accelerate vertically you have to add thrust to push the mass or you have to lose the mass of the object (not the fuel) to keep the acceleration. 

Why is rocket fuel not a part of the mass that needs thrusting? Burning fuel reduces the weight of the rocket. By f=ma, acceleration increases exactly as the expending of fuel reduces the mass of the rocket.

If you can't lose the mass of the object then you have to increase the thrust.

If the thrust is already greater than the weight of the rocket, it will continue to accelerate. Decreasing the mass accelerates it faster.

However, if you are already at full thrust then all you have left is to lose fuel mass to keep that thrust.

Thrust is a result of the potential chemical energy stored in the fuel and the efficiency of the engines. Losing mass doesn’t affect the amount of thrust (unless you burn all of your fuel).

The issue with this is, at best you keep a constant velocity and you can only keep it for each portion of atmosphere you plough through and push against.
A constant velocity cannot ever result in a deceleration because you are never accelerating, so a vertical deceleration that would allow the object an ever slowing forward vertical motion would not be relevant.

If you throttle back on a rocket engine until the thrust is less than the weight of the rocket, it will decelerate.

It all comes back to the rope climbing and snapping scenario.
If you were to jump on a rope and yank yourself up it and then it snaps, you will still advance a little due to your initial springboard start or accelerated start.

If after your springboard accelerated start you then start the climb and hit a constant velocity, you are required to use more energy with each movement just to keep that constant velocity, because you cannot lose your mass to aid you.
If that rope snaps at any time during any of that climb, you're going to stop dead and then accelerate downwards. You will never gain another inch UP.

That’s because rope climbing involves very short bursts of acceleration followed by short pauses. There’s always a jerkiness to it. Nobody climbs a rope at a steady speed.

No matter what you're up against in whatever way vertically, you will never keep accelerating unless you lose the mass of the object and the fuel of it at the same time.
You will never do it by losing just fuel. All you can achieve is a constant velocity at best.

Rockets will travel at a constant velocity if their thrust equals their weight. If thrust is greater than their weight, then they will accelerate upward.

Quote from: Jane on November 26, 2017, 09:43:25 AM

Ah, yep, right.
I'm kind of thinking of this in terms of storage, seems to be a good way to visualise a lot of it.
An object accelerating up is basically getting loaded up with 'up' velocities. A few downs from resistive forces (gravity/denpressure, air resistance etc) might get into it, but they're outweighed and the net velocity is up, and increasing. When it gets cut off from acceleration the ups on board don't vanish, but there's only a finite number of them and they'd quickly be outweighed by the downs, so the upward motion slowns as more and more ups get cancelled, and then there's nothing, and then there's donwards.
Meanwhile an object moving at a constant speed gets loaded with ups at the same rate the downs cancel them out, so there's no leftover force and as soon as ups stop being added the downs dominate and hence downward velocity.

Is that an accurate interpretation?

If I'm interpreting what you are trying to interpret from my point then yes, you appear to be on the right lines.

Quote from: Jane on November 26, 2017, 09:43:25 AM

Might just be my weird way of thinking about it, but it seems to be more about acceleration than momentum. It's less to do with energy, more to do with velocity being incremental; that's the only way a dead stop would work from what I can see, otherwise the object would be deccelerating at an infinite rate, even if just for an instant. The lack of any time to decelerate though works just fine with increments, it just goes from velocity A to velocity B.
That might be a lot of nonsense though.

You see, I class acceleration as one thing. I never use constant acceleration.
Acceleration is simply just that. If you're accelerating you are never constant, because your speed is ever changing and this is the only real way you can decelerate.

You can never decelerate from a constant velocity.

As you accelerate vertically you have to add thrust to push the mass or you have to lose the mass of the object (not the fuel) to keep the acceleration.
If you can't lose the mass of the object then you have to increase the thrust.
However, if you are already at full thrust then all you have left is to lose fuel mass to keep that thrust.

The issue with this is, at best you keep a constant velocity and you can only keep it for each portion of atmosphere you plough through and push against.
A constant velocity cannot ever result in a deceleration because you are never accelerating, so a vertical deceleration that would allow the object an ever slowing forward vertical motion would not be relevant.

It all comes back to the rope climbing and snapping scenario.
If you were to jump on a rope and yank yourself up it and then it snaps, you will still advance a little due to your initial springboard start or accelerated start.

If after your springboard accelerated start you then start the climb and hit a constant velocity, you are required to use more energy with each movement just to keep that constant velocity, because you cannot lose your mass to aid you.
If that rope snaps at any time during any of that climb, you're going to stop dead and then accelerate downwards. You will never gain another inch UP.


No matter what you're up against in whatever way vertically, you will never keep accelerating unless you lose the mass of the object and the fuel of it at the same time.
You will never do it by losing just fuel. All you can achieve is a constant velocity at best.



Obviously space rockets defy this because they kick themselves up their own arses, apparently. lol.

I have to say this is a classic post. Where to start is a difficult one. What it does demonstrate is that Sceptimatic hasnt the first clue about the laws of motion. Jane I hasten to add comes in a constant velocity second place. LoL possibly with a bit of deceleration.
Arguing the toss with Sceptimatic is a bit like chewing on a piece of cardboard, you can work out the rest if you care.
It also explains why he never produced anything on denpressure as the man is barely literate.
Wow...LOL....

"It all comes back to the rope climbing and snapping scenario"

It sure does...lol...What a fruitcake.