Motor Daddy
Valued Senior Member
Well I know I'm kickin' ass when people start screaming "lock the thread." They can't take the abuse anymore and just want the punishment to stop. It is destroying everything they believe in!!
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The Swing of a Pendulum
- Thread starter Motor Daddy
- Start date
Well I know I'm kickin' ass when people start screaming "lock the thread." They can't take the abuse anymore and just want the punishment to stop. It is destroying everything they believe in!!
Well I know I'm kickin' ass when people start screaming "lock the thread." They can't take the abuse anymore and just want the punishment to stop. It is destroying everything they believe in!!
Far from it.
You just keep showing us over and over how TOTALLY ignorant you are and I find it revolting! You seriously need to go back to tinkering with motors (or whatever it is that you're actually capable of) and quit bothering the intelligent people who actually understand physics!
The pendulum is what indicates acceleration. A rocket pendulum won't be swinging like a clock pendulum on earth but it will be more like a ball on a string. When there is an acceleration the weight will be left behind , the string will get tight and stretch a little. Once it stops accelerating the stretch will pull the weight ahead of its origin. So it is the motion of the weight that indicates acceleration, even if the relative motion of the two rockets or cars stay the same.No it isn't but if you think so then how is it that you claim there is a change in velocity between two rockets when the pendulum swings when the distance between the rockets stays the same at all times???
Not so much as a "change in velocity in the absolute frame" but the prior situation prior to the acceleration (shown by the action of the pendulum on board).What do you mean they accelerate at the same rate? The distance between the rockets remains the same at ALL times!
If you want to talk about acceleration in this scenario you MUST talk about it in the absolute frame! The pendulum is swinging due to the change in velocity in the absolute frame!
Do you not understand that you can't just change numbers like that without causing all sorts of changes in the universe?
Sure you can. No one but you cares what the initial speed of that object was. And the only reason YOU might care is if you want to use it to measure acceleration.
Each set of numbers means something. Each set of numbers means a specific reality. You are changing realities when you switch the numbers like that.
Really? When you change your speedometer from reading mph to kph, do you experience bone-crushing acceleration as you accelerate instantaneously to the larger number? Or can you just switch those numbers without changing the physical reality?
That's not what I said. I said if you assumed the initial velocity was 99 m/s and the velocity was 100 m/s after 1 second of acceleration then that is not an acceleration of 100 m/s^2.
Right, it's an acceleration of 1 m/s, assuming the same constraints. You can use such a method to measure any acceleration you like, using any initial speed you like.
How do you know if your acceleration is increasing velocity or decreasing velocity?
You don't. All you know is that you have CHANGED velocity. Whether you have increased or decreased it depends on which coordinate system you choose.
By convention you would have a fore and aft. Then motion of the pendulum would define acceleration or deceleration.
You haven't managed to discuss the detailed explanation I gave of the 2 rocket example, why should we move on to other things? I know you have to resort to the 'Gish Gallop' tactic of throwing out as many ignorant claims and swamping peoples ability or desire to respond to them all but that doesn't make them automatically valid. You have not actually shown any contradiction in any kinematics, you have simply asserted things without justification. I said I'd discuss the torque question when you can actually properly formalise it. That's part of the problem talking with people like you, you cannot actually properly define a scenario, due to your lack of mathematical capabilities, and thus you always leave arm waving 'wiggle room' to ignore people who explain your mistakes.Still workin' on that torque/HP problem? Damn, it's taking you quite some time. You must really be working hard on that one. If you need assistance let me know, I'm here to help!
Just as the car/rocket example I formalised using basic kinematics can you do likewise with your torque example. I said I'd discuss it with you if you could. You then saying "You're taking a long time!" is dishonest. And saying "I'm here to help" is laughable, since I've demonstrated I can formalise such scenarios while you're showing you cannot do it nor can you understand when others do it for you.
Until such time as you can formally construct the scenario pertaining to torque I do not see sufficient evidence that you understand it. Prove to me you understand it and can formalise that understanding and we'll talk. If you continue with this "You're taking a long time! Where's your response?" then I'll simply point you back to this post and report you for trolling. I'm asking you for a minimal level of understanding and honesty. I know you consider such things 'absolute nonsense' but that isn't my problem, it is yours.
Which no one has denied.I choose the ruler method. We extend a ruler between the rockets and it reads 100 meters at t=0. At t=1 it reads 100 meters. At t=2 is reads 100 meters, so the distance remains unchanged and for the purposes of this entire exercise the distance between the rockets remains 100 meters.
A car alone has no second object to be moving relative to but that doesn't matter. A single car which undergoes the acceleration described previously will still be able to measure that acceleration using accelerometers. It will then be able to say Initially I was in an inertial frame F. I underwent acceleration of magnitude a for time T. I am now in another inertial frame F'. The relative velocity difference between F' and F is therefore $$\Delta v = aT$$, F' is moving at speed v=aT compared to F. No need to know what, if any, absolute motion F and F' have. The car doesn't need a second object to compare itself with, it can do it all on its own since acceleration can be measured directly.No, I don't accept that. One car alone in space has no relative velocity to another object and you have no way of knowing the absolute velocity like I do. You have no measure of velocity or change of velocity with no relative velocity, do you??? If you do please share with the entire group exactly which velocity you are talking about if it isn't a relative velocity or a absolute velocity?
I have repeatedly explained this to you, talking about frames F, F' and F''. Obviously you didn't understand and yet you see no problem in the fact you get snarky and make assertions. You must know you don't understand, you must be aware how the mathematical expressions and methods I posted are things you don't understand. A rational person would then temper their responses, aware that perhaps there's a problem with their own understanding rather than everyone else, some of whom understand said mathematics.
If you ask the same question again, in a manner which is clearly meant to imply I haven't responded to the question when I have several times, then I'll start reporting you. I have no problem with people who don't know physics but want to understand it but someone who is wantonly ignorant and deliberately so I cannot abide.
AN, in order to know an acceleration you first must have measured a change in velocity.Do you know how accelerometers work? Hell, pendulums show it up. A pendulum not experiencing any horizontal acceleration will hang vertically down. When it is accelerated, say because it is in a car and the driver put his foot down more than it was initially, then it will tilt in the direction opposite to the acceleration. The angle it tilts allows you, again by trivial mathematics, to compute the magnitude of the acceleration. No need to know what speed the car was going initially, the acceleration is not dependent upon that. For a rocket it is different, as there's no gravity in space. Instead you can use a ball held in place by 6 springs (up/down/left/right/forwards/backwards orientations). The same thing occurs, under acceleration the ball is displaced from the centre and you can compute the magnitude and direction of the acceleration by some basic maths using the property of springs. No need to do position measurements or velocity measurements for the rocket, instead you measure where the ball is in the spring system and you're done.In order to measure a change in velocity you first must have measured a velocity. In order to have measured a velocity you first must have measured a change in distance over a duration of time. You have no basis for claiming you changed distance, velocity, or even changed velocity since all your measurements are based on the other rocket. You can't just claim an object accelerates at say 9.81 m/s^2 like a beach ball at rest on a beach. The ball is not getting any closer to the center of the earth so the ball's velocity is 0 m/s towards the center of the earth. Ten minutes later when you measure it again the distance to the center of the earth from the beach ball is exactly the same, so the ball's velocity is the same, no change over a duration of time. No change in velocity over a duration of time means a zero acceleration, ie the ball is not accelerating towards the center of the earth because the velocity of the ball towards the center of the earth is not changing, it remains at 0 m/s while at rest on the beach!
Since I know you don't understand this (despite it too being taught to children) I'll do a 1 dimensional spring/mass example in general.
A ball of mass M is held in place at x=0 by a pair of massless springs (you can do it with just 1 but it doesn't matter) with spring constant k. (bad diagram but it looks like |\/\/\/\/\/\O/\/\/\/\/\/\|) The springs are not in tension (ie being compressed or stretched) when the ball is at x=0. This spring/mass system is inside a box inside a car. The car accelerates at rate $$a$$. The spring/ball system moves, causing the left hand spring to be compressed and the right hand to stretch, |\/\/\/\/\O/\/\/\/\/\/\/\|. . The system moves to an new stationary equilibrium, with the ball at x=-L. Since the system is accelerating at rate a the ball must be and so it is experiencing a force F = Ma. The springs therefore must be applying this force. The spring which has been squashed by an amount L will push back with a force +kL. Likewise the stretched spring will pull back with a force +kL. Therefore the ball is experiencing a force F=+2kL. So we have Ma = 2kL, so a = 2kL/M = (2k/M)L. So if we know the strength of the springs, the mass of the ball and the distance it has moved we can compute the acceleration it experienced.
Nowhere have I had to say what the initial or final velocity of the car is. It could have started from stationary, with respect to the road, moving at 100kph or reversing a 54kph, anything. The acceleration measurement only requires me to measure the displacement of the spring mounted ball.
I know I keep saying it but it really is true, this stuff is covered by children. I remember doing this when I was 14 or 15! Balls on springs or pendulums swinging is the stuff of basic physics lessons!
As I just explained, the measurement of the acceleration really is very straight forward. The fact you didn't know this and also couldn't think up a way of doing it shows how narrow and flimsy your grasp of such things is. Must be somewhat depressing to have such a narrow view of things, struggling to grasp what children take for granted.No, because you did not measure a 10 m/s^2 acceleration you just claimed it. You have no measurements so how do you know what the velocity or increase/decrease in velocity was?
Do you understand that when you said the acceleration rate was 10 m/s^2 that that was an acceleration in the absolute frame, one which you have no means of measuring, so how would you know that??
I still maintain that this thread has run it's course and should be abandoned. It could be allowed to continue for 5 years and Motor Daddy would still be no wiser. I certainly understand the principle of trying to help someone learn something (I've spent considerable time and effort doing it myself) - but in his case he has NO desire to learn.
Just as with the old saying about a horse and water, you cannot teach something to someone who has no interest in learning. He's nothing more than a worthless troll.
Just as with the old saying about a horse and water, you cannot teach something to someone who has no interest in learning. He's nothing more than a worthless troll.
.. Do you know how accelerometers work? Hell, pendulums show it up. A pendulum not experiencing any horizontal acceleration will hang vertically down. When it is accelerated, say because it is in a car and the driver put his foot down more than it was initially, then it will tilt in the direction opposite to the acceleration. The angle it tilts allows you, again by trivial mathematics, to compute the magnitude of the acceleration. No need to know what speed the car was going initially, the acceleration is not dependent upon that. For a rocket it is different, as there's no gravity in space. Instead you can use a ball held in place by 6 springs (up/down/left/right/forwards/backwards orientations). The same thing occurs, under acceleration the ball is displaced from the centre and you can compute the magnitude and direction of the acceleration by some basic maths using the property of springs. No need to do position measurements or velocity measurements for the rocket, instead you measure where the ball is in the spring system and you're done.
Since I know you don't understand this (despite it too being taught to children) I'll do a 1 dimensional spring/mass example in general.
A ball of mass M is held in place at x=0 by a pair of massless springs (you can do it with just 1 but it doesn't matter) with spring constant k. (bad diagram but it looks like |\/\/\/\/\/\O/\/\/\/\/\/\|) The springs are not in tension (ie being compressed or stretched) when the ball is at x=0. This spring/mass system is inside a box inside a car. The car accelerates at rate $$a$$. The spring/ball system moves, causing the left hand spring to be compressed and the right hand to stretch, |\/\/\/\/\O/\/\/\/\/\/\/\|. . The system moves to an new stationary equilibrium, with the ball at x=-L. Since the system is accelerating at rate a the ball must be and so it is experiencing a force F = Ma. The springs therefore must be applying this force. The spring which has been squashed by an amount L will push back with a force +kL. Likewise the stretched spring will pull back with a force +kL. Therefore the ball is experiencing a force F=+2kL. So we have Ma = 2kL, so a = 2kL/M = (2k/M)L. So if we know the strength of the springs, the mass of the ball and the distance it has moved we can compute the acceleration it experienced ..
Balls on springs or pendulums swinging is the stuff of basic physics lessons! ..
Ha ! You must have twigged to that from what my learned self said in a post way back then ..
Wouldn't then, this movement of the pendulum due to a rocket boost, simply be showing it's inertia ? (post #85)
But I do think MD is making a deeper point - though I haven't been able to work out what it is - yet!
I still maintain that this thread has run it's course and should be abandoned. It could be allowed to continue for 5 years and Motor Daddy would still be no wiser. I certainly understand the principle of trying to help someone learn something (I've spent considerable time and effort doing it myself) - but in his case he has NO desire to learn.
Just as with the old saying about a horse and water, you cannot teach something to someone who has no interest in learning. He's nothing more than a worthless troll.
Nah! This is afterall, On the Fringe, Alternative Theories, etc, and I doubt Sf has run out of bandwidth.
I'm finding this discussion very interesting, enlightening and educational - irrespective of who is right or wrong.
And frankly, sceaming 'lock this thread', etc, kinda reminds me of the Emperors courtesans trying to silence the little boy ..
Motor Daddy
Valued Senior Member
The pendulum is what indicates acceleration. A rocket pendulum won't be swinging like a clock pendulum on earth but it will be more like a ball on a string. When there is an acceleration the weight will be left behind , the string will get tight and stretch a little. Once it stops accelerating the stretch will pull the weight ahead of its origin. So it is the motion of the weight that indicates acceleration, even if the relative motion of the two rockets or cars stay the same.
Correct, the pendulum indicates acceleration, which is change in velocity in the absolute frame. Acceleration is not a change in relative velocity, as evidence in the OP, the pendulum is swinging but the relative velocity stays the same. You have no way to measure that change of velocity because the relative velocity is not changing and you don't believe in an absolute velocity, so which velocity is it that you claim is changing?
Motor Daddy
Valued Senior Member
Not so much as a "change in velocity in the absolute frame" but the prior situation prior to the acceleration (shown by the action of the pendulum on board).
Acceleration is the rate of change of velocity in the absolute frame! You MUST have an absolute frame in order to measure an acceleration. By measure an acceleration I mean not just that it is indicated, but that you can tell me the rate of acceleration in units of m/s^2, so that I may check your work against the light sphere.
Motor Daddy
Valued Senior Member
Sure you can. No one but you cares what the initial speed of that object was. And the only reason YOU might care is if you want to use it to measure acceleration.
You didn't learn from your previous mistake did you? I'll ask you again, if after 1 second of acceleration the absolute velocity is 100 m/s, what was the acceleration? You seem to live in a world of no reality, but one in which you can make up your own reality. That's not how history works. History is a record of the past, and there is only one past.
Motor Daddy
Valued Senior Member
By convention you would have a fore and aft. Then motion of the pendulum would define acceleration or deceleration.
Oh, you want to play the direction game? Can I play? I'm really good at the direction game, because in order to play the direction game you must know the absolute velocities in order for all the pieces of the puzzle to fit properly.
And as I just explained, it is possible to construct a physical device from which you can measure acceleration, in whatever units of acceleration you desire and without need to know the motion of the vehicle/object the accelerometer is attached to.Acceleration is the rate of change of velocity in the absolute frame! You MUST have an absolute frame in order to measure an acceleration. By measure an acceleration I mean not just that it is indicated, but that you can tell me the rate of acceleration in units of m/s^2, so that I may check your work against the light sphere.
Would you like me to go through a similar mathematical formalisation and description for a pendulum based accelerometer? Pendulums and their behaviour under accelerations (or decelerations) are how seat belts work so it is another practical device in common use which shows you're mistaken.
Correct, the pendulum indicates acceleration, which is change in velocity in the absolute frame. Acceleration is not a change in relative velocity, as evidence in the OP, the pendulum is swinging but the relative velocity stays the same. You have no way to measure that change of velocity because the relative velocity is not changing and you don't believe in an absolute velocity, so which velocity is it that you claim is changing?
You didn't reply to my much earlier post, so I'll ask again, and hope you can clear this up for me.
If the two vehicles are moving at the same speed, always, no change whatsoever, aren't they then the same as one vehicle ? Couldn't we equally say they would behave the same as though they ere connected, say by a rigid frame, therefore one vehicle ? Where is the relative velocity then ?
And isn't the movement of the pendulum simply due to it's inertia ? Such as how AN detailed in post 227 above, being how accelerometers work ?
Consider a pendulum formed from a massless string of length L and a point mass at the end mass M in a constant gravitational field of strength g pointing vertically down (so for Earth g=9.81 m/s^2). The accelerometer the pendulum is in is attached to a car which accelerates with acceleration a, in the horizontal direction. The pendulum will be caused to tilt in the opposite direction due to inertia of the point mass (just as putting your foot down when driving pushes you into your seat). So what is the relationship between the angle of the pendulum and the magnitude of a?
To compute this we resolve forces about an equilibrium configuration. We want the sum of gravitational force $$M\mathbf{g} = (0,-Mg)$$, the acceleration $$M\mathbf{a} = (-Ma,0)$$ and the string tension $$\mathbf{T} = T(\sin \theta,\cos \theta)$$, where $$\theta$$ is deflection from the vertical in a clockwise manner. If the system is in its stationary tilted state then the sum should be zero. Therefore
$$M\mathbf{g} + M\mathbf{a} + \mathbf{T} = \mathbf{0}$$ which in terms of components gives
* Horizontal : $$0 - Ma + T\sin\theta = 0$$ which gives $$T\sin \theta = Ma$$
* Vertical : $$-Mg + 0 + T\cos \theta = 0$$ which gives $$T\cos \theta = Mg$$
Taking the ratio removes T and M and gives $$\tan \theta = \frac{a}{g}$$ and so $$\theta = \tan^{-1}(\frac{a}{g})$$ or $$a = g\tan \theta$$. A quick sanity check is to put in a=0, which means no acceleration and the pendulum should hang vertically. If a=0 then $$\tan \theta = 0$$ and so $$\theta = 0$$, as required.
Therefore if we know the strength of the gravitational field (g=9.81 for Earth's surface) and we know the angle the pendulum tilts then we can compute the acceleration the vehicle is experiencing. No need to know what the velocity is, relative or otherwise, since the only thing which matters is the acceleration, hence why it is an absolute concept not a relative one.
This is covered in high school physics class (teaches kids how to work with vectors and trigonometry), since you don't grasp any algebra I'll give a numerical example. The mass has M=1kg, g=9.81m/s^2 and the car accelerates such that the angle of the pendulum moves to $$\theta = \frac{\pi}{4}$$. Since $$\tan \frac{\pi}{4} = 1$$ we have $$a = g\tan \theta = g = 9.81$$. This makes sense because if a=g then the pendulum should be in a position symmetric between the horizontal and vertical, which is when $$\theta = \frac{\pi}{4}$$.
There, another high school home work problem example to teach you basic mechanics. Like I said, seat belt mechanisms work by this principle, when the deflection of the pendulum goes past a certain point the belt locks in place as this means the passenger is experiencing an acceleration sufficiently high to be dangerous (ie crashed into something). You can even experiment with this in your car. If you pull the belt with a slow but constant acceleration it will release but if you yank it it will lock in place, even if the speed you're yanking it to quite low.
Let me know when you can formalise the torque problem of yours in a similar manner to how I've formalised the pendulum and spring accelerometers. As you can see, I have nothing to fear from your assertions, they are refutable by anyone who didn't sleep through high school physics.
To compute this we resolve forces about an equilibrium configuration. We want the sum of gravitational force $$M\mathbf{g} = (0,-Mg)$$, the acceleration $$M\mathbf{a} = (-Ma,0)$$ and the string tension $$\mathbf{T} = T(\sin \theta,\cos \theta)$$, where $$\theta$$ is deflection from the vertical in a clockwise manner. If the system is in its stationary tilted state then the sum should be zero. Therefore
$$M\mathbf{g} + M\mathbf{a} + \mathbf{T} = \mathbf{0}$$ which in terms of components gives
* Horizontal : $$0 - Ma + T\sin\theta = 0$$ which gives $$T\sin \theta = Ma$$
* Vertical : $$-Mg + 0 + T\cos \theta = 0$$ which gives $$T\cos \theta = Mg$$
Taking the ratio removes T and M and gives $$\tan \theta = \frac{a}{g}$$ and so $$\theta = \tan^{-1}(\frac{a}{g})$$ or $$a = g\tan \theta$$. A quick sanity check is to put in a=0, which means no acceleration and the pendulum should hang vertically. If a=0 then $$\tan \theta = 0$$ and so $$\theta = 0$$, as required.
Therefore if we know the strength of the gravitational field (g=9.81 for Earth's surface) and we know the angle the pendulum tilts then we can compute the acceleration the vehicle is experiencing. No need to know what the velocity is, relative or otherwise, since the only thing which matters is the acceleration, hence why it is an absolute concept not a relative one.
This is covered in high school physics class (teaches kids how to work with vectors and trigonometry), since you don't grasp any algebra I'll give a numerical example. The mass has M=1kg, g=9.81m/s^2 and the car accelerates such that the angle of the pendulum moves to $$\theta = \frac{\pi}{4}$$. Since $$\tan \frac{\pi}{4} = 1$$ we have $$a = g\tan \theta = g = 9.81$$. This makes sense because if a=g then the pendulum should be in a position symmetric between the horizontal and vertical, which is when $$\theta = \frac{\pi}{4}$$.
There, another high school home work problem example to teach you basic mechanics. Like I said, seat belt mechanisms work by this principle, when the deflection of the pendulum goes past a certain point the belt locks in place as this means the passenger is experiencing an acceleration sufficiently high to be dangerous (ie crashed into something). You can even experiment with this in your car. If you pull the belt with a slow but constant acceleration it will release but if you yank it it will lock in place, even if the speed you're yanking it to quite low.
Let me know when you can formalise the torque problem of yours in a similar manner to how I've formalised the pendulum and spring accelerometers. As you can see, I have nothing to fear from your assertions, they are refutable by anyone who didn't sleep through high school physics.
You didn't learn from your previous mistake did you? I'll ask you again, if after 1 second of acceleration the absolute velocity is 100 m/s, what was the acceleration?
Insufficient information. If you started with that mass moving at 80 m/s, and ended up with it moving 100 m/s, then the acceleration was 20 m/s. You control how fast your own measurement mass is moving, so you get to choose the initial speed.
You seem to live in a world of no reality, but one in which you can make up your own reality. That's not how history works. History is a record of the past, and there is only one past.
Motor Daddy realizes he is losing again so he resorts to insults!
Motor Daddy
Valued Senior Member
You haven't managed to discuss the detailed explanation I gave of the 2 rocket example, why should we move on to other things?
When you can tell me how you arrived at the acceleration rate of 10 m/s^2 then we can continue that discussion. Explain to me the technique you used to measure the acceleration rate of 10 m/s^2, because I know you have no knowledge of the absolute frame and the only way you can know an acceleration rate of 10 m/s^2 is to have measured it in the absolute frame against the light sphere. If you persist in your blind traveling ways we too can play the direction game and you will lose, as I previously showed that relative velocity is like a blind squirrel looking for a nut. Once in a while he'll find one (because even a blind squirrel finds a nut once in a while) by pure luck, but other than luck, it's a needle in a hay stack.
I know you have to resort to the 'Gish Gallop' tactic of throwing out as many ignorant claims and swamping peoples ability or desire to respond to them all but that doesn't make them automatically valid.
It may look like an ignorant claim to you because claims are relative, and you don't understand it.
You have not actually shown any contradiction in any kinematics, you have simply asserted things without justification.
I've shown that relative velocity is not the velocity that is changing under acceleration.
I said I'd discuss the torque question when you can actually properly formalise it. That's part of the problem talking with people like you, you cannot actually properly define a scenario, due to your lack of mathematical capabilities, and thus you always leave arm waving 'wiggle room' to ignore people who explain your mistakes.
The scenario speaks for itself. Do you have a question about the scenario? The math is there, check it, it's flawless!!
Just as the car/rocket example I formalised using basic kinematics can you do likewise with your torque example.
That is formalized. HP=Torque*RPM/5252. You know hoe to calculate the circumference of a tire I assume? You know what RPM is? You know what a gear ratio is? You know what mass, distance, time, force, work, power, and acceleration , and velocity are I presume? Well get hot! Tell me why the car that has more HP at every MPH accelerates at a greater rate at every MPH! You have all the information in that example. We can go into the engine specifics if you like? Volumetric efficiency, bore, stroke, cam timing, duration, ICL, LSA...etc.. Just say the word
I said I'd discuss it with you if you could. You then saying "You're taking a long time!" is dishonest.
Why are you trying to dictate to me what is and what isn't a long time? I have a watch, and according to the watch, and my perception of the watch, you are taking a long time to answer the question I asked of you. There is nothing dishonest about that! You love to try to make claims of dishonesty, but you don;t actually know the true meaning of the word. That's not to say that you're not dishonest, that's not what I mean. You certainly are!
And saying "I'm here to help" is laughable, since I've demonstrated I can formalise such scenarios while you're showing you cannot do it nor can you understand when others do it for you.
You don't seem to understand the meaning of the word help. I will help if possible. You laughing gives me the impression that you seem to think that you know everything and I could never in a million years teach you anything. That is would be total BS and you know it, since I'm pretty sure I could run circles around in the physics of an internal combustion engine, just to give an example.
Until such time as you can formally construct the scenario pertaining to torque I do not see sufficient evidence that you understand it. Prove to me you understand it and can formalise that understanding and we'll talk. If you continue with this "You're taking a long time! Where's your response?" then I'll simply point you back to this post and report you for trolling. I'm asking you for a minimal level of understanding and honesty. I know you consider such things 'absolute nonsense' but that isn't my problem, it is yours.
That is formal. I gave you all the information you need. What information are you missing?
A car alone has no second object to be moving relative to but that doesn't matter. A single car which undergoes the acceleration described previously will still be able to measure that acceleration using accelerometers. It will then be able to say Initially I was in an inertial frame F. I underwent acceleration of magnitude a for time T. I am now in another inertial frame F'. The relative velocity difference between F' and F is therefore $$\Delta v = aT$$, F' is moving at speed v=aT compared to F. No need to know what, if any, absolute motion F and F' have. The car doesn't need a second object to compare itself with, it can do it all on its own since acceleration can be measured directly.
I can prove you wrong. What you are saying is that there is no absolute reality, and that is just flat out wrong. You are trying to say that it doesn't matter what the absolute velocity is, that all that matters is the acceleration of the two objects. I can draw two different scenarios of the relative velocity remaining the same, the acceleration of each rocket being the same, and with those numbers can make multiple realities. You are saying it doesn't matter which of those realities it is. Are you for real?? Math and physics is a tool used to describe reality, not reality describe physics and math!
If you ask the same question again, in a manner which is clearly meant to imply I haven't responded to the question when I have several times, then I'll start reporting you. I have no problem with people who don't know physics but want to understand it but someone who is wantonly ignorant and deliberately so I cannot abide.
Just because I don't agree with you doesn't make me ignorant. I asked you to walk through the scenario one step at a time so as to find the dependency between our methods and try to resolve the issue. You don't want to do that, maybe because you aren't capable of measuring reality and taking those measurements and using them in your theory to describe the reality that took place.
A ball of mass M is held in place at x=0 by a pair of massless springs (you can do it with just 1 but it doesn't matter) with spring constant k. (bad diagram but it looks like |\/\/\/\/\/\O/\/\/\/\/\/\|) The springs are not in tension (ie being compressed or stretched) when the ball is at x=0. This spring/mass system is inside a box inside a car. The car accelerates at rate $$a$$.
Based on what measurements?
The spring/ball system moves,
Does it move compared to the box or move compared to the light sphere, or both?
causing the left hand spring to be compressed and the right hand to stretch, |\/\/\/\/\O/\/\/\/\/\/\/\|. . The system moves to an new stationary equilibrium, with the ball at x=-L.
You're measuring the ball's position compared to the box. What is the box's position measured against?
Since the system is accelerating at rate a
How do you know the acceleration rate?
the ball must be and so it is experiencing a force F = Ma.
You better while you're ahead and stop talking about force. You first need to understand the torque/HP example before you know what force is.
The springs therefore must be applying this force. The spring which has been squashed by an amount L will push back with a force +kL. Likewise the stretched spring will pull back with a force +kL. Therefore the ball is experiencing a force F=+2kL.
If you want to measure force in that system that is fine, but those forces do not translate to the absolute frame, so all you "relative force" nonsense is just that, nonsense.
So we have Ma = 2kL, so a = 2kL/M = (2k/M)L. So if we know the strength of the springs, the mass of the ball and the distance it has moved we can compute the acceleration it experienced.
You only know the distance the ball moved compared to the box. You have no way of knowing what distance the ball moved in the absolute frame, which is according to the light sphere, which defines distance!
Nowhere have I had to say what the initial or final velocity of the car is. It could have started from stationary, with respect to the road, moving at 100kph or reversing a 54kph, anything. The acceleration measurement only requires me to measure the displacement of the spring mounted ball.
You still have not measured acceleration, you have simply created a new relative velocity. YOU DO NOT KNOW HOW FAR IN THE ABSOLUTE FRAME ANY OF THE OBJECTS TRAVELED!!!
I know I keep saying it but it really is true, this stuff is covered by children. I remember doing this when I was 14 or 15! Balls on springs or pendulums swinging is the stuff of basic physics lessons!
I need to go back to basics with your mentality because your theory makes unbelievable simply mistakes that any average person can comprehend. Create all the baffling BS you want and make it complex as you want, the mistake you make is simple and all the complexity is not going to change that!!!
As I just explained, the measurement of the acceleration really is very straight forward. The fact you didn't know this and also couldn't think up a way of doing it shows how narrow and flimsy your grasp of such things is. Must be somewhat depressing to have such a narrow view of things, struggling to grasp what children take for granted.
You have explained nothing except your method of deception. Crystal clear now! Tell me how you arrived at the 10 m/s^2 acceleration rate!!!
Motor Daddy
Valued Senior Member
You didn't reply to my much earlier post, so I'll ask again, and hope you can clear this up for me.
If the two vehicles are moving at the same speed, always, no change whatsoever, aren't they then the same as one vehicle ? Couldn't we equally say they would behave the same as though they ere connected, say by a rigid frame, therefore one vehicle ? Where is the relative velocity then ?
No you can not claim them as one vehicle, they are two separate objects, each with their own velocity and each with their own acceleration.
And isn't the movement of the pendulum simply due to it's inertia ? Such as how AN detailed in post 227 above, being how accelerometers work ?
If you are in a ship and you claim to be at rest then it is the pendulum that is moving away from the ship, not the ship accelerating. You only have relative velocity, you have no way of knowing absolute velocity so you have no way of knowing which is really moving and which isn't, or any combination of the two. You have no way of determining direction.