DawgTalkers.net Forums DawgTalk Palus Politicus Testing OldCold's Theory

OldCold has stated on several occasions that there are some very smart people on this board. I do not entirely agree with his concept in detail, so I offer a test. This is related to a recent thread, but serves as a standalone.

You got a spaceship going to a distant planet, which is 4 lightyears away. It has a drive which can accelerate to 20% of lightspeed. I say the trip will take 40 years, assuming a steady-state model and from the point of view of a stationary observer, is this correct?

My estimates are as follows: First, a fair number of people will refuse to do the math. Allowing for that, somewhere around 60-80% will say my math is wrong, but will be unable to explain why. Approximately 20-40% will accept that my math is correct, but will also be unable to explain why. Somewhere around 5% or less will verify the math is correct, and also be able to explain why this is so. About half of those will have to do some internet research in order to achieve this.

So, Oldcold, which category do YOU fall into?

Isn't it more like 20 years at 5 years per lightyear?

Assuming a steady speed of 20% of light speed, with no ramp up needed to reach that speed, it would make 1 light year take 5 years.

Thus 4 light years would take 20 years, right?

I am curious as to how you arrived at 40 years. Did I look at the equation wrong?

Just as an aside ..... light speed is 186,000 miles/second, so even 20% of that would leave us as mush in the cockpit. lol 669,600,000 miles per hour, divided by 5, means that 20% of light speed would be almost 134 million miles/hour.

Don't go to the bathroom, you're gonna miss a ton of scenery.

Once your on a object that is moving your mass becomes relative to the object if it has large enough mass and a ship that does that would be massive. It's the same reason you don't go flying off our planet that is moving at 67,000 miles per hour around the sun which is moving through our galaxy at over 40 thousand miles per hour.

It's also the same reason when you walk forward on a moving train that your speed in relative to the train and not the ground that it is racing over.

The trick will be in how long it takes you to get up to speed. If it's gradual enough you will never feel it. That might be where he gets 40 years from but he didn't give enough info to do do the math on that. However you can postulate that it took us many months to get voyager to travel at 75,000 miles per hour but that was on a slower moving mode of locomotion.

The trick is in knowing how long this ship takes to get up to it's top speed. You can't just assume it's instant when moving that kind of mass at that kind of velocity. Since no info was given regarding that I just assume there is no ramp up time and that it's simple math and not rocket science =)

Read it AGAIN. Read it CAREFULLY. Make CERTAIN that you understand what ALL the words mean. Process ALL AVAILABLE information.

I got ~24 years. Then again, our crew is using the warp speed technology. I assume Nelson's journey didn't take as long as they are not using the warp speed as the spaceship has no way to decelerate. I assume this is what he meant by steady state, although I'm not sure what that terminology means. Also the single observation leads to problems too, as we don't know the planet we're observing this from. Maybe Nelson wants us to observe this from Mars or another planet.

Do emotional IQ next.

HUH!

I am impressed. Most of you are actually dancing around the edges of the correct explanation. Now try and apply some analytic thought.

Some slightly helpful additional information -

A "Steady State" is just what sounds like and not complex at all. It means, a STATE which is STEADY.

The stationary observer is meant to greatly simplify the calculation. If you do not know the problem this is intended to solve, it is there simply to avoid a potential complication. You can safely ignore it, unless you know what it means, in which case the exact same simplification happens.

There is no math involved above a grade-school level.

A concept that may help is to ask "What is it that you are trying to achieve? What is the goal of this operation?" Next, "What are the necessary steps to achieve that goal?"

The people on the ship arrive in 20 years (per math above)

The Observer 'witnesses' the event a further 20 years later.

Ok so 'steady state model' means constant speed and does not refer to the steady state model. Glad you cleared that up.

And you said stationary observer to imply Earth. Got it. Now that's settled, it'll take the Earth twice as long to see them on the new planet.

That makes sense, but what if the stationary observer is on the destination planet, or halfway between earth and the destination, or anywhere else in between?

Or orbiting a black hole?

Based on the wording I take this to mean that the rate of acceleration is constant (based on time, not distance) from start to finish so therefore the average speed is 10% of light speed. 40 years.

How much time passed for the people on the ship?

btw: The calculation works just as well if there's constant acceleration until the midpoint of the trip and then constant deceleration after. Just in case you don't want to jump off the ship going 20% of the speed of light.

Redid the math after a nice shower and coffee and I got 28 years.

24 years for the journey going at 20% speed of light. And 4 years for the light from 4 light years to "reach" us.

MGH, wrong, twice, CHS, wrong, twice, and then twice more, Ytown, first post it does not matter, second if at a safe distance it also does not matter, if in a standard orbit, a VERY complex calculation which would need a LOT more data and is outside my knowledge base.

Haus is correct, which I knew he would be, although your first case does not apply because you have to stop when you get there. The second case is the proper procedure because in addition to stopping when you get there, you have the constant acceleration as a slight level of gravity for both physical and psychological benefit to the crew, during the entire voyage, without which, they would all get there dead.

The stationary observer eliminates all time dilation issues, location does not matter, seeing it is not relevant, they simply look at a clock.

On the ship itself, there would be a noticeable time dilation effect, I am honestly not certain if the relationship is constantly linear and if it is as simple as 10% lightspeed means 10% less time, but pretty sure it is fairly close to that.

X = the steady state

The steady state is what your looking for not how long it took ... u all ready told us that and then used wording to “deceive” us into thinking we had to look for your math error instead of solve for the actual rate we’re traveling at ...

Thanks ...

Love thinking ... u make me do that on a lot of fronts .. thanks for that also ...

Question: How do you guys plan to decelerate once in space?

Haus may not have needed to read it again, but I would bet that he did. I know he read it carefully. He made certain that he understood what all the words mean. He processed all available information.

He most likely did these things without any real conscious thought or effort. He did them because he has always done them, and likely has a very hard time understanding why anyone would do it any other way. He did them because long experience has taught him that doing it this way consistently arrives at the correct answer.

I would also guess that no additional research was required. I would further speculate that he does not need a "well done" from me because he knows he did it well, and also that I did not express "well done" to him because I EXPECT him to do it well, that is baseline performance. Now, if you want to run those black-hole-orbit time-dilation equations, THEN I would be suitably impressed.

You know you're talking about space, right? Having an observer makes it a time dilation issue.

To give you an elementary school example: The sun is 8 light minutes away from us. If we were to track the sun for a minute, we would become a single observer, while only observing the sun from 7-8 minutes ago.

Also, you never mentioned any abilities to decelerate in the OP. Decelerating in space is not as simple as pulling the emergency breaks. Personally, I stopped considering deceleration because it's lack of mention in the OP, written by an anglophile. I just thought it was a half baked idea with bad word usage, that in a real life scenario would break into pieces after hitting space dust at .1c and figured you just wanted to explode on impact at .1c

If i understand this correctly ...

20 times the speed of light is the fastest we can go ... because your rate of acceleration is constant .... that means 10 times the speed of light is your “average rate” of speed ... hence taking it from 20 years to 40 years ...

Is that correct? ...

Hopefully i understand how u got to the answer now that you’ve given the answer .. *L* ...

Please correct me if i’m wrong ... just curious as to if i understand this or if i’m Wrong again ...

Thanks man .. this was fun ..

When I was at a math class once - I was shown (mathematically) that for people who space travel at the high speed - they age less than someone on earth.

Is that the factor we need to figure?

Or is this word play? The space ship "Can" accelerate to 20% the speed of light ... but does it? And as others have mentioned how long does it take to accelerate and to decelerate? ... you stated the destination is 4 light years away. Is there an issue of traveling in a straight line to the destination?

It seems this isn't a math issue so much as a trick question.Substitute what you said to this:

"You got a spaceship going to a distant planet, which is 4 lightyears away. It has a drive which can accelerate to 100% of lightspeed. I say the trip will take 8 years, assuming a steady-state model and from the point of view of a stationary observer, is this correct?"

Is the statement still true? And if it is the 'trick' becomes more obvious.

Diam - Not 20 TIMES, but 20 PerCent, or one fifth of lightspeed. But yes, the average speed is 10%.

CHS - First statement is flat out wrong about time dilation, if you understand both the concept and what stationary means.

Your second paragraph is surprisingly accurate, but indicates that far from understanding how time dilation works, you do not understand what it means.

Your third paragraph indicates that you think after a long highway trip at 60mph, you pull into your driveway at 60 mph? Slowing down at the end of the trip is obviously necessary. The rest of your statements are reasonable concerning space dust, but problems with word usage are your own and certainly not mine, and Anglophile comes completely out of left field, but I am guessing you do not understand what that word means, either.

Exactly. You did a great job at expressing my thoughts. The funny thing is that it'll either take 44 years or 36 years depending on where the stationary observer is. I'm not even going to get into how something that decelerates half of it's speed is somehow a "steady-state model".

This is why rocket science is rocket science and elementary math is not.

my answer to everything is E=mc^2

Would you call this going at a steady state?

Also, the point of the matter is where the observer is matters, and claiming it doesn't is ignorant. Although, I was ignorant for following you on your 'time dilation' tangent, that no one else was talking about.

MGH - Second post, you are almost there.

Yes, time dilation will happen on the ship, but no, it is not a factor here, due to the stationary observer.

It is not wordplay, it does accelerate to 20%, there is no issue traveling in a straight line, Your substitute example is essentially correct.

The steady state is constant acceleration, for one-half the trip, then constant acceleration in the opposite direction, for one-half the trip. The reason for this is that first, you have to slow down at the end, and second, there is a constant level of gravity, which is acceleration, for humans and the garden you would need on the ship. Constantly varying gravity, or acceleration, is bad for both humans and the garden, steady state works best, and flipping the ship at the midway point accomplishes both that and slows you down.

This exact scenario is lifted almost verbatim from several NASA proposals for future long-distance space flight options, and is currently regarded as the optimum method for the process, barring the intervention of "magic" such as warp drive or artificially-induced gravity alternatives.

It is not my invention or creation. It is accurate, factual, and based on the work of the very best scientists we have. It is not based on wishful thinking, personal feelings, or ideology.

So this has nothing to do with space or speed.

If you travel at up to 100mph at a "Steady State" you essentially travel 50 miles by your definition.

So unless you know and recognize what 'Steady State' means then the question is unanswerable. . . .

Or have I misunderstood?

Ty sir ...

Not exactly sure what it is i understand ... but i feel like i undertand it better now .. *L* ...

How do you arrive at an average speed of 10% of lightspeed based on what you posted?

I could see, if the observer was on earth, it taking 24 years, 20 years out, 4 light years at 20% (1/5) of lightspeed, plus 4 years for the light from the distant planet to return to earth after the ship lands there.

Unless you expect your reader to magically infer anything except your 20% of lightspeed, then I see no way your reader should assume that, or that your answer is right. Frankly, because of the return time for the light to return to earth (the observer) from the planet 4 lightyears away, even at 1/10 of lightspeed, your answer would be wrong.

Time dilation is not really a factor at 20% of lightspeed. There is a marginal amount of time distortion, but according to what I can find, it is negligible until you actually reach lightspeed. At 20% of lightspeed, the time distortion is 100% = 97.979%. At 10% of light speed, using a 20 year time frame from the stationary observer, the difference is even smaller.

Time Dilation Calculator
http://www.emc2-explained.info/Dilation-Calc/#.W46bcnfu6aw

Using the same site, the time difference over 4 lightyears, at 20% of lightspeed, would be roughly 0.4 of a year. (4.8 months)

So .... 20 years to get there, minus the 4.8 months for time dilation, plus 4 years for the light to return to the stationary observer on earth, would be 23.6 years for the stationary observer, on earth, to see the image of the ship landing on the distant planet.

If I am reading your explanation correct, you are warping time X2 for the observer, but I am not sure how you get there from what you posted.

Anyway, it's an interesting puzzle, but I disagree with your answer.

I would also add that steady state is largely rejected these days.

I am just glad that I am going!

Haus's answer is the correct answer.

You also have to infer that the ship can also decelerate by accelerating backwards. I believe Nelson is using the hypothetical centari mission that's planned for 2069 (But no wishful thinking or ideology went into it or anything). It plans to use gold sails to get to alpha centari. Right now we're still using fuel based thrusters.

Depending on how one imagines spaceships, one could get a multitude of answers.

The concept of the stationary observer has been used for several decades for the specific purpose of removing time dilation effects from such calculations. For instance, without a comparison to the stationary observer, time on the ship is unchanged and exactly as it should be, which is actually totally accurate. The two change at a different rate because of the difference in their velocities. The observer is stationary relative to the ship, ALL velocities are relative.

I specifically stated that time dilation effects were not a factor, or at least this should have been clear.

You arrive at the 10% average by understanding that the only steady state available is the rate of acceleration, which when constant and starting and ending at close to zero while achieving 20% somewhere, leads directly to 20% exactly in the middle and flipping the ship around and accelerating in the opposite direction for the second half of the trip., with an overall average trip speed of 10%. The steady state of acceleration IS, also, a steady state of gravity on the ship as a secondary, but necessary, benefit.

You are certainly free to disagree but my answer is correct.

MGH, yes, you are confused,

Ytown, the concept of a steady state is and has been applied to many, many things, as a model of the universe it is somewhat outmoded but that particular concept does not apply and is not necessary for this example,

And, both last and least, CHS, ummmm... Decelerating IS accelerating backwards, no inference is needed, I would seriously suggest you refrain from posting in such topics until you have greatly improved your understanding of the various conditions and parameters involved, and it is Alpha CentaUri, using a solar sail, gold-plated foil on a mylar substrate, a concept which has been tested and found workable, as well as the nuclear powered ION drive, which is my own personal preference, and for which the necessary deceleration is greatly improved. The problem with the solar sail is that received power diminishes with distance and may need to be augmented with space-based lasers, and slowing down is a complex process, plus the return trip would have to work without the laser augmentation.

Nelson this is an excellent question to test my theory that you are nowhere near being the most intelligent person on this board. I'll be happy to respond, but this will be kind of long so do your best to follow.

First, Your question at best is flawed. When dealing with spacetime travel calculations the variables are innumerable. The variables created by the constant expansion of space in all directions (even into itself) and the speed at which even a stationary observer (which is all but impossible) is traveling relative to all of said variables makes proveable calculations nearly impossible. And this does nothing to explain the things you omitted in your question like acceleration and deceleration or if you intend to stop at or flyby the destination... hopefully you get my point.

Second, asking a sports fan board to actually calculate reliable space travel or verify your calculations based on the fragment of information showing no calculations whatsoever to be proven or disproven is laughable at best.

I for one am not a physicist nor am I a rocket scientist. So my attempt would obviously have to be researched and all conclusions be considered that of a layperson trying to solve an enormous mathematical calculation based on remarkably incomplete data. Not immediately seeing this before posting such a question proves to me that you calculations will be at best flawed as well.

I understand that your actual motivation for this post was to lend credibility to your own intelligence, while attacking mine and the others here. The irony to me is that you have no idea of what you are talking about other than things you have read on what appears to be scifi fan sites or spacetime conspiracy theory blogs.

From your question I have to assume we are talking about a Starshot spacecraft theory for travel to Alpha Centauri. Here is some information on this and the most brilliant minds of our time struggle to accurately decode all the math. Watch them all or don't bother responding to me.

Good description.



More.



Now Stephen Hawking on starshot.



Michio Kaku



So the consensus is that it should take just over 20 years to travel 4 light years without deceleration.

This video explains the stationary observer part of your question as envisioned from a 'steady chair' theory:



However, you specifically cited "assuming a steady-state model" which is just a theory that opposes the big bang model by hypothesising that the universe is eternal and infinite with no beginning and no end.

Sadly it has been mostly disproven but your usage suggest that you may be thinking of it in terms of a fixed unchanging distance between earth and the proposed destination. The funny part is, that even though that was the original thought behind the theory, it was later believed that even in steady state that the universe is still expanding but doing so by creating matter at a constant rate maintaining balance as it grows.

The only influence that I can see this having on the 'trip' is that the distance may grow greater during the voyage, thereby adding to the length of time it takes to travel the distance.

So taking into consideration that your very vague question lacks the mention of numerous variables and for which any definite answer would be considered unprovable with our current understanding of the subject. I can only answer this way:

- Your math, without seeing any actual calculations but taking into consideration universal expansion, acceleration and deceleration, and the basic math of the distance formula Distance = Speed x Time... Then your answer becomes plausible but in no way provable.

- The consensus of some very brilliant minds say that without all of the variables I've mentioned but rather just the strictest use of the distance formula would put you there in roughly 20 years. Again, opinions vary slightly and the answer is in no way 100% verifiable with our current knowledge.

Putting it bluntly, even the experts don't know for sure, so how could you expect a board full of laypeople to know if you are right or wrong?

That said, I think it's awesome that you are so insecure about your intellect that you thought posting a question like this and positing an answer that nobody could reasonably prove nor disprove would somehow prove that you are smarter than most here. Congratulations, you are truly pathetic. smdh

Please don't take me wrong, I think you are above average intelligence, just nowhere near as intelligent as you think you are.


EDIT - There were three responses when I started crafting this one for you Nelson! Who knew this woudl be such a viral thread.

I missed CHS post responding to MGH, #1499184, and this is really one of his best.

How exactly do you state that someone has expressed your thoughts exactly, and then go on to say that what they said is completely wrong?

Once again, I have words that I typically use to describe such actions, words and concepts that some here seem to frown upon, but IMO every single person here knows what those words are, and also that they are fairly and accurately applied.

I am truly amazed, even awed, at the processes involved which lead to such statements and beliefs. EXACTLY what I expected from that particular source.

I would very much like to know how the rest of you, and especially those associated with management, whether identified as such or not, characterize such a thought process?

Yes, I followed you down the time dilation rhetoric and that was my bad. However, the concept still applies. That the location of the single observer does matter, because it'll take time for the information to reach the observer.

Here is the problem with your "steady state model" and why it's impossible to decipher in your current word game. Just to reclarify this was the game.

"You got a spaceship going to a distant planet, which is 4 lightyears away. It has a drive which can accelerate to 20% of lightspeed. I say the trip will take 40 years, assuming a steady-state model and from the point of view of a stationary observer, is this correct?"

From the start you want us to consider this proposition from the stationary observer. You state that we, now the stationary observer, must assume a steady-state model. You are now asking your audience to see if your calculations are correct. Your audience tells you, you're not correct. Because they assume that a "steady-state model" means a constant speed or acceleration that the spacecraft goes.

However, you now inform us that we are wrong and you are correct.

You arrive at the 10% average by understanding that the only steady state available is the rate of acceleration, which when constant and starting and ending at close to zero while achieving 20% somewhere, leads directly to 20% exactly in the middle and flipping the ship around and accelerating in the opposite direction for the second half of the trip.,[b] with an overall average trip speed of 10%. [b]The steady state of acceleration IS, also, a steady state of gravity on the ship as a secondary, but necessary, benefit.

Again, your language game is built on the idea that an 3rd party observer has deduced that your 'steady-state model' is built on acceleration and not velocity or speed. This is because you expect the 3rd party observer to consider the needs of the subjects inside.

Secondly, with a 3rd party observer, assuming they could see you from 4 light years away, could see your entire trip. At which point they could easily deduct that you're not going at a steady velocity. However, it'd be near impossible to be able to tell you were exerting constant acceleration to yourself.

So good job. You got us on the word play part. If only your math checked out.

At this point I think you're either trying to be an ass for the sake of being an ass - or maybe your explanations aren't deliberately obtuse merely bad?>

I am sure you are correct with your math - but simply if the net effect of "steady state" is that you average half your ultimate top speed (both side of the peak top speed) then my analogy is correct and my statement that the "Math" in this question all hinges on knowing what "steady state" means is correct whether you like to admit it or not. It's not a math problem at all it's semantics question.

This is the best guess!