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thayln

Is my math correct?

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Ok, M17 in Sagitarius is 5,500 light years away. So, if you were going warp 5 it would take you 1,100 years to get there? Or is it a bigger exponent like 5 to the tenth power or something?

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accourding to the Trek encyclopedia, at warp 5 it would take 1 month to go 20 light years and 47 years to go 10,000 light years.

at warp 9.2 is goes down to 4 days and 6 years....if that helps.

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as i understand it, your not actually traveling at the speed of light or faster. but, traveling through warped space you are traveling distances faster than the speed of light. i think. if master q finds this he'll know.

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Warp speed is not a simple multiplicatation scale like warp 1 = light speed * 1 and Warp 5 = light speed * 5. It is (the TNG scale) best approximated by:

V = velocity in multiples of light speed

w = Warp Factor

 

V = w ^ (10/3)

 

This only actually works up until about warp 9 where the scale actually becomes more like:

 

V = w ^ [{(10/3)+a*(-ln(10-w))^n}+f1*((w-9)^5)+f2*((w-9)^11)]

 

ln is the natural log (e ^ x = y : to : ln y = x)

 

Where a is the subspace field density, n is the electromagnetic flux, and f1 and f2 are the Cochrane refraction and reflection indexes respectively. Under ideal conditions values of a = 0.00264320, n = 2.87926700, f1 = 0.06274120 and f2 = 0.32574600 can be expected within a "normal" area of deep interstellar space.

 

 

Based on the speed of Warp 5 (214 * speed of light), it would take about 25.7 years to travel there at warp 5.

 

 

A full table of Warp Speeds and Travel Times can be found here: Daystrom Institute Warp Scales

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If you have the correct scale for warp factors, then all you have to do is create a simple linear velocity equation.

 

It comes down to physics! Specifically kinematics! (A sub-branch of mechanics)

 

Well first off what is speed? Speed is the change in distance over the change in time. (Velocity is the same thing, but it is concerned with the direction also).

 

When we talk about speed or velocity we always talk about how much distance it covered in a specific amount of time. So we can say that . . . .

delta Velocity = (delta Distance) / (delta Time)

 

If you know the velocity (and if it is constant), then all you have to do is create an equation to go with it. So if you know that (haveing constant velocity), then you know that the velocity stays the same (during any time period).

 

We can make a general equation for this . . . .

 

Distance = Velocity * Time + Distance Naught

or d = Vt + (d sub o)

 

In fact we can think of that equation as “y=mx+b”!

Where y is distance, m is velocity, and b is the initial distance

And if you like you could even graph that to see the function

 

Because we know the initial distance is zero (& the time) and that we know what v equals.

 

 

 

I don’t know if you have the correct scale, but if you do the method that I’m saying then you’ll get the correct answer.

 

Also . . .

 

Besides having a position vs. time graph why not a velocity vs. time graph?

 

In the standard function when we graph it (position vs. time) we have a linear line that is diagonal (position vs. time graph)

 

Velocity vs. Time Graph

It will come out to be a flat horizontal line

This is because the slope of the position vs. time graph is velocity and because velocity is constant the graph will show a constant velocity. This is why it would come out to be a flat horizontal line. If you figure out the area under the line, then you can figure out the displacement.

 

 

 

Master Q

StarTrek_Master_Q@yahoo.com

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as i understand it, your not actually traveling at the speed of light or faster. but, traveling through warped space you are traveling distances faster than the speed of light. i think. if master q finds this he'll know.

Please remember cptwright it is true that in fact that you are not going faster then the speed of light because you are using subspace, **BUT** you are traveling distances that are light years away. The variable of subspace (and things of that nature) you don’t need to use to solve this probelm.

 

 

Master Q

StarTrek_Master_Q@yahoo.com

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Thanks, guys.  I really appreciate it. 

 

Course, now I've got to go take a migrane pill.  ;)  ;)  :innocent:

yeah, that usually happens with these kinds of questions. MIGRANEVILLE. ;) i only brouse through what the brains say, as to avoid the brainpain. ;) to the brains, of this board, i bow to your greatness. :laugh:;););)

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If you have the correct scale for warp factors, then all you have to do is create a simple linear velocity equation.

 

It comes down to physics! Specifically kinematics! (A sub-branch of mechanics)

 

Well first off what is speed? Speed is the change in distance over the change in time. (Velocity is the same thing, but it is concerned with the direction also).

 

. . . . . . .

I'm sorry, but that is getting a little off topic. offtopic.gif

 

Also, I think your making everyone's head hurt by making speed sound soo complex.

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If you have the correct scale for warp factors, then all you have to do is create a simple linear velocity equation.

 

It comes down to physics! Specifically kinematics! (A sub-branch of mechanics)

 

Well first off what is speed? Speed is the change in distance over the change in time. (Velocity is the same thing, but it is concerned with the direction also).

 

. . . . . . .

I'm sorry, but that is getting a little off topic. offtopic.gif

 

Also, I think your making everyone's head hurt by making speed sound soo complex.

Not really because to figure out distances or displacement you have to go according to the laws of physics.

 

Once you have the scales then you can solve them mathematically by solving the linear equation or find the area under the curve (or in this case the linear line).

 

I went "off topic" (as you put it) because I think that you can’t just give an answer and let it be done. To me it is all on concept. Once you know the concept, then you can work off that. Concept is everything! Of course I did not go detail on what the concepts of kinematics are, but I was leading to it.

 

If a linear line y=mx+b is too complex, then I don’t what to say (I’m of course referring to constant velocity Xeroc).

 

Speed is just how much distance you covered in such amount of time.

That’s all! How is that complex!? (LOL) That is the most basic dentition there can be. If I traveled 1 meter in 1 second, then my speed is 1 meter per second. You can’t get any simpler then that. Unless you can give better one.

 

That’s not off topic at all.

 

 

Someone needs to know what speed is before someone can really figure out a problem like that.

-->Like I have stated it is all about concept and in this case the concept is “what is speed”. If you don't really know that, then someone could not solve a probelm like we have now.

 

Xeroc, do you not need to know what 'speed' is before you can figure out a problem like that?

 

Of course you do!

You have to know what it is before you can figure out some kind of displacement (distance) of an object. And thats why I talked about it.

 

 

I do know that I should have gone and simplified what I said. I admit that, but in no way is that off topic.

 

 

No offense to you intended to you in anyway, but I think you need to get your facts straight. I hope that we can both agree to that there is no offense taken by either one of us and that you will state that in your reply. One of the problems that I see in a traditional physics courses is that they quote you a book of equations and nothing more. By a traditional physics course doing that you learn nothing. You must understand the building up to that equation. A teacher can’t spoon feed a student, but they can help the student develop and they can teach the concept before the equation really shows on paper. Thats why I went into what "speed" - you need to understand the concept before anything else.

 

 

Master Q

StarTrek_Master_Q@yahoo.com

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If a linear line y=mx+b is too complex, then I don’t know what to say (I’m of course referring to constant velocity Xeroc).

That is not complex.

 

The problem is, you make that into this:

If you have the correct scale for warp factors, then all you have to do is create a simple linear velocity equation.

 

It comes down to physics! Specifically kinematics! (A sub-branch of mechanics)

 

Well first off what is speed? Speed is the change in distance over the change in time. (Velocity is the same thing, but it is concerned with the direction also).

 

When we talk about speed or velocity we always talk about how much distance it covered in a specific amount of time. So we can say that . . . .

delta Velocity = (delta Distance) / (delta Time)

 

If you know the velocity (and if it is constant), then all you have to do is create an equation to go with it. So if you know that (haveing constant velocity), then you know that the velocity stays the same (during any time period).

 

We can make a general equation for this . . . .

 

Distance = Velocity * Time + Distance Naught

or d = Vt + (d sub o)

 

In fact we can think of that equation as “y=mx+b”!

Where y is distance, m is velocity, and b is the initial distance

And if you like you could even graph that to see the function

 

Because we know the initial distance is zero (& the time) and that we know what v equals.

 

 

 

I don’t know if you have the correct scale, but if you do the method that I’m saying then you’ll get the correct answer.

 

Also . . .

 

Besides having a position vs. time graph why not a velocity vs. time graph?

 

In the standard function when we graph it (position vs. time) we have a linear line that is diagonal (position vs. time graph)

 

Velocity vs. Time Graph

It will come out to be a flat horizontal line

This is because the slope of the position vs. time graph is velocity and because velocity is constant the graph will show a constant velocity. This is why it would come out to be a flat horizontal line. If you figure out the area under the line, then you can figure out the displacement.

 

 

 

Master Q

StarTrek_Master_Q@yahoo.com

Thst IS simple, but you make it sound COMPLEX.

 

 

Also, I said you were going off topic because this topic was not about the nature of speed, but about warp speeds. When I reply I try to stick with answering the person's question and not going off on tangents - related or not. I also think too little time is given in schools for the true derivation and substance of fomulas, etc. And a topic all about the true nature of them would be great! It just simply isn't what were talking about.

 

And, of course, no offence intended or taken.

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Here is a simpler version of what I said:

 

I admit that I probably should have simplified what I said in general, but to me (like I have stated before) when someone asks a question or learns something new you must focus on the concept of the problem in question. If it’s “how much distance will I travel?”, then one must look at the concepts of “speed” because traveling a certain amount of distance requires you to move and so requires the understanding of what “speed” is. It is not just “speed”, but applying what “speed” (or velocity) is gives you the focus or movement to figuring out displacement.

 

Speed is just how much distance in is covered in a certain amount of time.

So with that we can reason if you divide distance by time, then you can get speed.

Speed = distance / time

 

Now from that we can build up the displacement. Even from the simple understand of what speed is we can compute the distance by our simple equation “speed = distance / time”.

 

Let’s pretend we are doing an experiment!

In this experiment we are watching an object that has constant speed moving away from its origin.

 

If an object has constant speed this means that the speed of the object is always the same. So if I watch it one second from now or two seconds or three . . . . the speed of that object will be the same.

Does everyone see that and understand that?

 

Now from that we could graph a simple position vs. time graph.

 

This is what we have on our x and y axis:

X: time (like 1 second, 2 seconds, 3 seconds, . . )

Y: position (like 1 meter, 2 meters, 3 meters, . . . )

 

Now let’s say that we figured out that our object is moving 1 meter per second. If we graphed that then it would look like a straight linear line!

Why?

Because at one second it is one meter away

At two seconds it is two meters away

And so on

 

From this we can create a general equation! Now think about the equation “y=mx+b”. We know that we are traveling one additional meter each second and that we started at our origin.

So we can say a general equation would be: position = speed * time + original distance

 

Now let’s make an equation for our experiment!

Position = 1 (m/s) * time (sec) or Position = time (sec)

 

From this we can look at many things!

 

 

Does everyone understand?

(If you don’t, then please say so)

 

 

Master Q

StarTrek_Master_Q@yahoo.com

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