please explain Interplanetary Travel rules to a dummy!

[venture]

all I own traveller wise is the traveller hardback book.Its been some years since ive even looked at it (1984 to be specific).I was rereading the rules to maybe start up a campaign and forgot how much the rules for space movement confused me...could someone please explain them step by step in laymen terms and for someone not great in math,MKS?what is it explain,parsec how big is it ...i saw it mentioned once in the book....help me please as this is the only psrt of the rules(so far)that confounds me.is there better(easier more simple)space travel(and battle)rules in later editions(mega traveller maybe).
thanks to you all

 

[venture]

all I own traveller wise is the traveller hardback book.Its been some years since ive even looked at it (1984 to be specific).I was rereading the rules to maybe start up a campaign and forgot how much the rules for space movement confused me...could someone please explain them step by step in laymen terms and for someone not great in math,MKS?what is it explain,parsec how big is it ...i saw it mentioned once in the book....help me please as this is the only psrt of the rules(so far)that confounds me.is there better(easier more simple)space travel(and battle)rules in later editions(mega traveller maybe).
thanks to you all

 

[ravells]

I'm no expert but this is what I understand:

Starships in Traveller have two forms of propulsion: A manoeuvre drive and a Jump drive.

The manoevre drive is used to travel 'shorter' distances, typically between planets within a system. It's basically a sort of jet rocket.

the Jump drive is used to travel between systems. We do not know how a jump drive works we just know that it does work. Subsector maps in Traveller are laid out in a hexagonal grid. Each hexagon is 1 parsec = about 3.25 light years. Jump drives (depending on their power) are able to jump between 1 and 6 parsecs (i.e. 1 - 6 hexes) in a single 'move'. Regardless of how many parsecs you travel in single jump, the travelling time is always one week. So if you had a Jump 1 drive and needed to travel 3 hexes it would take you 3X1 week = 3 weeks. If you had a Jump 2 drive it would take you 2 weeks and if you had a Jump 3 drive or better it would take you 1 week.

That's it in a nutshell, I guess.

 

[ravells]

I'm no expert but this is what I understand:

Starships in Traveller have two forms of propulsion: A manoeuvre drive and a Jump drive.

The manoevre drive is used to travel 'shorter' distances, typically between planets within a system. It's basically a sort of jet rocket.

the Jump drive is used to travel between systems. We do not know how a jump drive works we just know that it does work. Subsector maps in Traveller are laid out in a hexagonal grid. Each hexagon is 1 parsec = about 3.25 light years. Jump drives (depending on their power) are able to jump between 1 and 6 parsecs (i.e. 1 - 6 hexes) in a single 'move'. Regardless of how many parsecs you travel in single jump, the travelling time is always one week. So if you had a Jump 1 drive and needed to travel 3 hexes it would take you 3X1 week = 3 weeks. If you had a Jump 2 drive it would take you 2 weeks and if you had a Jump 3 drive or better it would take you 1 week.

That's it in a nutshell, I guess.

 

[RainOfSteel]

In regard to the parsec.

Go to Google and type in "1 parsec in light years".

The answer it returns is: 3.26163626.

 

[RainOfSteel]

In regard to the parsec.

Go to Google and type in "1 parsec in light years".

The answer it returns is: 3.26163626.

 

[The Oz]

For travel inside a star system (using the maneuver drive), look at page 54 of the hardcover book.

Page 54 has the "Typical Travel Times" table. On this table you can find the time it would take to travel anywhere from 1,000 kilometers to 1,000,000,000 (one billion) kilometers, using a ship with from 1 to 6 G's of acceleration. Either look up the distance (in kilometers) along the left edge of the table and read across until you find the time it takes to travel that distance at the acceleration the ship has. Or you can start by looking at the example distances in the right column, find the one that fits your situation, and then read across to find the travel time for the acceleration available.

 

[The Oz]

For travel inside a star system (using the maneuver drive), look at page 54 of the hardcover book.

Page 54 has the "Typical Travel Times" table. On this table you can find the time it would take to travel anywhere from 1,000 kilometers to 1,000,000,000 (one billion) kilometers, using a ship with from 1 to 6 G's of acceleration. Either look up the distance (in kilometers) along the left edge of the table and read across until you find the time it takes to travel that distance at the acceleration the ship has. Or you can start by looking at the example distances in the right column, find the one that fits your situation, and then read across to find the travel time for the acceleration available.

 

[Plankowner]

The distances involved assume a constant acceleration to the halfway point and then decelerate to the end point. The math isn't too bad, but the unit conversions can get tricky.

I did the math a few decades ago and figured out a pretty easy conversion for long distances (interplanetary distances).

1 AU = 150,000,000 Kilometers (150Mkm) = 93,000,000 miles.

T = 2*squareroot(D/a)

Where T is time in DAYS
D is distance in AU
a is acceleration in G's

It isn't exact, but it is pretty close.

If you don't want to turn around and decellerate, but want to keep on accelerating the entire distance (ramming speed), then the formula becomes:

T = squareroot(2D/a)

So, it takes 141% longer when you want to be stopped at the end of the journey (typical so you can go into orbit, not fly by).

Typically, if you are travelling from planetary orbit to the 100D limit to jump, you are looking at a trip of 6-10 hours or so at 1G acceleration (with turnaround).

 

[Plankowner]

The distances involved assume a constant acceleration to the halfway point and then decelerate to the end point. The math isn't too bad, but the unit conversions can get tricky.

I did the math a few decades ago and figured out a pretty easy conversion for long distances (interplanetary distances).

1 AU = 150,000,000 Kilometers (150Mkm) = 93,000,000 miles.

T = 2*squareroot(D/a)

Where T is time in DAYS
D is distance in AU
a is acceleration in G's

It isn't exact, but it is pretty close.

If you don't want to turn around and decellerate, but want to keep on accelerating the entire distance (ramming speed), then the formula becomes:

T = squareroot(2D/a)

So, it takes 141% longer when you want to be stopped at the end of the journey (typical so you can go into orbit, not fly by).

Typically, if you are travelling from planetary orbit to the 100D limit to jump, you are looking at a trip of 6-10 hours or so at 1G acceleration (with turnaround).

 

[Icosahedron]

I'm not quite sure what you are looking for, Venture, but the common measures in space are:

The AU (astronomical unit) This is the distance from Earth to the Sun, 93 million miles or about 8 light-minutes.

The Light-Year, or the distance light travels in one year, at a speed of 186,000 miles per second.

The Parsec, a distance of about three and a quarter light years.

In space, things don't slow down, there is no friction from roads or air, so anything that moves will continue to move in a straight line until something stops it. Therefore spacecraft have to fire their drives to accelerate and fire them again to decelerate. The quickest method of travel is therefore to accelerate to the halfway point, then decelerate the other half to bring the ship to rest at its destination.

Also, if a ship is moving at 1 inch per turn, toward the top of the table (lets call it north) it will continue to move in a straight line at 1 inch per turn without any input from the drives.
This is called a 1 inch vector. Firing the drives with a 1 inch reverse thrust will bring it to rest. Or, a further 1 inch forward thrust will increase the movement vector to 2 inches per turn.

If the moving ship fires its drives with a 1 inch vector to the left (west) instead, this will be added to the existing movement vector, and instead of moving 1 inch forward, it will move about an inch and a half at an angle of 45 degrees to the left. (ie northwest) The exact movement can be found by placing two one-inch arrows (vectors) at right angles, and the actual movement of the ship completes the triangle.

From now on, the ship's movement vector will be 1.41 (ish) northwest, until the ship fires its drives again. You can fire the drives in any direction and at any thrust short of your maximum to move your ship where you want it. Just lay the tail of the thrust vector to the head of the existing movement vector and your new vector is found by completing the triangle.

Hope that helps. More if needed.

 

[Icosahedron]

I'm not quite sure what you are looking for, Venture, but the common measures in space are:

The AU (astronomical unit) This is the distance from Earth to the Sun, 93 million miles or about 8 light-minutes.

The Light-Year, or the distance light travels in one year, at a speed of 186,000 miles per second.

The Parsec, a distance of about three and a quarter light years.

In space, things don't slow down, there is no friction from roads or air, so anything that moves will continue to move in a straight line until something stops it. Therefore spacecraft have to fire their drives to accelerate and fire them again to decelerate. The quickest method of travel is therefore to accelerate to the halfway point, then decelerate the other half to bring the ship to rest at its destination.

Also, if a ship is moving at 1 inch per turn, toward the top of the table (lets call it north) it will continue to move in a straight line at 1 inch per turn without any input from the drives.
This is called a 1 inch vector. Firing the drives with a 1 inch reverse thrust will bring it to rest. Or, a further 1 inch forward thrust will increase the movement vector to 2 inches per turn.

If the moving ship fires its drives with a 1 inch vector to the left (west) instead, this will be added to the existing movement vector, and instead of moving 1 inch forward, it will move about an inch and a half at an angle of 45 degrees to the left. (ie northwest) The exact movement can be found by placing two one-inch arrows (vectors) at right angles, and the actual movement of the ship completes the triangle.

From now on, the ship's movement vector will be 1.41 (ish) northwest, until the ship fires its drives again. You can fire the drives in any direction and at any thrust short of your maximum to move your ship where you want it. Just lay the tail of the thrust vector to the head of the existing movement vector and your new vector is found by completing the triangle.

Hope that helps. More if needed.