['77 Rules] Travel Times

[Deleted]

Equation and quote per '77 rules and quoted without permission ...

Interplanetary Travel: Worlds orbiting the same star are accessible by interplanetary travel, including by scheduled liners and by ship's boats, pinnaces, cutters and even by lifeboats. Generally, however, interplanetary travel takes long periods of time. Since most stellar systems have only one major world, interplanetary travel is infrequently used.

The typical travel times list indicates the time required to travel a specified distance (assuming 1 G constant acceleration, turnaround at midpoint, and 1 G constant deceleration). Specific distance travel times can be calculated by the referee or by characters, using the travel  formula shown, where T equals travel time, d equals the total distance travelled, and a equals the acceleration used. Constant acceleration, turn-around, and constant deceleration are assumed. It is suggested that the units used be 1000 miles, 10 minute periods, and 1 G (2000 miles per [10 minutes] 2). These units are those used in the space combat section later in this booklet.

Okay. So ... given this is the only information one might have for calculating in-system travel (travell?) times, how might this work?

Assuming 238,900 miles from Earth to the moon (mean distance, rounded) and 2 gee acceleration.

T = 2 x (square root of [238,900/2]) and using PEMDAS

T = 2 x (square root of [119450])

T = 2 x (346)

T = 691

691 ... what?

I'm sorry to be a deadglow here, but can someone shed light on how to make this equation work?

[True Black Raven]

Apr 3, 2018 15:40:30 GMT -5 @piper said:

Equation and quote per '77 rules and quoted without permission ...

Interplanetary Travel: Worlds orbiting the same star are accessible by interplanetary travel, including by scheduled liners and by ship's boats, pinnaces, cutters and even by lifeboats. Generally, however, interplanetary travel takes long periods of time. Since most stellar systems have only one major world, interplanetary travel is infrequently used.

The typical travel times list indicates the time required to travel a specified distance (assuming 1 G constant acceleration, turnaround at midpoint, and 1 G constant deceleration). Specific distance travel times can be calculated by the referee or by characters, using the travel  formula shown, where T equals travel time, d equals the total distance travelled, and a equals the acceleration used. Constant acceleration, turn-around, and constant deceleration are assumed. It is suggested that the units used be 1000 miles, 10 minute periods, and 1 G (2000 miles per [10 minutes] 2). These units are those used in the space combat section later in this booklet.

Okay. So ... given this is the only information one might have for calculating in-system travel (travell?) times, how might this work?

Assuming 238,900 miles from Earth to the moon (mean distance, rounded) and 2 gee acceleration.

T = 2 x (square root of [238,900/2]) and using PEMDAS

T = 2 x (square root of [119450])

T = 2 x (346)

T = 691

691 ... what?

I'm sorry to be a deadglow here, but can someone shed light on how to make this equation work?

Not looking at the rules and just taking a shot in the dark solely based on the quote I am thinking 691 minutes which equals 11 hours and 31 minutes.

[Deleted]

Apr 3, 2018 16:24:13 GMT -5 True Black Raven said:

Not looking at the rules and just taking a shot in the dark solely based on the quote I am thinking 691 minutes which equals 11 hours and 31 minutes.

Okay, I see how you arrived at that figure. I got similar results. But, then?

To check my work I looked at the Typical Travel Times table in Book 2, page 1 and see the typical travel time for 100,000 miles at 1-gee to be 2.3 hours. So ... twice as for at 2-gee (a bit less than the actual distance) is still way less than 11+ hours. Even if one triples that figure it is still less.

To triple-check. I cross-referenced with the '81 version of the rules. Converting miles to kilometers for the mean distance from Earth to the moon I get 384,472 clicks. Then, looking at the Typical Travel Times tables in Book 2, page 10? I see 129-149 minutes for a similar distance (300,000 and 400,000, respectively). So ... from a bit over 2 hours to less than 3 hours. These figures seem to more or less line up with table values in '77 Traveller.

I'm out to sea on this one. I know I'm a proponent of run it your way but I would at least like to understand the rules as written before I head off into my own territory.

[True Black Raven]

Apr 3, 2018 18:46:43 GMT -5 @piper said:

Apr 3, 2018 16:24:13 GMT -5 True Black Raven said:

Not looking at the rules and just taking a shot in the dark solely based on the quote I am thinking 691 minutes which equals 11 hours and 31 minutes.

Okay, I see how you arrived at that figure. I got similar results. But, then?

To check my work I looked at the Typical Travel Times table in Book 2, page 1 and see the typical travel time for 100,000 miles at 1-gee to be 2.3 hours. So ... twice as for at 2-gee (a bit less than the actual distance) is still way less than 11+ hours. Even if one triples that figure it is still less.

To triple-check. I cross-referenced with the '81 version of the rules. Converting miles to kilometers for the mean distance from Earth to the moon I get 384,472 clicks. Then, looking at the Typical Travel Times tables in Book 2, page 10? I see 129-149 minutes for a similar distance (300,000 and 400,000, respectively). So ... from a bit over 2 hours to less than 3 hours. These figures seem to more or less line up with table values in '77 Traveller.

I'm out to sea on this one. I know I'm a proponent of run it your way but I would at least like to understand the rules as written before I head off into my own territory.

I thought I would take a stab at it, would have been surprised if I was right. Hopefully one of our experts will explain it to us.

[Deleted]

Apr 3, 2018 20:35:18 GMT -5 True Black Raven said:

]I thought I would take a stab at it, would have been surprised if I was right. Hopefully one of our experts will explain it to us.

Hey, I appreciate you giving it a try! I figure it’s something simple I’m overlooking but I can’t figure out what.

[Deleted]

Okay, here is a good breakdown of the math from over on Google+. It uses metric units, but the math checks out. It’s been a bit too long since physics class, I made a fundamental error setting up the equation.

ETA: another poster plugged in English measurements. Just in case you don’t like metric.

[Admin Pete]

Apr 3, 2018 23:41:20 GMT -5 @piper said:

Okay, here is a good breakdown of the math from over on Google+. It uses metric units, but the math checks out. It’s been a bit too long since physics class, I made a fundamental error setting up the equation.

ETA: another poster plugged in English measurements. Just in case you don’t like metric.

That link gives me this

404. That’s an error.



The requested URL was not found on this server. That’s all we know.

[dizzysaxophone]

link worked for me. First commentator forgot to multiply the result of the square root by 2 though. Once you've done that you get an answer that lines pretty perfectly with the information in book 2. Y'all got me itching to play some traveller noow!

[Deleted]

Apr 4, 2018 1:03:45 GMT -5 dizzysaxophone said:

link worked for me. First commentator forgot to multiply the result of the square root by 2 though. Once you've done that you get an answer that lines pretty perfectly with the information in book 2. Y'all got me itching to play some traveller noow!

Yes, I noticed that. But he put me on the right track so I didn’t actually say anything. 

[Deleted]

Apr 4, 2018 1:01:49 GMT -5 Admin Pete said:

Apr 3, 2018 23:41:20 GMT -5 @piper said:

Okay, here is a good breakdown of the math from over on Google+. It uses metric units, but the math checks out. It’s been a bit too long since physics class, I made a fundamental error setting up the equation.

ETA: another poster plugged in English measurements. Just in case you don’t like metric.

That link gives me this

404. That’s an error.



The requested URL was not found on this server. That’s all we know.

Hmm ... it works for me.  

[Admin Pete]

Apr 4, 2018 1:06:03 GMT -5 @piper said:

Apr 4, 2018 1:01:49 GMT -5 Admin Pete said:

That link gives me this

Hmm ... it works for me.  

The first link is the link in the thread for me, the second link is what I get when I pull your post up from Classic Traveller


plus.google.com/u/0/%20CameronDuBeers/posts/cCvkLTtW3b3  This gives me 404 
plus.google.com/u/0/+CameronDuBeers/posts/cCvkLTtW3b3       is what I copy paste from the post, But when I click either link from this thread I get 404. Weird.

[Deleted]

Apr 4, 2018 2:15:51 GMT -5 Admin Pete said:

The first link is the link in the thread for me, the second link is what I get when I pull your post up from Classic Traveller

And now, oddly enough? Neither works for me. It did last night but not today.

Test case: plus.google.com/u/0/+CameronDuBeers/posts/cCvkLTtW3b3

[Deleted]

Apr 4, 2018 11:09:36 GMT -5 @piper said:

Apr 4, 2018 2:15:51 GMT -5 Admin Pete said:

The first link is the link in the thread for me, the second link is what I get when I pull your post up from Classic Traveller

And now, oddly enough? Neither works for me. It did last night but not today.

Test case: plus.google.com/u/0/+CameronDuBeers/posts/cCvkLTtW3b3

Okay ... the link works from the "Recent Posts" screen view, but not the individual thread. What the Gehenna?

[True Black Raven]

Well the important part is solved, now if we knew what was going on with Google+ links.

[Deleted]

Okay, mathematics has never been my forté. I've taken (and passed) algebra, calculus, statistics, physics; but it's been a while and an informal poll while I was taking a few of those classes revealed I spent twice as long on the homework as my friends.

So, I tried to type up an example of the travel times equation for my house rules document. I'm happy to share it here ... but if anyone sees an issue with the math or how I communicated it please let me know. This in an example Brian Renninger and Winchell Chung gave me over on the G+ Classic Traveller group.

[Mighty Darci]

Have an exalt @piper!

[erisred]

Apr 7, 2018 14:15:19 GMT -5 @piper said:

Okay, mathematics has never been my forté. I've taken (and passed) algebra, calculus, statistics, physics; but it's been a while and an informal poll while I was taking a few of those classes revealed I spent twice as long on the homework as my friends.

So, I tried to type up an example of the travel times equation for my house rules document. I'm happy to share it here ... but if anyone sees an issue with the math or how I communicated it please let me know. This in an example Brian Renninger and Winchell Chung gave me over on the G+ Classic Traveller group.

This is correct, with a bit of rounding, if you're flying to Luna and landing there. The final x2 is assuming acceleration to the halfway point followed by deceleration to a stop.

If you're accelerating all the way your time to get to that distance would be 73.8 minutes.

Here is a shortcut if you're using Standard rather than Metric measurements:

1g = ~22mph.

You can use that and save a lot of work, as in...

(rounding everything off to whole numbers)

T = 2(sqrt(D/a)

D = 240,000
A = 2g = (2x22mph)

T = 2(sqrt(240000/44))
T = 148 minutes for landing and 74 minutes for a fly-by.

EXAMPLE:

Time (rounded to the nearest min) to 10 dia and 100 dia for an 8,000 mile planet (ie, Earth) at 1 g.

1. D = 10x8000 = 80,000
2. T = sqrt(80,000/22)
   
   
3. T = 60 min

1. D = 100x8000 = 800,000
2. T = sqrt(800,000/22)
   
   
3. T = 191 min (3 hr 11 min)

IIRC the Traveller Book has a table with times to 10 and 100 dia for all the UWP diameters. Like I said, IIRC, so it might not be there, it might be I did that and wrote that table into my copy of the book. Mine is currently packed away or I'd look.

[mormonyoyoman]

I was good in math and I STILL hated these travel questions.