Map of the Verse discussion
POSTED BY: jewelstaitefan
UPDATED: Tuesday, April 30, 2024 08:33
VIEWED: 49136
PAGE 7 of 18
Another thing that seems weird.
In The Message, Book talks about how Lt Womack is from the Silverhold Colonies, and how very far away it is. Yet on the Map of the Verse, they are relatively very close, in the same system of Red Sun.
Anybody else consider this a contradiction?
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A similar post to this was already done twice, but this is updated with the figures from the White Papaer, so no further revisions should be needed. The previous ones were based on conclusions and measured estimates.
Based upon the White Paper figures, I will recalculate speed and distance figures from an above post.
In the Red Sun system, the orbit of Jiangyin has a radius of 0.525Au, and the orbit of Greenleaf has a radius of 6.900Au.
That gives a distance of 6.375Au when they are nearest, and about 7.725 when farthest from each other (including arcing around the Red Sun itself).
We can assume in Safe that they will use max speed/acceleration when trying to get dying Book to Med Facilities, and this is supposed to be 10 hours. So 5 hours of acceleration,. and 5 hours of decceleration.
The min accel rate would be 0.255 Au/hour sq, and the fastest would be about 0.309 Au/hour sq.
A rudimentary time/distance table would be like this:
Time Min & Max distance/speed
1 hour. 0.1275 @0.255 & 0.1545 @0.309
2 hours. 0.5100 @0.510 & 0.6180 @0.618
3 hours. 1.1475 @0.765 & 1.3905 @0.927
4 hours. 2.0400 @1.020 & 2.4720 @1.236
5 hours. 3.1875 @1.275 & 3.8625 @1.545
6 hours. 4.5900 @1.530 & 5.5620 @1.854
7 hours. 6.2475 @1.785 & 7.5705 @2.163
8 hours. 8.1600 @2.040 & 9.8880 @2.472
9 hours. 10.3275 @2.295 & 12.5145 @2.781
10 hours. 12.7500 @2.550 & 15.4500 @3.090
15 hours. 28.6875 @3.825 & 34.7625 @4.635
18 hours. 41.3100 @4.590 & 50.0580 @5.562
20 hours. 51.0000 @5.100 & 61.8000 @6.180
So it's about 18-20 hours to accelerate until traveled 50Au, then the same to decelerate back to zero, for a total of 36-40 hours to travel 100Au.
please post correction if I have erred.
The speed of light would be approx 7.2Au per hour. Extending the table without consideration of limiting formulae for approaching speed of light, 23 hours would get to 81.7Au distance at velocity 7.1Au/hr at fastest, and 28 hours would get to 99.96Au distance at velocity 7.14Au/hr at slower range of max speed.
White Paper mentioned the average speed of the Exodus to the verse was 1/3 speed of light, but not sure if this is a limitation to extend to 2517.
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Quote:
Originally posted by jewelstaitefan:
The min accel rate would be 0.255 Au/hour sq, and the fastest would be about 0.309 Au/hour sq.
That's a maximum acceleration of ~12840483.8 m/s/s, or 1310253.4g, where 1g is acceleration due to gravity at the Earths surface, roughly 9.8m/s/s.
Solving for a distance of 100AU and assuming constant acceleration then deceleration, that's a trip time of roughly 1.62 Hours.
At that level of acceleration you'd be experiencing relativistic effects too, say on board ship time of about 0.6 hours.
You seem to be scaling your distance traveled linearly. Since the ship is accelerating, distance traveled will actually expand exponentially.
The simple equation for distance travelled per acceleration would be Newtons:
distance = acceleration * time^2
When Jiangyin and Greenleaf are at their furthest, there'll be a star in the way so the greatest travel distance would be somewhat higher than their straight line distance, but ignoring that for a rough figure, I'll take a median between the high and low:
6.375 + 7.725 / 2 = 7.05AU
We need it in metres, which comes out too:
1,054,664,985,621.488m
Time is 10 Hours, 36000 seconds.
Solving the above equation to get acceleration:
d=a*t^2
a=d/t^2
But we're decelerating half way, so we only need to workout half the journey.
Which gives us:
a=527332492810.744 / 18000^2=1627.569422255m/s/s
We want to double that number, because it's only half way, we want to do all our acceleration for half the time, so the final figure is:
~3255.2m/s/s
That's all very rough, but it's more or less what you want. For accelerating half way then decelerating, this will give you a ball park figure:
d=( (a/2)*((t/2)^2) )*2
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Not sure if we are on the same page, saying the same thing. But our calculations seem to differ.
Quote:
Originally posted by citizen:Quote:
Originally posted by jewelstaitefan:
The min accel rate would be 0.255 Au/hour sq, and the fastest would be about 0.309 Au/hour sq.
That's a maximum acceleration of ~12840483.8 m/s/s, or 1310253.4g, where 1g is acceleration due to gravity at the Earths surface, roughly 9.8m/s/s.
Solving for a distance of 100AU and assuming constant acceleration then deceleration, that's a trip time of roughly 1.62 Hours.
Let's just use the 0.255Au per hour per hour rate which I specified was the minimum accel rate (based upon my calculations - it can be altered if proven to be incorrect). Consider that the assumed given for this problem. After one hour of acceleration, by definition, the craft is at a velocity of 0.255Au per hour. This is the definition, the defined acceleration rate (and decel rate). It will have traveled approx 0.1275Au in that first hour while accelerating from zero velocity up to 0.255Au per hour. Correct so far?
If still accelerating in the second hour, by the end of that second hour, the velocity will have achieved 0.51Au per hour, as defined, right? In that second hour of accelerating from 0.255Au per hour up to 0.5100Au per hour, it traveled approx 0.3825Au, right? Adding that 0.3825 with the 0.1275 from the first hour, the craft will have traveled about 0.5100Au in the first 2 hours of acceleration from zero velocity, correct?
If still accelerating in the third consecutive hour, the velocity will have achieved 0.765Au per hour (0.255Au per hour faster than the 0.51Au per hour it was an hour before), as defined, right? During this 3rd hour, the travel distance while accelerating from 0.51Au per hour up to 0.765Au per hour will proximate 0.6375Au. Added to the 0.51 distance of the first 2 hours, the travel distance of 3 consecutive hours of acceleration would be about 1.1475Au, correct?
I don't see where, if accelerating at 0.255Au per hour per hour, we could achieve a distance of 50Au and also decelerate at the same rate for another 50Au, and complete 100Au in only 1.62 hours.
Possibly your conversion is off? Or both conversions? Not sure, but I don't see your conclusion to fit the given that I stated.
Quote:
At that level of acceleration you'd be experiencing relativistic effects too, say on board ship time of about 0.6 hours.
You seem to be scaling your distance traveled linearly. Since the ship is accelerating, distance traveled will actually expand exponentially.
The simple equation for distance travelled per acceleration would be Newtons:
distance = acceleration * time^2
I'm sorry, this equation does not make real world sense to me. In one unit of time, the square will have been one, and therefore the distance traqveled in your equation is the same as the acceleration rate. Unless the first instant of this time period saw the velocity jump from zero to the acceleration rate velocity (meaning the entire first unit of time was spent AT the velocity, instead of spending that first unit of time accelerating up to that velocity). If it was to accelerate up to 1 Au per unit of time in it's first unit of time, how could it have traveled a full 1 Au in that first unit of time, having started at zero and not exceeded the velocity it needed to achieve by the end of that first unit of time.
This seems incorrect.
Quote:
When Jiangyin and Greenleaf are at their furthest, there'll be a star in the way so the greatest travel distance would be somewhat higher than their straight line distance, but ignoring that for a rough figure,
The shortest distance is when they are closest, assuming nothing in the way, and that is 6.900Au - 0.525Au = 6.375Au. When farthest apart straightline distane - thru Red Sun - is 6.9Au + 0.525Au = 7.425Au. Since Serenity is not capable of traversing the interior of the Red Sun (at least not with it's suspect buffer panels), I conjured substituting a quarter o
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Quote:
Possibly your conversion is off? Or both conversions? Not sure, but I don't see your conclusion to fit the given that I stated.
Possibly, I did it quickly while at work, I'll rerun the numbers later when I have time.
Quote:
I'm sorry, this equation does not make real world sense to me. In one unit of time, the square will have been one, and therefore the distance traveled in your equation is the same as the acceleration rate.
Acceleration provides an exponential increase in distance traveled for linear time, while you're supposition is linear distance for linear time:
Quote:
This discussion illustrates that a free-falling object which is accelerating at a constant rate will cover different distances in each consecutive second. Further analysis of the first and last columns of the data above reveal that there is a square relationship between the total distance traveled and the time of travel for an object starting from rest and moving with a constant acceleration. The total distance traveled is directly proportional to the square of the time. As such, if an object travels for twice the time, it will cover four times (2^2) the distance; the total distance traveled after two seconds is four times the total distance traveled after one second. If an object travels for three times the time, then it will cover nine times (3^2) the distance; the distance traveled after three seconds is nine times the distance traveled after one second. Finally, if an object travels for four times the time, then it will cover 16 times (4^2) the distance; the distance traveled after four seconds is 16 times the distance traveled after one second. For objects with a constant acceleration, the distance of travel is directly proportional to the square of the time of travel.
http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1e.html
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I did not intend to average them. We do not know which is correct, where the planets were in relation to each other, so either value could be correct, and any value in between, and I did not want to discount any valid values until other evidence presented itself. The values can be calculated as possible ranges of calculations, and all remains valid.
I know, but I only wanted to run the numbers once, and was really only wanting to show you how to do it, and the pertinent equation, and thought an average of the two extremes would be more useful.
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I do not know why we need it in meters. But if that works for you, fine. How many Au is traveled in the first hour, by your reckoning?
I'm converting to metres and seconds because those are the units the equation is meant for. You can probably use Hours and AUs if you want, but it makes unit conflicts much more likely.
I'll get back to you in a little while.
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Ok, so I wasn't deriving the equation properly, chalk it up to running through them quickly at work. The equation is actually:
d=ut+(1/2)at^2
d = distance
u = initial velocity
a = acceleration
t = time
Since the initial velocity, ut is factored out (multiplication by zero = zero). That leaves:
d=(0.5)at^2
Similar to what I had, but I forgot to introduce the 0.5 when I solved the equation. Sorry. That means you're right for the initial hour of acceleration, it'll be half the number found in a. Distance travelled still increases exponentially.
Anyway the equations you want, that will also be easier than constructing a table, are:
a=2d/t^2
t=sqrt[2d/a]
d=(0.5)at^2
Looking back at it now, your table seems close to right, so I may just have misread last time.
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Quote:
Originally posted by citizen:
... be easier than constructing a table, are:
equations are nice, but I wanted to provide a reference table for others to be able to refer to quicky, without mucking with equations and formulae. As you have adroitly demonstrated, even those who think they are calculating correctly can mistake 1.62 hours for 50 hours.
With so many errors and/or inconsistencies already present in the Map of the Verse, I was hoping to reduce confusion, not multiply confusion, and a table seemed to help rectify this aspect of the confusing parts.
Quote:
your table seems close to right,
I agree.
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Quote:
Originally posted by jewelstaitefan:
As you have adroitly demonstrated, even those who think they are calculating correctly can mistake 1.62 hours for 50 hours.
Well I derived the equation wrong, with the correct equations that appear in my previous post, they're use will be less error prone than using a table.
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Holy Cow!
Just noticed that Mr. Universe is given the location of Comm Station in ring 2 (Kalidasa orbital ring) in the position Kalidasa plus 150 degrees. If I understand correctly, that is almost the opposite side of the verse from Kalidasa (opposite would be 180 degrees).
If that is the case, then in BDM Serenity goes from Lilac in Blue Sun sytem, to Beaumonde in Kalidasa system, back to Haven in Blue Sun system (both trips don't seem very long), and then the Traingin House and back to Haven (in Blue system) and Miranda in Blue system, and then to Mr Universe, the opposite side of the verse from Kalidasa!! In script (not on screen) River says it will take 4 hours. After luring the reavers, with Alliance/Operative inside the ion cloud of Mr. Universe's place, it did not seem they traveled for several days from Reaver Armada to Mr Universe, to me at least.
Anybody think that is reasonable?
Remember that in OMR it was supposed to take 5-6 days to get from Triumph in Red Sun system to Beaumonde in Kalidasa (implying they were on opposite sides of the verse). And Blue Sun ystem is much farther than Red Sun when on opposite sides of the verse from Kalidasa (or Mr Universe 150 degrees away).
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Based upont the revised information above after the White Paper, we can now review the itineraries of the Verse.
Let's start with BDM Serenity.
Lilac to Beaumonde to Haven to Training House to Haven to Miranda to Mr. Universe's place.
Beaumonde in 15th orbital ring around Kalidasa, which is 121Au from White Sun. Beaumonde is 12.850Au from Kalidasa Sun, so Beaumonde's farthest distance from White Sun is 133.850Au.
Lilac, second moon on New Canaan in 2nd orbital ring around Blue Sun, which is 180Au from White Sun. New Canaan is 2.025Au from Blue Sun, so the nearest distance to White Sun that New Canaan and Lilac get is 177.975Au.
Therefore the shortest distance that Lilac and Beaumonde can get is 44.125Au, or about 24-30 hours at max Serenity acceleration.
Haven, 1st moon on Deadwood in 7th orbital ring around Blue Sun. Deadwood is 14.025Au from Blue Sun, so nearest disance to White Sun would be 165.975Au. Shortest distance from Beaumonde to Haven would be 32.125Au, or about 20-26 hours at max Serenity accel/decel.
Miranda orbiting Burnham around Blue Sun isn't much distance from Haven. Miranda is 0.037Au from Burnham, and Burnham is 23.000Au from Blue Sun (8th orbital ring). So nearest that Miranda gets to Haven is about 8.975Au and farthest apart is about 37.06Au.
Mr. Universe is at Comm Ring 2 Station 2e, in the same orbital ring as Kalidasa (121Au from White Sun) 150 degrees of orbit arc from Kalidasa. At best, this puts it about 190Au from Miranda when Kalidasa is closest to Blue Sun. This also means when the Reaver Armada comes through the ion cloud, it has followed Serenity across the entire verse to surprise the Operative, which doesn't seem to mesh with on-screen.
If Mr. Universe is much closer, then Kalidasa is much farther away, like 120Au if halfway out of closest position to Blue Sun. That 120Au would make it over 40 hours both from Lilac to Beaumonde, and also from Beaumonde to Haven.
So, does this seem reasonable? Or is the Beaumonde distance out of wack? Does it seem reasonable that it took a day to get from Lilac to Beaumonde, and another day to get back to Haven (a short ways from Lilac)?
I didn't get that impression from film and scripts, how about you?
The travels to and from Training House are not a problem, nor are trips to Miranda.
Lemming has admitted that not all of the Map of the Verse is vetted within the series and BDM travel itinerary, so should we decide that this placement of Beaumonde in the Kalidasa System is one of these anomolies?
It would seem that if Beaumonde was in the Blue Sun system, or at least the same system as either Lilac or Haven, then the travel tiems would seem more in line with what the BDM showed.
Anybody else?
One wrinkle to consider is the upgrades to Serenity after TLB and before BDM. Apparently the proceeds from the fencing of the Lassiter paid for all the Serenity upgrades, like more blinking lights around the cockpit, and the new mule. Did they include speed/performance upgrades? Is their travel time quicker now? If so, it's also possible to consider one of the moons of Dragon's Egg as a training House location.
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