Time equation applied to Mote in God's Eye

[ShapeShifter]

I'm working my way through the wonderful Mote in God's Eye, and in Ch. 18 came to a reference to a calculation of the time it would take to travel a certain distance. I thought that I'd apply the ole CT equation, and -- though I didn't expect to get an exact correspondence -- was wondering why my computation was so far off from the authors'.

T = 2 * square root of (D/A).
T = 2 * sq. root of (3^10/9.08665) [I thought I'd try to be more accurate than the book's 10 m/sec^2.]
T = 110,619.1419 sec
T = 30.73 hrs.

Now this is the quote:

The rock was thirty million kilometers distant from MacArthur, about a twenty-five-hour trip each way at one gee.

I'm "only" off by five -- close to six -- hours. I don't think I did the CT equation wrong. So, how did the authors figure this one out?

 

[Cryton]

Ask the author.

 

[Whipsnade]

I'm "only" off by five -- close to six -- hours. I don't think I did the CT equation wrong. So, how did the authors figure this one out?

Shapeshifter,

Neither the rock, the "Beehive Asteroid" if I remember the book correctly, or MacArthur are stationary. Both objects are moving along vectors.

Because both the rock and the ship are moving, the boat is leaving one moving point to intercept another moving point. When the boat leaves MacArthur, it shares the battlecruiser's vector. It must then adjust that vector to one that not only intercepts the rock but also matches the rock's own vector.

I don't know how familiar you are with Mayday, one of the vector-based ship combat games from CT, but I'll use it to illustrate what I'm attempting to explain.

Each moving object in the game is represented by three counters; Past Position, Present Position, and Future Position. Taken together, those three counters model the object's current vector.

In order to intercept an object, I only need to match my Future Position marker with the object's Future Position marker. With the Future Position markers occupying the same hex, the objects and I will be in the same location after the next movement phase. However, because our vectors are different, the object and I will be separated again after the following movement phase.

In order to board an object however, I need to match both my Future and Present Position markers with the object's Future and Present markers. When that occurs, both the object and I are moving along the same vector; we're moving at the same velocity in the same direction.

Vector matching is why the trip made by MacArthur's boat takes longer than you calculated. The trip isn't simply one between Point A and Point B. Instead, it's one between a moving Point A and a moving Point B.


Regards,
Bill

 

[ShapeShifter]

Thanks, Whip. I suspected there was something amiss with the idea of two stationary points. Space would be far less migraine-inducing if only things would just stay put!

 

[Andrew M]

Shapeshifter,

Neither the rock, the "Beehive Asteroid" if I remember the book correctly, or MacArthur are stationary. Both objects are moving along vectors.

Because both the rock and the ship are moving, the boat is leaving one moving point to intercept another moving point. When the boat leaves MacArthur, it shares the battlecruiser's vector. It must then adjust that vector to one that not only intercepts the rock but also matches the rock's own vector.

I don't know how familiar you are with Mayday, one of the vector-based ship combat games from CT, but I'll use it to illustrate what I'm attempting to explain.

Each moving object in the game is represented by three counters; Past Position, Present Position, and Future Position. Taken together, those three counters model the object's current vector.

In order to intercept an object, I only need to match my Future Position marker with the object's Future Position marker. With the Future Position markers occupying the same hex, the objects and I will be in the same location after the next movement phase. However, because our vectors are different, the object and I will be separated again after the following movement phase.

In order to board an object however, I need to match both my Future and Present Position markers with the object's Future and Present markers. When that occurs, both the object and I are moving along the same vector; we're moving at the same velocity in the same direction.

Vector matching is why the trip made by MacArthur's boat takes longer than you calculated. The trip isn't simply one between Point A and Point B. Instead, it's one between a moving Point A and a moving Point B.


Regards,
Bill

What Bill is saying here, is that you not only have to get there, you have to slow to a stop when you arrive. This is all dealt with in the equations you have used, but it assumes that the start and end points are not moving in the same direction at the same speed.

Is there anything in the text that suggests they might be getting closer? In this case Bill is spot on.

Is there anything in the text to suggest that this is the usual travelling time? In that case, the distance is going to be stable. They may be rotating around some central point, but this will not shorten the travel time and consequently, Bill is spouting bull.

 

[Whipsnade]

What Bill is saying here, is that you not only have to get there, you have to slow to a stop when you arrive.

Andrew,

Not quite.

The equation ShapeShifter used does take into consideration that the object starts and ends at rest. It's the same constant acceleration, turn over, constant deceleration travel time equation that has been published in Traveller since 1977.

What ShapeShifter guessed was amiss, that the boat didn't start or end at "rest", makes the equation much more hairy.

Is there anything in the text that suggests they might be getting closer?

I remember examining the same question decades ago. If memory serves, the midshipman accompanying the mission to the Beehive asteroid originally calculates an intercept course at something other than one gee. He suggests a couple of reasons for this, fuel savings is one IIRC, but the officer in charge explains why the idea isn't as good as the middie thinks.

Anyway, way back then I ran the numbers using the middie's alternate acceleration and got a result that suggests the Beehive was moving towards MacArthur. Not on an intercept course mind you, but moving as to decrease the distance between the two points.

In this case Bill is spot on.

Because we have no idea of either the ship's or asteroid's vector, we have no knowing how the distance between the two objects is changing. Even if they were rapidly moving closer, the necessity of matching vectors would still make calculating travel times without knowing the vectors in question impossible.

They may be rotating around some central point...

The Beehive will be in orbit around the Mote. Not the planet, that's Mote Prime, but Mote the star. We cannot even guess at what MacArthur's vector is though.

... but this will not shorten the travel time and consequently, Bill is spouting bull.

The idea that you cannot use a formula that presumes stationary start and end points in this situation isn't "bull" and, seeing as Niven graduated with a degree in mathematics, I'd guess he didn't simply pluck the quoted distance and travel time out of thin air either. He may have failed to realize that his Ringworld was unstable over long periods of time, but he wouldn't screw up a three-dimensional vector matching problem.

The preparatory work, or "bible", created for the novel is known to be rather extensive. Niven wrote much later that he and Pournelle had to change the orbit of New Chicago and the subsequent details in order to retain a single descriptive line in the text.

So, if we assume Niven isn't a boob and that he didn't simply make up the numbers in the novel, the Beehive Asteroid and MacArthur must be on vectors which are currently shortening the distance between them.


Regards,
Bill

 

[aramis]

Actually, if the book gives two different intercept times and the accelerations, one could do the math and figure out the relative vector's length...

 

[Whipsnade]

Actually, if the book gives two different intercept times and the accelerations, one could do the math and figure out the relative vector's length...

Wil,

Sadly, the book doesn't.

I checked a copy and found, that while the middie suggests thrusting at 1.5 gees so that they can spend more time examining the "Stone Beehive" before the Motie embassy ship arrives, there's no mention of a specific arrival time. There's only the earlier arrival inferred from the middie's suggestion.

I do remember toying with the problem way back in 1982 after a certain course at Naval Nuclear Power School gave me the "tools" do to so, but, while I don't remember the exact answers I derived, I do remember that I didn't calculate much more than what Niven/Pournelle told us in the book.


Regards,
Bill

P.S. I checked another part of the book detailing the Motie Engineer the humans meet first in the system hoping to find a few more numbers. The Engineer had been "prospecting" in the Stone Beehive asteroid and she began thrusting towards the two human ships as soon as her redesigned sensors showed her the huge amounts of metal within the warships' Langston Fields. MacArthur detects her and moves to meet her, but no time periods are mentioned for either of the trips involved.

 

[Andrew M]

I checked a copy and found, that while the middie suggests thrusting at 1.5 gees so that they can spend more time examining the "Stone Beehive" before the Motie embassy ship arrives, there's no mention of a specific arrival time. There's only the earlier arrival inferred from the middie's suggestion.

The 25 hour journey time is correct for 1.5 gees. [To be precice 25.0846467 hours]. It makes sense that Larry Niven would make this kind of mistake after all the careful bookwork. He did notoriously get the rotation of the Earth backwards in one story. As a maths graduate myself I can empathise with this entirely.

NB I missed this on first reading, but the quoted text said twenty-five hours each way. -- oops. this isn't relevant is it? --

 

[Andrew M]

The general fromula if you start with relative velocity V is:

(AT + V)^2 = 4AD + 2V^2​

(The '^2' means take the square.)

To find the time to travel:

AT = root[4AD + 2V^2] - V​

and the relative velocity from the travel time and distance:

V = AT - root[2(AT)^2 - 4AD]​

By my calculations the initial velocity would have to be just shy of a million kph !!