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Author Topic: Check my math re: transit times?  (Read 779 times)

Weirdo

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Check my math re: transit times?
« on: 07 May 2021, 11:22:14 »
So I'm trying to get a feel for the strategic range of a Lyonesse gunboat. It has a five-ton fuel tank, but if we want to reserve 100 Fuel Points for liftoff, landing, and combat, this reduces it to 3.75 tons. 1.84 tons per 1G burn day means a maximum transit time of roughly 2.04 days, which according to the Interplanetary Flight Times Table in StratOps means that our shuttle can manage a flight from Terra to Venus, or maybe to Mars if they're getting close to minimum separation. (This is for a one-way trip.) I'm pretty comfortable with these numbers.

But can you save fuel by accelerating at lower Gs? I'm going to make the very dangerous and unfounded assumption that flying at 0.75 Gs consumes 0.75 times the fuel compared to 1G. That boosts our total burn time from 2.04 days to 2.72 days. Because Cray loves to show off his mathematitude, StratOps gives us a formula of

Distance = 0.25 x Accel x Time2

Where distance is in meters, Accel is in meters/second/second(9.8 at 1G), and time is in seconds. These are going to involve very big numbers very quickly, so I'm going to round off when typing this, but won't do so in my actual math. Converting the Accel to 0.75G and my burn time to seconds gives me

Distance = 0.25 x 7.35 x 234,6832 = 101,288,279,773.16 meters, or roughly 101.3 million kilometers compared to the 76 million kilometers at 1G. This seems like quite a jump, until I look at the table and see that's still not enough to reach Mars at typical separation, though flying at 0.5G should be able to manage the 150 million kilometers needed.

Does my math check out?
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Elmoth

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Re: Check my math re: transit times?
« Reply #1 on: 07 May 2021, 12:00:36 »
Haven't checked the math, but the reasoning sounds good. Except that a lower acceleration should consume LESS fuel.

You can save fuel another way as well: these are Small Craft. So they fit in the ASF/SC spaces of a dropship. Lunch from a dropship and you do not need to consume any of your precious fuel in liftoff and can use those for actual black navy operations.

Weirdo

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Re: Check my math re: transit times?
« Reply #2 on: 07 May 2021, 13:09:30 »
When escorting or traveling with units that have small craft cubicles, yes. But plenty of DropShips don't have such cubicles, and I was performing a mental exercise into just what kind of in-system flights the Lyonesse could fly along. And even if it can't manage the full trip to a jump point, it can still provide cover for the first day or so of such a trip when ships are at their most vulnerable (once you get fast enough, interception becomes extremely difficult without a high-speed engagement, and boarding becomes almost impossible), with similar shuttles launched from stations at the jump point providing cover for the last bit of decel at that end.

In addition, I was also curious about the Lyonesse's utility as a long-range strike craft. Any space-capable craft can lift off from a ground site and hit any other location on the same world with ease, but small craft gunships allow planetary craft to launch strikes against stations at Lagrange points(or vice versa), or between the widely scattered stations and moons of a well developed gas giant system, something that might push the limits of an ASF's endurance.
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idea weenie

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Re: Check my math re: transit times?
« Reply #3 on: 07 May 2021, 19:02:44 »
Going slower will use less fuel and take longer, you are exactly right.

The short version is you take your acceleration in Gs, invert it, take the square root of it, and multiply that by what the trip time would be if you traveled at 1G.  You then multiply that trip time by the acceleration in Gs to get the fuel usage (compared to 1G).  So if something takes 3 days at 1G (3 Burn-days), and instead you go at 1/4 G, the inverse is 4, the square root is 2, so it will take 6 days.  6 days multiplied by 1/4 G is 1.5 Burn-days of fuel.

Similarly, you can accelerate to get to the destination faster, at a higher cost in fuel.  Accelerating at 2Gs, the inverse is .5, the square root of that is ~.707, so that 3-day trip at 1G is now only ~2.12 days.  However, you are burning twice as much fuel, so you use up ~4.24 Burn-days of fuel.

The fun question becomes how many days of life support you have on board.  Each extra person-day assuming Bay Quarters masses 50 kilos*.  Are you saving enough fuel to make up for the increased life support needs?  Or would putting additional fuel into the cargo space be a better idea?

* Bays get 20 person-days of life support per ton of cargo allocated to life-support.  1 ton = 1000 kg.  1000 kg / 20 person-days = 50 kg per person-day.

Quarters get a better deal (5 kg per person-day), but require the up-front mass.  If you are traveling for 106 days or less, see if you can use Bays and the 50 kg/person-day life support.  If 107 days or more, Steerage Quarters are more efficient.

You can also perform the reverse calculation.  If you have a distance that requires 5 Burn-days to cover, and your vessel can only carry 3 Burn-days of Fuel, you do the following:
Take the burn-days of fuel you have available (3)
Divide that by the number of burn-days needed (3/5 = .6)
Square it (.6^2 = .36)
That is the number of Gs you will be accelerating under (.36 Gs)

Working backwards:
Gs accel: .36
Inversion: 2.7778
Square root: 1.6667
Original trip time @ 1G: 5 days
Adjusted trip time: 8.3333 days
Trip time * accel = Burn-days used = 3 Burn-days

(I figured that by using math for Burn-days, the absolute value of the exponents involved would be smaller)

Hope this helps

Weirdo

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Re: Check my math re: transit times?
« Reply #4 on: 07 May 2021, 19:25:43 »
You can also perform the reverse calculation.  If you have a distance that requires 5 Burn-days to cover, and your vessel can only carry 3 Burn-days of Fuel, you do the following:
Take the burn-days of fuel you have available (3)
Divide that by the number of burn-days needed (3/5 = .6)
Square it (.6^2 = .36)
That is the number of Gs you will be accelerating under (.36 Gs)

THIS is the formula I wish was in StratOps! Thank you!
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Daryk

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Re: Check my math re: transit times?
« Reply #5 on: 09 May 2021, 09:00:16 »
The math is in StratOps, just not in that form.

As for the original question, the Lyonesse has quarters the last I checked (because Small Craft crew HAVE to have quarters now), so consumables don't really enter in to it.

Even 0.1G is pretty speedy by today's standards, and relatively sips fuel.  At 0.1g, 3.75 tons of fuel will last you:

3.75 tons / 1.84 tons per 1G-burn-day / 0.1 G = 20.38 days

20.38 days at 0.1G = about 10 AU one way, but you only get a high speed pass.  Using the turnover method, you can reach 5 AU without coasting (which is a whole other ball of snakes)

Basically, a Lyonesse can get just about anywhere you want in a system, given enough time.

Weirdo

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Re: Check my math re: transit times?
« Reply #6 on: 09 May 2021, 10:30:42 »
That's definitely good news for the long-range strike role, though most of the things you'd escort with a Lyonesse are too busy to putter along at 0.1G, so the long escort is still likely just at the start and/or stop of a journey.

Hmmm... Patrol boats for Belter communities...
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Daryk

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Re: Check my math re: transit times?
« Reply #7 on: 09 May 2021, 10:58:16 »
Exactly!  Belter communities are far apart, prefer low signature transits, and are used to operating in zero-g for long periods of time.

Elmoth

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Re: Check my math re: transit times?
« Reply #8 on: 10 May 2021, 00:27:42 »
Or you can have lyonesses and aquarius doing patrols at several points in the route, and keep switching charges. No reason why the lyonese that brings you to final destination now the same that covered your liftoff or mid transit.

Weirdo

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Re: Check my math re: transit times?
« Reply #9 on: 10 May 2021, 09:48:38 »
That's the idea behind one Lyonesse(or group thereof) covering the first day or so of transit out from the planet, and another group launched from stations at the destination to cover the final decel leg of the journey.
"Thanks to Megamek, I can finally play BattleTech the way it was meant to be played--pantsless!"   -Neko Bijin
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Daryk

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Re: Check my math re: transit times?
« Reply #10 on: 10 May 2021, 15:49:14 »
I do believe that's exactly what they were designed for...  ^-^

cray

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Re: Check my math re: transit times?
« Reply #11 on: 13 May 2021, 20:58:48 »
It has a five-ton fuel tank, but if we want to reserve 100 Fuel Points for liftoff, landing, and combat, this reduces it to 3.75 tons.

As I recall, the TW landing and takeoff rules specify fixed tonnages for takeoff and landing. You might want to budget 2 tons.

Quote
But can you save fuel by accelerating at lower Gs? I'm going to make the very dangerous and unfounded assumption that flying at 0.75 Gs consumes 0.75 times the fuel compared to 1G.

That's not dangerous and unfounded at all. Burn-days are technically "G-burn-days," or the fuel consumed in 1 day at 1G. A small craft uses 1.84 tons per G-burn-day, so it'd use 1.38 tons per day at 0.75Gs.

Quote
That boosts our total burn time from 2.04 days to 2.72 days. Because Cray loves to show off his mathematitude,

I was copying from DropShips & JumpShips, which gives more math and pre-calculated tables. I boiled it down in StratOps due to word count limits. And I totally wing it in Kerbal Space Program. I don't think I've pre-calculated anything in KSP in all my years of playing.

Quote
StratOps gives us a formula of

Distance = 0.25 x Accel x Time2

...

Distance = 0.25 x 7.35 x 234,6832 = 101,288,279,773.16 meters, or roughly 101.3 million kilometers compared to the 76 million kilometers at 1G. This seems like quite a jump, until I look at the table and see that's still not enough to reach Mars at typical separation, though flying at 0.5G should be able to manage the 150 million kilometers needed.

Does my math check out?

With 3.75 tons of fuel and 1.38 tons per day at 0.75Gs, I get 241,791 seconds, which would moderately impact the calculation. Otherwise, the math looks correct.

A takeaway should be that all the squares and square roots mean that you're not going to get a linear relation between acceleration and fuel consumption for a fixed distance.
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VhenRa

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Re: Check my math re: transit times?
« Reply #12 on: 24 May 2021, 07:03:35 »
Incidentally, I found a calculator that indicates if you have that much fuel, to reach the Terran Jump Point would take about 21.45 days.

1.02 days of thrust, 19.41 days of coasting followed by another 1.02 days to decelerate prior to arrival.

Daryk

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Re: Check my math re: transit times?
« Reply #13 on: 24 May 2021, 18:24:21 »
How much thrust? 1G? ???

Giovanni Blasini

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Re: Check my math re: transit times?
« Reply #14 on: 24 May 2021, 19:20:46 »
How much thrust? 1G? ???

Accelerating at 1G for 1.02 days works out to a velocity of 44064 m/s, which is around 0.02545 AU/day.  Coasting for 19.41 days would allow you to coast around half an AU.  So that seems...off.
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Daryk

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Re: Check my math re: transit times?
« Reply #15 on: 24 May 2021, 19:50:17 »
Agreed... my questions still stands...

VhenRa

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Re: Check my math re: transit times?
« Reply #16 on: 26 May 2021, 11:29:37 »
Accelerating at 1G for 1.02 days works out to a velocity of 44064 m/s, which is around 0.02545 AU/day.  Coasting for 19.41 days would allow you to coast around half an AU.  So that seems...off.

That is clearly off.

A day is 86400 seconds.

86400 seconds, times 9.80665 m/s = 847,294 m/s. [AKA: 847.294 km/s]

So about half an AU a day after accelerating.
Agreed... my questions still stands...

Yes, 1G.
« Last Edit: 26 May 2021, 11:34:53 by VhenRa »

Daryk

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Re: Check my math re: transit times?
« Reply #17 on: 26 May 2021, 17:22:12 »
Thanks for the clarification!  :thumbsup:

Giovanni Blasini

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Re: Check my math re: transit times?
« Reply #18 on: 26 May 2021, 19:06:43 »
Yeah, I think i see where I did my math error now, dividing 2*a into delta-x instead of multiplying 2*a by delta-x when figuring out your velocity.
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