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CLAC LAC launching

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Re: CLAC LAC launching
Post by Jonathan_S   » Mon Jan 26, 2015 1:36 am

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Annachie wrote:How much distance is between the top and bottom wedges on an SD/CLAC, and if needed could they be pushed further apart or brought closer together by the generating ship?
They're angled; but at the aft end of the wedge (where the gap is narrowest) the two are about 40 km apart. Up front they're about 190 km apart. (So presumably, abreast of the ship they'd be about 115 km apart)

See http://infodump.thefifthimperium.com/en ... gton/100/0 for more, and a diagram.

But as far as I know you can't alter that spacing significantly. (The planes become a little more parallel under heavy acceleration; but not all that much)
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Re: CLAC LAC launching
Post by Joat42   » Mon Jan 26, 2015 11:28 am

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Jonathan_S wrote:
Jonathan_S wrote:How are you getting 4 hours? At only 1 g after a single hour the LAC would be 64,504 km away! And it needs to be less that 1,000 km to safely raise its wedge (even an SD has only about a 300 km wide wedge, so it's only 150 km to the edge of it. But then you need room for your wedge plus safety margin between them)[snip]

Lord Skimper wrote:My Math very well could be wrong I was unable to find a calculator that can tell me if one accelerates at 9.8 m/s2 for 10 seconds how far will one have gone?

everytime I figure it out for 1 hour I don't get 64,000 KM.
Like fallsfromtrees said, 10 seconds is 490 m.

You can do the calculation directly in google's search bar (and it'll even track the units for you so you can tell you didn't screw it up)
.5*9.8m/s/s*(10s)^2 = 490 m
.5*9.8m/s/s*(30s)^2 = 4.4 km
.5*9.8m/s/s*(60s)^2 = 17.64 km
.5*9.8m/s/s*(5*60s)^2 = 441 km
.5*9.8m/s/s*(10*60s)^2 = 1,764 km
ect.

(You could change the last two to "5min" and "10min"; google would still handle it correctly. Just don't do "5m" or it'll think "meters" not "minutes"; but you can tell something went wrong because the units on the results aren't the expected kilometers)

Use Wolfram Alpha instead, it takes natural language questions:
http://www.wolframalpha.com/input/?i=distance+of+9.8m%2Fs2+acceleration+for+1+hour
or:
http://www.wolframalpha.com/input/?i=time+to+travel+1000km+with+9.8m%2Fs2+acceleration

You can query it for all kinds of mathematical problems.

---
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Re: CLAC LAC launching
Post by Lord Skimper   » Mon Jan 26, 2015 3:49 pm

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Cool thanks.

So 30G would take 1 minute 22.5 seconds to go 1000 km.

If the CLAC was going 30 G it would take 2 hours 17.5 minutes to drop off all 100 ships spacing them 1000 km apart assuming they don't move themselves. Until their wedge is up.

Minotaur CLAC. 100 Shrike based LAC.

The Minotaur would be 324,000 Km away from the first ship thus dropped of if it hasn't moved.

Assuming I did this right.

Okay I did that wrong
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Re: CLAC LAC launching
Post by Joat42   » Mon Jan 26, 2015 4:59 pm

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Lord Skimper wrote:Cool thanks.

So 30G would take 1 minute 22.5 seconds to go 1000 km.

If the CLAC was going 30 G it would take 2 hours 17.5 minutes to drop off all 100 ships spacing them 1000 km apart assuming they don't move themselves. Until their wedge is up.

Minotaur CLAC. 100 Shrike based LAC.

The Minotaur would be 324,000 Km away from the first ship thus dropped of if it hasn't moved.

Assuming I did this right.

Okay I did that wrong


If the CLAC is going at 300G, it can launch 2 LACs every 26 seconds (port and starboard), these accelerate away from each other staggered after clearing the CLACs wedge. This takes ~22 minutes for 100 LACs, assuming the LACs doesn't use any other propulsion to open the distance until they can use their wedge.

Edit: Actually, it will go faster since the CLAC is accelerating and may already have built up the starting speed. :oops:

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Jack of all trades and destructive tinkerer.

Anyone who have simple solutions for complex problems is a fool.
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