CLAC LAC launching

Joat42 wrote:

SharkHunter wrote:The reaction thrusters that was used to boost a shuttle out of the wedge maxed out at about 20G in HoE briefly, which was also apparently quite a bit more violent than a normal small ship launch.

In terms of clearing the ship's field, CLACs only accelerate until they reach a desired velocity, so when launching, are likely up to a desired speed but don't need to accelerate further. Even if they're under some acceleration, IIRC the grav plates can do about 20:1, so divide it down and at any accel less than say 200G, the LAC launch would be no more violent than that of a fighter off of an aircraft carrier or less, even without impellers.

Uhm, my recollection is that the grav-plates can handle up to 50G, then it starts to bleed through.

Found it:

Echoes of Honor wrote:Warships had much more powerful internal grav plates than shuttles or other small craft, but without the sump of a grav wave for their compensators to work with, the best they could do was reduce the apparent force of a hundred and fifty gravities by a factor of about thirty.

So at what would have been 30G, that's a 1G bleed through, etc. At 200G that would be an effective G of less than 7; I've done a bit more than 6G at times, and while that is not fun for more than parts of a minute, it's not all that tough. Any fighter jet pilot since WWII has likely pulled that much or more repeatedly; current generation fighters regularly perform in the 8-10G range. Easily handled by personnel trained for it, if even required in the Honorverse

Lord Skimper wrote:
Wedge up or down?

To clear the wedge stopped (acceleration zero not stopped totally) the LAC at 1 G would take 4 hours 10 minutes.

wedge down it would still need to go out the CLAC side, turn around and clear the LAC wedge size of at least 1000 metres at 1 G thrusters 100-200 seconds to clear the CLAC and each other LAC on it's side. 50 LAC per side for a Minotaur it could take 15 plus minutes to over an hour to launch all the LAC. Wedge up even longer.

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)

but even if it would take over an hour wouldn't it be easier (and quicker) for the CLAC to move again once the LACs are, say, 100 m out. At 500+ g that's going to zip clear of the LACs in seconds... (and the kilt opening of the wedge would be large enough not to risk the LACs)

Jonathan_S wrote:

Lord Skimper wrote:
Wedge up or down?

To clear the wedge stopped (acceleration zero not stopped totally) the LAC at 1 G would take 4 hours 10 minutes.

wedge down it would still need to go out the CLAC side, turn around and clear the LAC wedge size of at least 1000 metres at 1 G thrusters 100-200 seconds to clear the CLAC and each other LAC on it's side. 50 LAC per side for a Minotaur it could take 15 plus minutes to over an hour to launch all the LAC. Wedge up even longer.

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)

but even if it would take over an hour wouldn't it be easier (and quicker) for the CLAC to move again once the LACs are, say, 100 m out. At 500+ g that's going to zip clear of the LACs in seconds... (and the kilt opening of the wedge would be large enough not to risk the LACs)

For that matter, relatively quick LAC deployment like this seems to be what we actually see in the books. If you've got theoretical calculations that conflict with actual data, you reject the calculations or the numbers you feed into them, not the world.

JeffEngel wrote:For that matter, relatively quick LAC deployment like this seems to be what we actually see in the books. If you've got theoretical calculations that conflict with actual data, you reject the calculations or the numbers you feed into them, not the world.

Nah. The eqns. are fine. The button puncher on the calculator may be in question. The data imputed into the equations is downright barmy in his scenario. 1g...

We know exactly what RHN grav plates and RMN grav plates can do per the books. Why would anyone believe LAC's would not have them? Only have to have localized grav plates in a LAC residing under the 10 crews chairs. Unless of course it is less mass intensive to add the grav plates instead of over designing the hull for multi G's of acceleration.

JeffEngel wrote:For that matter, relatively quick LAC deployment like this seems to be what we actually see in the books. If you've got theoretical calculations that conflict with actual data, you reject the calculations or the numbers you feed into them, not the world.

Well I am not that good at relating that relatively quick in the written word to a concept of real time. Remembering what RFC has said later about how long it took to transition a wormhole and get out of its affects and what I pictured. I was close to an order of magnitude off, in my mind. When I went back and reread the sections after that post, "Duh what was I thinking."

Funniest part of this whole topic. Why not compare launching a Pinnace they routinely use a 100 g's to enter or leave a boat bay. or Honor's departure of the repair ship in HAE.

Just 2 quick thoughts that occurred to me when this topic first started.

Have fun,
T2M

Jonathan_S wrote:

Lord Skimper wrote:
Wedge up or down?

To clear the wedge stopped (acceleration zero not stopped totally) the LAC at 1 G would take 4 hours 10 minutes.

wedge down it would still need to go out the CLAC side, turn around and clear the LAC wedge size of at least 1000 metres at 1 G thrusters 100-200 seconds to clear the CLAC and each other LAC on it's side. 50 LAC per side for a Minotaur it could take 15 plus minutes to over an hour to launch all the LAC. Wedge up even longer.

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)

but even if it would take over an hour wouldn't it be easier (and quicker) for the CLAC to move again once the LACs are, say, 100 m out. At 500+ g that's going to zip clear of the LACs in seconds... (and the kilt opening of the wedge would be large enough not to risk the LACs)

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.

The LAC need to be 1000+ Metres apart from each other as well as from the CLAC wedge. Assuming a 2KM wedge for a LAC, twice that for safety. There are 50-100 LAC. I suppose they could all mass launch and then have the CLAC clear the area, accelerating at 350G starting at 0 relative to the LAC. Then have them accelerate at 1 G or 30 G or whatever the limit is for G for the Thrusters counter grav etc.... in every different direction from each other.

In Shadows of Freedom the CLAC barrels into the system at 350 G looking like a SD causing the BC to flee. How could the LAC be launched at that Acceleration? at 30 G Counter Grav that would be about 12 G.

"Most of her captains had thought she was out of her mind when she first proposed using reaction thrusters to generate an intercept vector. It simply wasn't done. The maximum acceleration a ship like Farnese could attain on her auxiliary thrusters, even if she ran them up to maximum emergency power, was on the order of only about a hundred and fifty gravities, which was less than a third of what her impeller wedge could impart. Worse, those thrusters were fuel hogs, drinking up days' worth of reactor mass in minutes. And to add insult to injury, without a wedge, there was no inertial compensator. Warships had much more powerful internal grav plates than shuttles or other small craft, but without the sump of a grav wave for their compensators to work with, the best they could do was reduce the apparent force of a hundred and fifty gravities by a factor of about thirty." Echoes of Honor


Are the Reaction Thruster going to power the LAC for just a few seconds or how do they burn reactor mass when their is no reactor?

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?

The formula for calculating distance at a given acceleration for a period of time is:
d = 1/2at^2. So for an acceleration of 9.8 m/sec2 for 10 seconds, the distance is 1/2 * 9.8 *100 or 490 meters.

Lord Skimper 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]

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)

And when you do it for 60min (1 hr) you get 63,504 km.

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?