Jump to content

Can someone ELI5 the concept of diminishing return in the "moar boosters" idea?


SlabGizor117

Recommended Posts

There IS also a limit of diminishing returns to Asparagus, if less strongly pronounced than with standard efficiency.

You have your basic payload of 5 tons and 10 tons of boosters that can lift it to orbit. 33% payload.

You want to triple your payload. 15 tons plus 30 tons of boosters? Nope, you need beams, struts etc to connect the new boosters to the new craft. That's an extra ton. So, 14 tons of payload, 31 tons of boosters+infrastructure. 32%

You want to scale your concept up? You triple that again, three units of 31 tons. to lift 3*14 = 42 tons? Nope. The units are larger, so you need more struts and longer beams to hold them together. Extra two tons per unit for six tons of junk. That's six tons off your payload, 36 tons vs 99 tons of boosters. Only 26% efficiency.

Triple that again, with 3 tons of beams per unit. 102*3=306 tons of boosters and infrastructure. 9 tons off your tripled payload. 99 tons payload, Your efficiency is down to 24%. And that's optimistically speaking, because this thing becomes a real nightmare to hold together, the forces between the new joints really ripping it apart.

That's why multiple smaller launches and orbital assembly are often preferable to lifting huge structures whole. Moar boosters means moar struts, and their number grows geometrically.

Link to comment
Share on other sites

There IS also a limit of diminishing returns to Asparagus, if less strongly pronounced than with standard efficiency.

You have your basic payload of 5 tons and 10 tons of boosters that can lift it to orbit. 33% payload.

You want to triple your payload. 15 tons plus 30 tons of boosters? Nope, you need beams, struts etc to connect the new boosters to the new craft. That's an extra ton. So, 14 tons of payload, 31 tons of boosters+infrastructure. 32%

You want to scale your concept up? You triple that again, three units of 31 tons. to lift 3*14 = 42 tons? Nope. The units are larger, so you need more struts and longer beams to hold them together. Extra two tons per unit for six tons of junk. That's six tons off your payload, 36 tons vs 99 tons of boosters. Only 26% efficiency.

Triple that again, with 3 tons of beams per unit. 102*3=306 tons of boosters and infrastructure. 9 tons off your tripled payload. 99 tons payload, Your efficiency is down to 24%. And that's optimistically speaking, because this thing becomes a real nightmare to hold together, the forces between the new joints really ripping it apart.

That's why multiple smaller launches and orbital assembly are often preferable to lifting huge structures whole. Moar boosters means moar struts, and their number grows geometrically.

OTOH, asparagus staging (or any staging) doesn't approach a maximum dV.

Link to comment
Share on other sites

So you are 5 my little friend, which means you probably can count to 5. Do this:

1) Take a probe core. Attach a booster to it. Call it R-1.

2) Put one R-1 on a launchpad and launch it. Apparently R-1 has x TWR, y m/s of dV and goes up to z meters before falling back.

3) Now put two R-1's on the launchpad and launch them at the same time. As you can see, they still have x TWR, y m/s of dV and go up to z meters before falling back. Which is kinda expected because they still are good ol' R-1's.

4) Now put two R-1's on the launchpad and duct tape them to each other. Call it R-1 Heavy. Launch it. Apparently R-1 Heavy still has x TWR, y m/s of dV and goes up to z meters before falling back. Which is kinda expected because duct tape isn't that magic.

5) Now make R-1 Heavy one probe core less heavy and call it R-2. Launch it. Apparently R-2 has slightly (not much) more than x TWR, slightly (not much) more than y m/s of dV and thus goes slightly (mot much) higher than z meters up before falling back. Which is kinda expected because these days, with microelectronics and stuff, one tiny probe core doesn't weight that much.

Link to comment
Share on other sites

OTOH, asparagus staging (or any staging) doesn't approach a maximum dV.

Well, if you keep adding asparagus stages, you do run into the problem of having the outside stages being nearly instantly drained of fuel. In that respect, traditional asparagus staging does hit a maximum dV. The only way to offset this is to increase the number of boosters per each stage (ie. 4 inner rockets are fed by 8 rockets, which are fed by 16 rockets), but this obviously gets out of hand very quickly..

Vertical inline staging, or non-asparagus radial staging, does not have that problem of course - and is theoretically unlimited (as you say).

Link to comment
Share on other sites

There is no maximum delta-V you can achieve theoretically (though from a practical standpoint, your rocket will eventually get so big that the game wouldn't be able to handle it).

Here's another example to see the math. For simplicity, let's assume that except for the payload and fuel, everything else has 0 mass (e.g. engines have infinite TWR and 0 mass).

Let's say that you have a 10-ton payload, and with 10 tons of fuel, you get X m/s of delta-V.

Warm-up questions: if you have a 5 ton payload, how much fuel do you need to get X m/s of delta-V? (5 tons -- half the payload, half the fuel) What if you have a 20 ton payload? (20 tons -- twice the payload, twice the fuel)

So now, we have a 20 ton rocket that can get X m/s of delta-V. What if we want to get another X m/s of delta-V? We'd need 20 tons of fuel. As above, 20 tons of fuel + 20 ton payload gives us X m/s. We then get another X m/s from the 10 ton fuel + 10-ton payload.

He's a chart:

10-ton payload + 10 tons of fuel = X m/s (20 tons total)

20-ton payload + 20 tons of fuel = X m/s (40 tons total)

40-ton payload + 40 tons of fuel = X m/s (80 tons total)

80-ton payload + 80 tons of fuel = X m/s (160 tons total)

160-ton payload + 160 tons of fuel = X m/s (320 tons total)

320-ton payload + 320 tons of fuel = X m/s (640 tons total)

640-ton payload + 640 tons of fuel = X m/s (1280 tons total)

1280-ton payload + 1280 tons of fuel = X m/s (2560 tons total)

The mass of the rocket doubles every row, but we're only adding X m/s of delta-V each time. The above rocket masses a whopping 2560 tons to give a 10-ton payload 8 * X m/s of delta-V. You can keep going to get more delta-V, but your rocket must get ridiculously huge to squeeze out more delta-V. And note that we assumed that engines, fuel tanks, decouplers, struts, etc. all had zero-mass! This is that logarithm in (https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation) in action.

Edited by Empiro
Link to comment
Share on other sites

This thread is quite old. Please consider starting a new thread rather than reviving this one.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...