So one Atomic Rocket Motor is better than four?

[Rusty6899]

Just started practising delta-v calculations.

By my estimate you initially used a Rockomax Jumbo-64 fuel tank, and possible a Mk1 cockpit.

[DeMatt]

The idea that TWR isn't important in space is an opinion only batted around by those who let MechJeb play the game for them.

The idea that a thrust-to-WEIGHT ratio even exists in free-fall situations is a gross misconception.

[Rusty6899]

I tried to argue a while back that thrust-weight ratio was irrelevant in orbit, and was challenged by numerous parties. It's thrust to mass ratio that I can see being important, but I think I went on to ignorantly hypothesise that Newtonian Mechanics could potentially be used for delta-v calculations.

I'm pretty sure that you do "have" a TWR in orbit, as you aren't technically weightless, but you are pretty much always going to be accelerating perpendicular to a body, so it doesn't come into play.

[Kerbart]

If it had zero mass it would shoot off at lightspeed

No, because the fuel and fueltank still have mass.

And if both didn't have mass either you'd be going nowhere instead as the thrust of the engine is derived from spewing out mass.

[Franklin]

The idea that a thrust-to-WEIGHT ratio even exists in free-fall situations is a gross misconception.

You still have mass in space, and a gravitational "weight". There's never a case in KSP where you're not within one body or another's gravitational pull.

[FenrirWolf]

When you're in space, TTW simply means how fast your engines accelerate your ship. Means the same thing on the ground too, actually. You just have to be sure your acceleration beats the opposite acceleration due to gravity.

[TheDarkStar]

No, because the fuel and fueltank still have mass.

And if both didn't have mass either you'd be going nowhere instead as the thrust of the engine is derived from spewing out mass.

The math would break down only if the fueltank and engine were massless while the fuel had mass.

[Vanamonde]

The idea that TWR isn't important in space is an opinion only batted around by those who let MechJeb play the game for them.

This is ~450 tons heading out for Tylo at 0.1g. Is there any reason to be in a bigger hurry than that?

[Tank Buddy]

That's partially incorrect, the Tsiolkovsky rocket equation is about the relationship between mass of fuel and the mass of anything without fuel. You burn the same amount of fuel, but you are pushing the weight of the LV-Ns, which just happen to be very heavy. The only thing that changes in the equation is the dry mass, not Isp, so you are burning fuel at the same efficiently, but you are doing so in a less efficient manner.

If you don't care about long burns or TWR, go with one engine. Also, in some instances you will get more Delta V when using a lighter, but less efficient engine. But if you're burning a certain amount of fuel at a lesser efficiency you will have less Delta V.

[Franklin]

This is ~450 tons heading out for Tylo at 0.1g. Is there any reason to be in a bigger hurry than that?

Aha. To be able to get there before the game leaves alpha, and to be able to circularize orbit once you do.

[Dante80]

Aha. To be able to get there before the game leaves alpha, and to be able to circularize orbit once you do.

Which is perfectly doable with the design shown above. If you are using/need more thrust than that, you are probably doing it wrong (ie, less efficiently).

Edited November 18, 2013 by Dante80

[Beowolf]

If you had a theoretical zero-mass engine, it would cause no change in delta-V. Of course, such a thing is impossible, but it's nice to dream.

Not impossible! Drilling a small hole into a monopropellant tank creates an instant zero-mass engine.

In fact, it's slightly negative mass if you count the material that was removed by the drill bit.

[Dante80]

Thats not an engine, its a leak. XD

Edited November 19, 2013 by Dante80

[WafflesToo]

Thats not an engine, its a leak. XD

I would call it an improvised nozzle ;p

[Steve-S]

If it had zero mass it would shoot off at lightspeed

Technically no. Because you see E=MC^2. So the energy in the fuel is equivalent to the MASS of the fuel times the speed of light. 0 Mass = 0 energy. It would just sit there being smug. Even at 100% energy use with 0 loss, which, IIRC, is theoretically impossible.

[Edit]: Should've read the whole post first. Oh well.

Edited November 19, 2013 by Steve-S
read thru to end.

[Vanamonde]

Aha. To be able to get there before the game leaves alpha, and to be able to circularize orbit once you do.

A transit to another planet takes tens or hundreds of days, so what difference does it make if the burn takes 1 minute or (with this ship pictured) 17 minutes? Also, that ship orbited Tylo, then returned to Kerbin, so it's certainly capable of just about maneuver a ship might need to perform. Takeoff and landing are about the only times where thrust/weight is other than a matter of convenience and patience.

[Superfluous J]

A transit to another planet takes tens or hundreds of days, so what difference does it make if the burn takes 1 minute or (with this ship pictured) 17 minutes? Also, that ship orbited Tylo, then returned to Kerbin, so it's certainly capable of just about maneuver a ship might need to perform. Takeoff and landing are about the only times where thrust/weight is other than a matter of convenience and patience.

If I was in NASA? None. But I'm not in NASA. I'm playing a video game. For me, a 1 minute burn instead of a 17 minute burn means I can spend 16 extra minutes playing the game instead of watching the blue maneuver marker. 4 such burns just gave me an hour to play the game I wouldn't otherwise have had. Sounds like a pretty good reason to pack on some more thrusters to me.

[Vanamonde]

I'm playing a video game.

Okay, though the thread started off as a discussion of efficiency/delta-V rather than the patience of the player, and I was arguing from that perspective.

[RRoan]

A transit to another planet takes tens or hundreds of days, so what difference does it make if the burn takes 1 minute or (with this ship pictured) 17 minutes?

Oberth.

10 character limit is tyranny.

[Specialist290]

Oberth.

10 character limit is tyranny.

Something I've been wondering about in the back of my mind -- has anyone tried quantifying the gains from higher TWR versus the losses from a lower mass ratio on two otherwise-identical craft performing the same burn in terms of fuel actually used? I'd like to see at what point (if any) they reach equilibrium, myself.

[Vanamonde]

What is the counter-argument here? That loading a ship with a pile of redundant engines is a good idea?

[Superfluous J]

I believe the counter argument is that there is more to consider than total dV of a craft. That is the #1 concern, true, but if you have say 4000 dV to play with and adding an extra engine doubles your TWR (halving your burn times) but cuts your dV to 3700... that's a trade off I personally will take.

To get the above numbers, I used a payload I've actually used and then added a small fuel tank and the LV-N engine to it to see the dV and TWR. I then swapped that engine out for a bi-coupler and 2 engines. I rounded the dV and TWR in both cases, but the dV loss is actually better than what I said. It's closer to 250 than it is 300. TWR went from .15 to .29 which is essentially double.

[SSSPutnik]

No, because the fuel and fueltank still have mass.

And if both didn't have mass either you'd be going nowhere instead as the thrust of the engine is derived from spewing out mass.

Technically your engine would shoot off at lightspeed, anything with zero mass does this.

[capi3101]

Something I've been wondering about in the back of my mind -- has anyone tried quantifying the gains from higher TWR versus the losses from a lower mass ratio on two otherwise-identical craft performing the same burn in terms of fuel actually used? I'd like to see at what point (if any) they reach equilibrium, myself.

We can do that real quick. Let's say we've got a twenty tonne lander that we want to send to Eve. Let's further assume we're a steely-eyed missile man that can get that there and back for the exact amount of delta-V the delta-V map indicates; 1030 to get there, 1310 to get into orbit, 1030 to get back (I guess we'll forego aerobraking), so we need a transfer stage with 3370 m/s. We're starting from 100 km above Kerbin; gravity at that altitude is 7.207 m/s². Let's say our "default case" is going to be three LV-Ns. So we have twenty tonnes payload, let's say a Clamp-o-Tron Senior (0.2 t), three LV-N's (6.75) attached via BZ-52s (0.12), a long girder segment and a girder adapter (.85) that attached it to the booster for launch, and our fuel. We have 27.92 tonnes deadmass, so it works out as:

Md = (y-1)x /(9-y), where y= e^(dv / (I_sp * g_o)) and x is the deadmass (This is Tsiolkovsky backwards, making an assumption of a 9:1 mass ratio for fuel tanks).

y= e^(dv / (I_sp * g_o)) = e^(3370 / (800 * 9.81)) = 1.53635

Md = (y-1)x /(9-y) = (1.53635-1)27.92 /(9-1.53635) = 2.00638

M = 9Md = 18.05739 tonnes.

So we need 18.05739 tonnes of fuel to get 3370 m/s for the default case. If there's .5625 tonnes in an FL-T100, we need 32-33 FL-T100s; 32 FL-T100s equals one X200-32. Let's just go with that to keep it simple - we have eighteen tonnes of fuel wet, two tonnes fuel dry, 27.92 tonnes deadmass, 800 is the I_sp.

Tsiolkovsky: dV = ln((27.92+18)/(27.92+2)) * 800 * 9.81 = 3361.88 m/s

TWR = 180 / (27.92+18) * 7.207 = 0.544

Okay. Now for an array of values:

One Engine (-5.43 tonnes deadmass, -2LV-N, -2 BZ-52, no girder assembly needed): dV = ln((22.49+18)/(22.49+2)) * 800 * 9.81 = 3945.897 m/s, TWR = 60 / (22.49+18) * 7.207 = 0.206

Two Engines (-2.29 tonnes deadmass, -1 LV-N, -1 BZ-52): ln((25.63+18)/(25.63+2)) * 800 * 9.81 = 3585.303 m/s, TWR = 120/(25.63+18) * 7.207 = 0.382

Three Engines (default case): dV = ln((27.92+18)/(27.92+2)) * 800 * 9.81 = 3361.88 m/s, TWR = 180 / (27.92+18) * 7.207 = 0.544

Four Engines (+2.29 tonnes deadmass, +1 LV-N, +1 BZ-52): dV = ln((30.21+18)/(30.21+2)) * 800 * 9.81 = 3165.016 m/s, TWR = 240 / (30.21+18) * 7.207 = 0.691

Six Engines (+6.87 tonnes deadmass, +3 LV-N, +3 BZ-52): dV = ln((34.79+18)/(34.79+2)) * 800 * 9.81 = 2833.879 m/s, TWR = 300 / (34.79+18) * 7.207 = 0.789

You can put those data in a scatterplot and add a trendline. The formula comes out to y = -0.0006x + 2.3861, where y is TWR and X is delta-V. R² = 0.9791, so that formula would be an accurate way of calculating delta-V given a set TWR or vice versa 97.91% of the time for this particular craft given our set of initial conditions. It doesn't look like it would ever reach a state of "equilibrium"; ultimately as the TWR goes higher and higher, the delta-V will fall below acceptable limits. Of course, that formula isn't accurate - with zero engines, I have zero Isp and therefore zero delta-V (and zero TWR); that formula doesn't take that case into account. But what do you expect at 11 PM local, anyway?

Edited November 19, 2013 by capi3101

[Tigermisu]

capi, I only understood 3.14% of your post, but it was beautiful.