OHM Easy Eris Explorer

[OrbitHangar]

Author: nicholander

This is a simple mod for people who are having trouble completing (With out cheating, of course!) Piper's fantastic mod "Eris Explorer". What it does it simple, it modifies the Eris Explorer's config file so it has only 1 engine, not three (Well, techincally), and the isp is tripled while the thrust the divided by 3. It also removes 2 of the engine effects, so now the two other engines are "Back-up" engines, incase the main one breaks down and is unable to operate.



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[N_Molson]

ISp does not work that way, and is unaffected by the number of engines. Its something qualitative, not quantitative.

 

[Nicholander]

Wait, what?!? So you're saying that the number of engines will not affect the isp at all? So how would having just one operational engine affect the delta v? Sorry for the mistake.

 

[ISProgram]

This is a concept that confuses me too. The thrust of your engine can be really low, but as long as it has a decent Isp, your resulting delta-V will be decent.

This concept goes hand in hand with ion propulsion, with a low thrust (<1kN) but a high Isp (1000s of s).

The reason you want significant thrust is to get somewhere quick (ion prop can't do that) or to have a high T:W to get into orbit, to minimize gravity losses.

 

[jedidia]

Wait, what?!? So you're saying that the number of engines will not affect the isp at all? So how would having just one operational engine affect the delta v? Sorry for the mistake.

If all engines have the same ISP, then that is the overall ISP. If they have different ISPs, then calculating the DV becomes a nightmare.

As such, multiple engines of the same type do not affect ISP. They do affect DeltaV, but simply by adding mass... Three engines of the same type have the same ISP but three times the mass, ergo they get less DeltV from the same ammount of propellant. But they have more thrust (overall thrust is simple addition of the thrust of all engines involved), ergo more acceleration.

The reason you want significant thrust is to get somewhere quick

Common misconception. The reason you want high thrust is that you can change Velocity quickly, which makes trajectory calculation a lot easier... in fact we don't have a fully developed tool in orbiter that gives reliable results for very low thrust drives (whose burn times are usually measured in hours for orbital operations, and in days or even weeks for interplanetary operations).
This does not mean that you can get somewhere faster. In case of a close target like the moon, this can hold true, but for any interplanetary operations the higher maximum velocity of a high ISP more than makes up for the long time required to reach that velocity, resulting in potentially drastic decrease in travel times.

You can read the write-u on thrust vs. ISP I did a while ago for IMS if you want a bit more detail. It's a fun read, or so people tell me.

 

[Urwumpe]

If all engines have the same ISP, then that is the overall ISP. If they have different ISPs, then calculating the DV becomes a nightmare.

Not really. You only need to know what ISP means.

ISP is the impulse you get per kg propellant consumed by your engine. And impulse is in its simplest definition force multiplied by time. Which means you can also interpret it as :

Thrust force by fuel consumption.

[math]I_{sp} = \frac{F \times t}{m}[/math]

So, for your whole rocket stage, it is just the force you get by the fuel that you burn.


Now I that example, ISP was tripled by force was divided by three. This means, that the mass flow or fuel consumption was changed a lot:

[math]\frac{m}{t} = \frac{F}{I_sp} [/math]

and was changed relative by:

[math]\frac{\frac{1}{3}}{3} = \frac{1}{3 \times 3} = \frac{1}{9}[/math]

So, in the same time, your spacecraft now consumes 1/9th of the fuel.

Now, for the maneuvers in space, only specific impulse matters, by rocket equation:

[math]\Delta v = I_{sp} \times \log \left ( \frac{m_0}{m_b} \right )[/math]

Since you did not change the propellant mass but only tripled your ISP, your spacecraft has now three times the DV. As long as gravitation does not have much effect in your maneuvers, the longer burn time has no negative effect at all.

 

[jedidia]

I think we have a slight misunderstanding there. What I meant is that, if you have engines with different ISPs running at the same time, the calculation of your total DV becomes rather involved...

 

[Urwumpe]

I think we have a slight misunderstanding there. What I meant is that, if you have engines with different ISPs running at the same time, the calculation of your total DV becomes rather involved...

It doesn't - it still stays the same... of course the formulas CAN look pretty annoying if you decide to go all crazy.


But practically it is still:

ISP = Total thrust / Total propellant flow


Total propellant flow can look a bit annoying then, if you calculate it as

[math]\dot{m} = \frac{F}{I_{sp}}[/math]

for every engine in the formula, but if you know that quantity already, its much simpler.

 

[Nicholander]

Okay, this is all a bit confusing. But all the 3 engines have the same isp, and if you removed by functionality of 2, the fuel tank would be feeding 1/3 of the fuel to the engines, since there is now only 1 engine, right? So would that increase the delta-v, and if so how much? Would it be the same as tripling the isp?

 

[statickid]

Turning off two engines would increase burn times needed to achieve the same ∆v...

UNLESS you REPLACE the engine(s) with a far superior one with a much higher ISP and retain the same fuel mass, then I believe you would be increasing available ∆v and reducing the difficulty of the mission

 

[Urwumpe]

Okay, this is all a bit confusing. But all the 3 engines have the same isp, and if you removed by functionality of 2, the fuel tank would be feeding 1/3 of the fuel to the engines, since there is now only 1 engine, right? So would that increase the delta-v, and if so how much? Would it be the same as tripling the isp?

No, its independent.

Delta V is defined by the rocket equation:

[math] \Delta v = I_{sp} \cdot \log{ \frac {m0}{m1} }[/math]

As you can see, this depends only on three factors: ISP, total mass before the maneuver and total mass after the maneuver

So, only your ISP change has an effect there. The number of engines or the thrust of the engines is not part of the equation.

Having a lower thrust increases the burn time, which in turn increases what is called "gravity losses". Everytime you are thrusting against the force of gravity, you need more DV to compensate for the gravity pulling you back. The longer you burn like that, the more you need to compensate. But this gravity loss also depends on the flight path angle of your burn: If you burn horizontal, you have no gravity losses. if you burn towards the source of gravity, you even gain a bit of extra DV.

If you would burn straight up for a short time (like for sounding rockets), the gravity losses are well approximated by [math]\Delta v_g = g \cdot t[/math]

 

[Nicholander]

So, your saying that the fuel consumption has no effect at ALL on the delta-v? So if you changed the number of engines, nothing would be changed except the thrust? And if for example, a rocket with 1 engine would have exactly the same delta-v as a rocket with 9999 of the same engine? That doesn't make any sense to me. And this is a space probe which never lands on or lifts off from anything, so gravity loss wouldn't apply, right?

 

[Urwumpe]

So, your saying that the fuel consumption has no effect at ALL on the delta-v? So if you changed the number of engines, nothing would be changed except the thrust? And if for example, a rocket with 1 engine would have exactly the same delta-v as a rocket with 9999 of the same engine? That doesn't make any sense to me. And this is a space probe which never lands on or lifts off from anything, so gravity loss wouldn't apply, right?

Exactly. You ALWAYS have small gravity losses in real spaceflight because you are never perfectly perpendicular to the force of gravity. But then we are rarely speaking of more than 1-2% difference to the ideal instant impulse maneuver.

The more thrust you have and the faster you thus burn the fuel at this impulse, the closer you are to the "all dv expended in a single short instant and the velocity is changed instantly". But the needed minimum change in the velocity vector does never change at all, regardless how long you need for changing it.

 

[sputnik]

Correct. Number of engines, or thrust/weight, has no direct effect on the available Delta-V of a spacecraft.

One of the buried, usually-unstated assumptions in astrodynamics, examining Hohmann transfers, etc., is "impulsive burns". That is, the delta-V change happens in an infinitesimally short time.
In the real world, burns take time, but as long as it's a short period as compared to the orbital period, you express this difference as a gravity loss and get on with life.

If your intent was to replace three engines with one engine of equivalent thrust, triple the thrust. Isp should not be changed.

 

[Nicholander]

Oh. So the point of this mod is defeated because it is completly unrealistic. Or, maybe the engine used on the Chapman probes has more isp, than the 300 that this has, I'll check and update this when I can.

 

[Urwumpe]

Oh. So the point of this mod is defeated because it is completly unrealistic. Or, maybe the engine used on the Chapman probes has more isp, than the 300 that this has, I'll check and update this when I can.

Well, remember: If the probe has more total dv, you have more margin for error you have.

 

[jedidia]

And if for example, a rocket with 1 engine would have exactly the same delta-v as a rocket with 9999 of the same engine?

Well, in theory... if the mass stayed the same. In reality, the one-engine design would have more, simply because it has less mass. But if you took the design with 9999 engines and only used one of them, then yes, it would have the same delta V as if you were running all of them.

ISP is directly related to, and very often used sinonimously for, exhaust velocity. Look at it this way: Two horses can pull more than one... that's thrust. But two horses can't run faster than one. That's exhaust velocity, which determines how much propellant you have to spend per m/s of dV.

 

[N_Molson]

If you are new to this stuff, I really recommend you to learn and understand [ame="http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation"]Tsiolkovsky rocket equation[/ame] posted by Urwumpe above :

Note that Isp (in seconds) can be also expressed as Ve (Velocity exhaust, in meters/second). Ve = Isp * 9.81...

It will really give you the essential basics in rocket propulsion. Begin by that, and keep more advanced stuff for later, else it's really easy to run into misconceptions, as said above.

 

[rseferino]

The specific impulse is a measure of engine efficiency. no rocket. An engine may have a specific impulse of 325 seconds, but a stage of 5 engines will not have 1625 seconds. Increase thrust. 1 engine can produce 100 tons of thrust but if the rocket weighs 400 tons not take off. 5 engines generate 500 tons, enough to take off. 1 engine might have the same delta-V to 999 engines but it would take 999 times longer no matter how fuel used.

On the other hand, I've flown this mission and (spoiler ) I'm editing my video on the Eris.

 

[Nicholander]

Well, that was disappointing. The Chapman Probe's engine has an isp of 336.9, just 36.9 higher than the original engine. And it turns out it only gets about 200 more m/s of delta-v, not very helpful. Is there any other (non-super-low-thrust) engine which would get, say at least 700 more m/s of dv? Keep in mind the original has only about 1.1 km/s of delta-v.

EDIT: Also, I'm familiar with the rocket equation, but I did not know that it was unaffected by the thrust and the number of engines.