Delta V of decouplers?

[ToukieToucan]

I am planning a relay sat system and would like to know the Dv of decouplers, does anyone know how much Dv they or how to calculate it?

[THX1138]

Doesn't it depend on how much mass they're pushing? It's not much, anyway; maybe even negligible?

[Mastikator]

In my experience a single capsule with parachute and heatshield can gain about 1m/s from a decoupler of equal size.

So basically nothing. Just use rockets, tiny rockets work fine.

[tomf]

If you are worried about your perfect orbit being spoiled by decouplers you could turn the ejection force down to 0 on the decoupler.

[ToukieToucan]

14 minutes ago, tomf said:

If you are worried about your perfect orbit being spoiled by decouplers you could turn the ejection force down to 0 on the decoupler.

but then I can't turn the main craft to preform another burn.

[OhioBob]

We really can't calculate it.  The decoupler specs gives the ejection force, but what we really need to know is the impulse, i.e. force and duration.

[SchweinAero]

5 minutes ago, OhioBob said:

We really can't calculate it.  The decoupler specs gives the ejection force, but what we really need to know is the impulse, i.e. force and duration.

So I just conducted an experiment on this.

The most powerful decoupling part is the TR-XL Stack Separator with an ejection force of 600. When detaching a 1,000 kg mass from both sides simultaneously, it gives both masses a velocity change of 6.0 m/s. With one side against the ground, it likewise launches one ton at 6.0 m/s.

The Rockomax Brand Decoupler, which is marked as having an ejection force of 250, only detaches one side, and the velocity change of a similar mass in a similar configuration is 2.5 m/s relative to the decoupler.

I think it's safe to say that the "ejection force" number specifies a change in velocity imparted on a 10-kilogram mass per side detached.

 

[OhioBob]

4 minutes ago, SchweinAero said:

So I just conducted an experiment on this.

The most powerful decoupling part is the TR-XL Stack Separator with an ejection force of 600. When detaching a 1,000 kg mass from both sides simultaneously, it gives both masses a velocity change of 6.0 m/s. With one side against the ground, it likewise launches one ton at 6.0 m/s.

The Rockomax Brand Decoupler, which is marked as having an ejection force of 250, only detaches one side, and the velocity change of a similar mass in a similar configuration is 2.5 m/s relative to the decoupler.

I think it's safe to say that the "ejection force" number specifies a change in velocity imparted on a 10-kilogram mass per side detached.

 

Good experiment.

So it looks like each unit of "ejection force" is actually 10 Ns of impulse imparted to each half.  That is,

Δv = Impulse / mass = (600 * 10) / 1000 = 6 m/s

Have you tried the same experiment with unequal masses, say 1000 kg on one side and 2000 kg on the other?  I would think in that case the change in velocity of the 1000 kg mass would be 6 m/s and the change in velocity of the 2000 kg mass would be 3 m/s.

[ToukieToucan]

1 minute ago, OhioBob said:

Good experiment.

So it looks like each unit of "ejection force" is actually 10 Ns of impulse imparted to each half.  That is,

Δv = Impulse / mass = (600 * 10) / 1000 = 6 m/s

Have you tried the same experiment with unequal masses, say 1000 kg on one side and 2000 kg on the other?  I would think in that case the change in velocity of the 1000 kg mass would be 6 m/s and the change in velocity of the 2000 kg mass would be 3 m/s.

And doesn't the base part from which the 1000 kg object gets ejected get a kind of recoil of the decoupler?

[Sharpy]

delta v is a function of impulse and mass. We have the impulse of the decouplers, knowing the mass we can easily find the delta-V

Δv = I/m

The last column of the Decoupler page table contains their impulse in Kilogram Force * Seconds, so divide the value by mass of the ejected object and you get the change of velocity - assuming the decoupler remains stationary (bound to the ground)

Now if the decoupler and whatever it's still attached to is ejected backwards, this is a bit harder. From conservation of momentum, you must solve:

m1 v1 = m2 v2

and

(v1+v2)(m1+m2) = I

Solve this and you're getting the velocity:

v1 = I / (1+m1/m2)(m1+m2)

v2 = I / (1+m2/m1)(m1+m2)

That way you're getting delta-V of both parts.

Edited August 6, 2016 by Sharpy

[SchweinAero]

6 minutes ago, OhioBob said:

Have you tried the same experiment with unequal masses, say 1000 kg on one side and 2000 kg on the other?  I would think in that case the change in velocity of the 1000 kg mass would be 6 m/s and the change in velocity of the 2000 kg mass would be 3 m/s.

Just did, and that's exactly what happens. 6 m/s on one side and 3 m/s on the other. I don't know if recoil works as expected, but based on probe launchers that use multiple decouplers to accelerate themselves, that seems to be the case.

Thanks for that math, @Sharpy. Your physics are solid.

[Red Iron Crown]

53 minutes ago, OhioBob said:

We really can't calculate it.  The decoupler specs gives the ejection force, but what we really need to know is the impulse, i.e. force and duration.

The wiki helpfully lists their ejection impulse, example in the sidebar.

[Corona688]

1 hour ago, ToukieToucan said:

I am planning a relay sat system and would like to know the Dv of decouplers, does anyone know how much Dv they or how to calculate it?

Zero, if you turn their ejection force down to nothing in the VAB.

Edited August 5, 2016 by Corona688

[OhioBob]

1 hour ago, Sharpy said:

The last column of the Decoupler page table contains their impulse in Kilogram Force * Seconds

 

45 minutes ago, Red Iron Crown said:

The wiki helpfully lists their ejection impulse, example in the sidebar.

Thanks, guys.  I was just looking at the specs in the VAB, didn't realize the Wiki listed the impulse.

[Sharpy]

Just don't confuse Impulse with Specific Impulse. One is the total amount of accelerating force exerted, the other is a property of engine.

[John FX]

IIRC decouplers used to be the fastest way to Mun

Ah, here it is.

 

[Tweeker]

1 hour ago, Corona688 said:

Zero, if you turn their ejection force down to nothing in the VAB.

This is the way to go, don't over think this.

[Sharpy]

2 minutes ago, John FX said:

IIRC decouplers used to be the fastest way to Mun

 

No longer the case, new aero makes it impossible.

 

[John FX]

1 minute ago, Sharpy said:

No longer the case, new aero makes it impossible.

Of course.

That`s why I said it used to be, because it isn`t any more.

[Tatonf]

About a year ago I designed a challenge that got no one interested in where I calculated the Isp of the decouplers. However it's a bit hard to read because it was made on the ancient forum, now the editing is destroyed.

I'll repost it here :

TT-38K Radial Decoupler : 11.4

Hydraulic Detachment Manifold : 2.9

TT-70 Radial Decoupler : 6.5

Structural Pylon : 2.3

TR-2V Stack Decoupler : 1.1

R-2C Stack Separato : 0.8

TR-18A Stack Decoupler : 6.3

TR-18D Stack Separator : 4.6

Rockomax Brand Decoupler : 1.6

TR-XL Stack Separator : 3.6

TR-38-D Stack Separator : 0.5

So yeah the TT-38K Radial Decoupler will probably give you the best delta-V. Check out my challenge by the way !

Edited August 5, 2016 by Tatonf

[Snark]

Moving to Gameplay Questions.

[OhioBob]

On ‎8‎/‎5‎/‎2016 at 2:28 PM, Sharpy said:

and

(v1+v2)(m1+m2) = I

Solve this and you're getting the velocity:

v1 = I / (1+m1/m2)(m1+m2)

v2 = I / (1+m2/m1)(m1+m2)

That way you're getting delta-V of both parts.

I don't think this is right.  I believe it should be,

I = Δv₂ m₂ - Δv₁ m₁

Δv₁ = – (I / 2) (1 + m₂ / m₁) / (m₁ + m₂)

Δv₂ = (I / 2) (1 + m₁ / m₂) / (m₁ + m₂)

Let's work through an example from @SchweinAero's experiment.  We have,

m₁ = 2000 kg
Δv₁ = -3 m/s
m₂ = 1000 kg
Δv₂ = 6 m/s

Plugging these numbers into the equations we get,

I = (6)(1000) - (-3)(2000) = 12000 Ns

Δv₁ = – (12000 / 2) (1 + 1000 / 2000) / (2000 + 1000) = -3 m/s

Δv₂ = (12000 / 2) (1 + 2000 / 1000) / (2000 + 1000) = 6 m/s

(Earlier we talked about the impulse equaling 6000 Ns, but that was 6000 applied to each part for a total of 12000.)

 

Edited August 6, 2016 by OhioBob
made a slight rearrangement of the equations

[ToukieToucan]

19 minutes ago, OhioBob said:

I don't think this is right.  I believe it should be,

I = Δv₂ m₂ - Δv₁ m₁

Δv₁ = – I (1 + m₂ / m₁) / (2 (m₁ + m₂))

Δv₂ = I (1 + m₁ / m₂) / (2 (m₁ + m₂))

Let's work through an example from @SchweinAero's experiment.  We have,

m₁ = 2000 kg
Δv₁ = -3 m/s
m₂ = 1000 kg
Δv₂ = 6 m/s

Plugging these numbers into the equations we get,

I = (6)(1000) - (-3)(2000) = 12000 Ns

Δv₁ = – 12000 (1 + 1000 / 2000) / (2 (2000 + 1000)) = -3 m/s

Δv₂ = 12000 (1 + 2000 / 1000) / (2 (2000 + 1000)) = 6 m/s

(Earlier we talked about the impulse equaling 6000 Ns, but that was 6000 applied to each part for a total of 12000.)

 

I = Δv₂ m₂ - Δv₁ m₁

Sorry for being newb but you already know I, right?

[Sharpy]

3 minutes ago, ToukieToucan said:

I = Δv₂ m₂ - Δv₁ m₁

Sorry for being newb but you already know I, right?

Yes, you just use that one equation to pull both v1 and v2 to substitute in the other.

[OhioBob]

1 hour ago, ToukieToucan said:

I = Δv₂ m₂ - Δv₁ m₁

Sorry for being newb but you already know I, right?

Yeah, what @Sharpy said.  I just plugged the numbers into all three equations to show the internal consistency.  You are correct, we would know the impulse and use it to compute Δv₁ and  Δv₂.