# Talk:Decoupler and separator

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## Unit of ejection “force”

Hi, I wanted to verify that the unit of ejection “force” is really kNs. So I did a little test and shot a little probe consisting of: RC-001S Remote Guidance Unit, Z-1k Rechargable Battery Bank, TR-18D Stack Separator, Z-1k Rechargable Battery Bank, RC-001S Remote Guidance Unit

So each side of the separator has the same mass (150 kg) and I then activated the stack separator I got a speed relative to the separator of 16.7 m/s and 33.4 m/s relative to each side. The momentum is then p=m·v=150 kg·16.7 m/s=2505 kg m/s (or Ns) which is only one 10th of the ejection “force” given by the part.cfg of 250. Now did I do a mistake in my calculations, or is the unit really given in 10 Ns or is it really a force and not a momentum (although that doesn't really make sense). — xZise [talk] 07:20, 5 October 2013 (CDT)

I did additional measurements with all separators and I always get the given value in 10 Ns: For the TR-2C Stack Separator I have 1 m/s (→ 150 kg·1 m/s = 150 Ns, specified is 15) and the TR-XL Stack Separator got wopping 40 m/s (destroyed half my setup) (→ 150 kg·40 m/s = 6000 Ns, specified is 600). I'll test how decouplers work. Although I guess the number specified is only for one side. — xZise [talk] 13:41, 5 October 2013 (CDT)

Now with more additional measurements including decouplers, I don't see any pattern. My data is available on that spread sheet. For separators it appears that those 10 Ns unit might work, but I'm not sure anymore if this is even the correct value. When using decouplers there are two parts and between both was a impulse of “250” causing them to float away. The momentum of the impulse should then be the product of the velocity between both and then mass of both. But it appears that as soon as one side is heavier than the other it doesn't work. I also uploaded images from my first setup, including the two measurements: http://imgur.com/0mS5eMv,kyZpH4s,IFsewuc . — xZise [talk] 05:43, 6 October 2013 (CDT)

Your physics is wrong in your last comment. Momentum is conserved, so the total momentum stays zero in the rest frame before the separation, the two parts will have opposite momenta. So the momentum (impulse) delivered by the separator is

$I_{sep}=p_{1}=m_{1}v_{1}=-p_{2}=-m_{2}v_{2}$ .

The data in your spreadsheet are consistent with the delivered momentum being constant for the separators with units of 10 Ns. dvarjas 00:03, 16 September 2014 (PDT)

And the velocity is calculated relative the frame before the separation? So v1 != -v2 (unless they mass the same). I've updated the spread sheet and calculated the separate velocities by using the following formula (derived from yours): $v_{1}=-{\frac {m_{2}v}{m_{1}+m_{2}}}$ . For that I really get about a factor of 10. — xZise [talk] 08:21, 16 September 2014 (CDT)
Where $v$ is the relative velocity after the separation. Yes, I totally agree. dvarjas 08:18, 16 September 2014 (PDT)
So a decoupler is like a separator which is 'virtual' between the decoupler and whatever is connected to the decoupler. And both sides get pushed by the 'virtual separator' with a momentum of 10 Ns·'ejection force'. — xZise [talk] 05:44, 17 September 2014 (CDT)