30th January 2013, 09:20 PM
Calculating Nose Cone Drag Separation?
I am trying to wrap my head around drag separation, particularly as it applies to nose cones.
Definition: Drag Separation - The unintended (or in a few cases, intended) separation of bodies while in flight that occur after motor burnout, due to the inertia of one section having more force to separate the bodies than another section has force to keep the bodies together.
As I understand it, the heavier a nose cone is, the more prone it will be to drag separate with high thrust motors.
Can someone show me the way to calculate out these forces? I did a cursory search and didn't find much.
With this knowledge, it would be easier to make a determination on when to think about the use of shear pins.
30th January 2013, 10:46 PM
ALWAYS use shear pins. The simple solution. You could find the aerodynamic drag on all the components, nosecone, rocket with fins and then force exerted on the nosecone shoulder by the airframe and figure out if you need them also. Or you could just always use them.
30th January 2013, 11:03 PM
"Definition: Drag Separation - The unintended (or in a few cases, intended) separation of bodies while in flight that occur after motor burnout, due to the inertia of one section having more force to separate the bodies than another section has force to keep the bodies together"
It's not inertia that causes drag separation, it's ... drag.
After burnout, drag and gravity are working to slow the rocket. Gravity affects all parts of the rocket equally, but drag doesn't. The nose of a rocket is more streamlined than the fins. So, the drag force on the back end of the rocket is greater than on the front. It's as if the air is trying to pull the rocket apart. If the friction (or shear pins) holding the parts together is less than the difference in drag, the parts can be pulled apart.
If you can calculate the drag on the nose section and the drag on the fin section then compare the difference to the force required to separate the sections of the rockets, then you can determine if drag separation will occur.
30th January 2013, 11:17 PM
Sometimes, just using shear pins is not enough.
If you trust all of its calculations sufficiently, you can use OpenRocket's component analysis to compute what percent of the drag on the rocket comes from forward of the break. Then, figure out what percentage of the mass is forward of the break.
The drag separation force (disregarding internal pressure) will be:
(deceleration of rocket(t)*mass of front) - (% of drag from front(velocity(t))*total drag force(t))
For a given deceleration of the rocket, the total force needed to keep the front part attached will come from two sources: the friction at the break and the drag on the front. Subtracting the drag at the front will yield the amount of friction at the break.
Note that for supersonic rockets, the distribution of drag force will vary greatly with Mach number, so be sure to record a few values: the highest Mach numbers will be when the magnitude of the drag force is highest, but it might not be the point when the drag separation force peaks.
I wish you could set a separation point in OpenRocket, so that it could output drag from forward and aft of the separation point in graphs. Then I could set up a formula to plot drag separation force versus time.
Temporarily potential impulse transformed into no-longer-potential-anymore impulse since 2013-01-01: A lot.
30th January 2013, 11:18 PM
Not completely true. There was a nice derivation of this on RP before it went down, but in a nutshell you can calculate the separation force as follows:
Originally Posted by jadebox
Fsep = a [ M / (1+R) - M1 ]
a = max deceleration
M = total mass of rocket (mass NOT weight)
M1 = mass of lower section
R = drag ratio
If Fsep is +, then there is a drag separation force, if - then there isn't.
Thus we can define the Derimiggio-Newton 1st law of Drag Separation:
If M/(1+R) < M1, then the sections won't drag separate period.
I've computed this for typical rockets and it convinced me that in many cases you could have a totally loose coupler and the sections just couldn't drag-separate. This is one reason why I have never bothered to use shear pins at the booster-payload joint.
Of course, at the nose cone the larger forces are not from drag but from shock cord snapback, so shear pins there are usually needed.
30th January 2013, 11:36 PM
I realized that what I said wasn't exactly right as soon as I clicked submit. Gravity accelerates all the parts of the rocket equally, but the force is proportional to the mass of each object (otherwise, we'd all weight the same and I wouldn't need to go on a diet).
Originally Posted by mikec
But, I was right in the sense that, absent drag, there would be no "drag separation." In a vacuum, for example, the parts wouldn't separate after burnout regardless of the difference in mass.
Last edited by jadebox; 1st February 2013 at 10:59 PM.
30th January 2013, 11:47 PM
Ok, this is the kind of stuff I was looking for: the math.
Originally Posted by mikec
Is [a] measured in m/s/s?
Is [Mx] = W/g? (where W is in kg and g 9.81 m/s/s)
Is the [R] the Cd or is it something else?
Do you have a handy sample example?
31st January 2013, 12:15 AM
First off, to give credit where it's due, this derivation is by John Derimiggio, not me. He had it written up well (with ASCII free-body diagrams)
Originally Posted by GregGleason
but that site is no longer available (it was a club website, not RP, IIRC).
Units are in any consistent format, like all MKS. Mass is kg, acceleration in m/s2, Cd is dimensionless.
The drag ratio is the ratio between the Cds of the upper and lower sections by themselves; you can look at the Cd breakdown in OpenRocket to get a handle on this.
An example: for my L3 rocket, upper airframe drag is 0.12 and lower is 0.38. M1 is mass of
lower section, total mass of rocket at burnout is 9 kg, let's say,
and upper mass is 1 kg. So R is 0.32, M is 9 kg, M1 is 8 kg, and max
drag decel on an M1315 is 50 m/s2. So fsep is 50*(9/(1+0.32) - 8) = -59.1
and there is no drag separation.
Last edited by mikec; 31st January 2013 at 12:33 AM.
Reason: fixed mass values
1st February 2013, 10:55 PM
I'll see if I can plug in the values and come up with something.
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