Drag Separation Question

Hi,

I am interested in determining if my 38mm minimum diameter rocket will suffer from drag separation. I have attached it for anyone interested to take a look at it. I have it set up as a two stage rocket and I am trying to determine the drag coefficient for each of them, and from knowing how the rocket will decelerate on an Aerotech J350, I'd like to see if drag separation might be an issue.

Thanks,
-Gene

P.S. My apologies in advance if I didn't set the model up correctly.

View attachment Genes_2_stg.rkt

glauria said:

I have it set up as a two stage rocket and I am trying to determine the drag coefficient for each of them...

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You don't have to do that, in Openrocket anyway. Just open component analysis/drag characteristics and add up the nose cone and upper BT for the top section Cd and the lower BT and fins for the lower. Upper is about 0.2, lower is about 0.4 if the surface is polished (usually optimistic.)

Max deceleration is about 22-23 m/s2 on the J350.

mikec said:

You don't have to do that, in Openrocket anyway. Just open component analysis/drag characteristics and add up the nose cone and upper BT for the top section Cd and the lower BT and fins for the lower. Upper is about 0.2, lower is about 0.4 if the surface is polished (usually optimistic.)

Max deceleration is about 22-23 m/s2 on the J350.

Click to expand...

Cd isn't as important as force/momentum after burnout. If the nosecone is very heavy, and the body is very light, the nosecone will not slow down nearly as quickly, even though it is pushing the most air out of the way (look up almost any V2 launched at ROC to see that in action). For most rockets it's not a problem, if you think it might be, just put some shear pins in and you'll be fine no matter what.

How many G's at burnout?

Apply this to your upper section.

If that figure, once obtained, is greater than the forces holding the upper and lower sections together, it will drag separate.

To prevent this, you may need shear pins. How many and what size... well... that depends on how much you're trying to hold (figure determined above) with a reasonable margin of error.

Keep in mind that adding these may necessitate a larger ejection charge to break them...


Later!

--Coop

See https://www.rocketryforum.com/showt...Nose-Cone-Drag-Separation&p=473406#post473406 for the equations.

Thanks guys,

Those were very helpful comments. Indeed, the momentum is the major factor here. I did a quick check using the equations posted by Mike and found that for the mass of each section, the forces are about equal for the J350 motor. The speed is just over Mach 1. I don't think shear pins will be required, but probably a good idea.

One thing about the equation that I noticed. If the determining factor whether drag separation would occur is if the separation force is positive or negative, does the deceleration even matter since it just is a multiplication of the expression in the parentheses?

glauria said:

One thing about the equation that I noticed. If the determining factor whether drag separation would occur is if the separation force is positive or negative, does the deceleration even matter since it just is a multiplication of the expression in the parentheses?

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If the quantity calculates to be negative then the magnitude of the decelleration doesn't matter. If the quantity is positive, the decelleration estimate will help you size your shear pins.

glauria said:

I did a quick check using the equations posted by Mike and found that for the mass of each section, the forces are about equal for the J350 motor.

Click to expand...

What is the mass of your upper and lower sections? It wasn't clear from the file you posted because you had a mass override. I find it hard to believe that you're anywhere close to needing shear pins on this unless you've really weighted the forward section down.

Hi Mike,

I'm learning how to use Openrocket. My original file that I sent had all sorts of errors because I had to override the weights because the default values were inconsistent. I believe my newer model is a more accurate representation.

I'll share what result I obtained:
Total weight of the rocket: 1254 g
Lower section weight: 767 g (Includes the weight of an Aerotech 38mm/480 case & closures, but no propellent.)
Rocket reaches about Mach 0.98 with the J350.
Cd upper at Mach 0.98: 0.15
Cd lower at Mach 0.98: 0.9
Upper/Lower ratio: 0.167
Maximum deceleration: ~ 88m/s^2
And the drag force is... (88/9.8)*(1.254/1.16 - 0.767) = 2.82 N

I divided by 9.8 because everything is given as weight, not mass. I think this is correct. If not, the force is about 10 stronger.

Here is the updated model file:


In any event, since the result is positive, it looks like drag separation will be an issue. If I got force correct by correctly converting weight to mass, the 3 N force doesn't seem too strong, and perhaps I don't need shear pins. If the force is indeed more like 30 N, then it looks like shear pins are necessary.

View attachment Genes_2_stg.rkt

Thank-you John, especially for providing the formula!

I was wondering though, it looks like both rocksim and openrocket gives the value of weight for mass. Is this correct? I think if one wants the mass, they have to divide the weight by 9.8, right? When I use your formula, I am dividing by 9.8 to achieve the right value of force. Would this be correct?

Thanks,
-Gene

Might want to include a second set of fins for the upper section ;-)

glauria said:

When I use your formula, I am dividing by 9.8 to achieve the right value of force. Would this be correct?

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grams and kg are units of mass. You don't have to divide by 9.8. You've got an acceleration times a mass giving kg-m/s2 -- the unit of force, aka the newton.

I don't see anything immediately wrong with your calculation. The drag ratio is pretty high at mach 1, partly due to base drag for the bottom section. But I don't think the max deceleration is 88 m/s2, I get more like 25.

Thanks Mike,

If got the 88 m/s^2 by directly reading the acceleration off of the plot in openrocket. If you run my latest uploaded model I think you'll get the same 88 m/s^2 value.

However, I might be getting confused in regards towards mass and weight. For example, a 38mm diameter 48" length of blue tube is listed to have a weight of 295g. If I select the same thing in open rocket it reports a mass of 305g. So to me, it looks like weight and mass are being used interchangeably which I think is incorrect. If I remember from high school physics, weight is a measure of force, but mass is not. So, what's the mass of the blue tube? 305g or 31.1g?

glauria said:

Thanks Mike,

If got the 88 m/s^2 by directly reading the acceleration off of the plot in openrocket. If you run my latest uploaded model I think you'll get the same 88 m/s^2 value.

However, I might be getting confused in regards towards mass and weight. For example, a 38mm diameter 48" length of blue tube is listed to have a weight of 295g. If I select the same thing in open rocket it reports a mass of 305g. So to me, it looks like weight and mass are being used interchangeably which I think is incorrect. If I remember from high school physics, weight is a measure of force, but mass is not. So, what's the mass of the blue tube? 305g or 31.1g?

Click to expand...

Yes common error for many who do not understand the mass versus weight. Even our food labeling is wrong.

The mass of the blue tube is 305g, the weight of the blue tube is 2.9 Newtons.

jderimig said:

Yes common error for many who do not understand the mass versus weight. Even our food labeling is wrong.

The mass of the blue tube is 305g, the weight of the blue tube is 2.9 Newtons.

Click to expand...

Thanks for setting this straight for me John,

Yes, this can get really confusing when people use weight and mass interchangeably. This is incorrect and can be quite confusing. It looks like the separation force that I calculated then is indeed about 30 N, which is quite significant and surprising.

glauria said:

If got the 88 m/s^2 by directly reading the acceleration off of the plot in openrocket. If you run my latest uploaded model I think you'll get the same 88 m/s^2 value.

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Nope. Still about -20 m/s2. Green curve, labels to right.

mikec said:

Nope. Still about -20 m/s2. Green curve, labels to right.

View attachment 172127

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Boy Mike, I must be doing something wrong with the simulation. Here's my graph: I'm just learning how to use open rocket, so I must have an incorrect parameter somewhere.

glauria said:

I'm just learning how to use open rocket, so I must have an incorrect parameter somewhere.

Click to expand...

Post your .ork file so we don't have format conversion to deal with.

The better way to do this is with Tools -> Component Analysis, which gives you the percentage of total drag from each exterior component.

CarVac said:

The better way to do this is with Tools -> Component Analysis...

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Yes. See post #2. But you still need the maximum deceleration to know the magnitude of the separation force.

Here it is,

View attachment Genes_dual_deploy.ork

glauria said:

Here it is,

Click to expand...

Your massive deceleration is coming from your chute deploying early. Set the parachute to deploy at apogee and you'll get the result I posted earlier, more or less.

This is a believable result: you're showing a little propensity for drag separation, but it could easily be addressed by friction-fitting the coupler. Some people would use shear pins anyway, but I wouldn't for this size rocket.

mikec said:

Your massive deceleration is coming from your chute deploying early. Set the parachute to deploy at apogee and you'll get the result I posted earlier, more or less.

This is a believable result: you're showing a little propensity for drag separation, but it could easily be addressed by friction-fitting the coupler. Some people would use shear pins anyway, but I wouldn't for this size rocket.

Click to expand...

That's good news. It would be good to avoid to have to use them because I assume that the ejection charge would have to be a lot stronger to shear them as compared to using a tighter friction fit of the coupler.

Thanks to everyone for their very helpful input. I learned some useful physics here.

glauria said:

it would be good to avoid to have to use them because I assume that the ejection charge would have to be a lot stronger to shear them...

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Not really. The problem is that for most materials you'd have to reinforce the tube somehow to keep the shear pins from ripping up the tube. For Blue Tube, I've used styrene rod, which is a little easier to shear but less predictable than nylon.

See the BP calculator at https://www.info-central.org/?article=303 to see the effects of shear pins, though as I said I wouldn't use them in this particular case.

Very good Mike,

This whole exercise was to prove that I didn't need them any way. I'm just going to go with the friction fit. I'll use a tight fit and do some ground testing first to make sure everything deploys properly. I'll start with the smaller motors first and then work my way up to the larger ones. I am using a Missile Works RRC3 altimeter, so I'll be able to download the launch data to see how well it agrees with the simulations.