As part of my L3 documentation, I'm trying to show a lot of the mathematical proofs behind why things are done. As for fins, we all know well about stability calibers and CG+CP. However, I don't see as much about how high a fin has to be to be effective. I've read the rules of thumb that they should have a one body diameter semispan to make sure they reach beyond the boundary layer of the laminar air flow. That makes sure they reach into the free stream of the air and can function to keep the rocket pointed in the right direction.

However, I wanted to calculate how wide that boundary layer really is.

The calculation for boundary layer thickness is:

ẟ / x = 5 / (√R)
ẟ = boundary layer height
x = length from nosecone tip to fin
R = Reynold’s number

Fortunately, Open Rocket can calculate the Reynold’s number:

At launch, it appears that the Reynold's number is 1,000,000.

x = 355 cm
R = 1,000,000

ẟ / 355 cm = 5 / (√1,000,000)
ẟ = 355 cm * (5 / (1,000)
ẟ = 1.8 cm

So that means the boundary layer is only 1.8 cm.

I did some reading and see that aerodynamic rockets do have very high Reynold's numbers of 10^{5} to 10^{8} range, so I'm in the right range.

Mark NAR#: 100890 / TRA#: 16580
L1 - Binder Design Excel Plus 38 - CTI I242 2,540'
L2 - Binder Design Excel Plus 38 - CTI J330 3,672'
L3 - Scratch built 8" PAC-3 MSE Patriot - Loki M1969 6,338'

I am curious, what is the diameter of the rocket? read something a while back where some amateur rocket groups(circa 1960ish) were using 1/2 diameter as a minimum semi-span guide.
Rex

I am curious, what is the diameter of the rocket? read something a while back where some amateur rocket groups(circa 1960ish) were using 1/2 diameter as a minimum semi-span guide.
Rex

I could be remembering incorrectly. It might be semispan equaling the radius. This is for my L3 rocket with an 8" diameter.

Mark NAR#: 100890 / TRA#: 16580
L1 - Binder Design Excel Plus 38 - CTI I242 2,540'
L2 - Binder Design Excel Plus 38 - CTI J330 3,672'
L3 - Scratch built 8" PAC-3 MSE Patriot - Loki M1969 6,338'

okay thanks. was curious to see how the boundary compared to diameter. it might have been Brinley's book that I saw that in...and they would have been fudging things to ensure safety/stability.
Rex

Laminar flow is Reynolds number dependent, low Reynolds numbers < 1e4 indicate laminar flow, turbulent flow is somewhere above 1e6. In between is the transition region.

Reynolds number is rho*V*L/viscosity. You want to choose a length L that captures the flow you are interested in, in the case of a fin I would probably choose mean chord length. For the whole rocket I might choose the length of the rocket or the diameter.

What you are calculating is the boundary layer thickness. The Reynolds number will tell you if the boundary layer is laminar or turbulent.

Last edited by djkingsley; 26th December 2016 at 10:38 PM.

Isn't Reynold's number tightly coupled to velocity? In other words, "it depends on how fast I'm going"?

This is what I would assume. Boundary layer issues due to short fins are an issue at high Mach, this is where fins can loose their effectiveness. Launch issues are weather cocking and fin stall etc.

QRS: 124 AMRS: 32 L2 RSO
Highest: 13,647 feet. Fastest: Mach 1.55.
Largest Motor: CTI 1115J530 IM
Current Projects: HPR X Wing

Isn't Reynold's number tightly coupled to velocity? In other words, "it depends on how fast I'm going"?

Reynolds number relates a reference length on the airframe to the fluid flow around the fin. The use of the Reynolds number allows calculations of fluid dynamics to be measured and scaled effectively. https://en.wikipedia.org/wiki/Reynolds_number

Reynolds number depends on the fluid viscosity. The reference length is normally chosen as the root chord length for wings on aircraft, so I would assume fin root chord for rockets.

The higher the Reynolds number the faster a wing would need to travel to have the fluid flow across it transition from laminar flow to turbulent.

TRA 13430, Level 3

"Everybody's simulation model is guilty until proven innocent" (Thomas H. Lawrence 1994)

Here's the turbulent boundary layer approximation if you want to plug that into another one of your spreadsheets!

I'm also interested in the analytical effectiveness of fins on rockets like the Ventris or similar that have an expanded forward section.

The reducing-transition screws with the original boundary layer, but how much? Does it rejoin and match the forward thickness or do something funky?
If you try to go fast, then there's this weird balance between short for anti-flutter, but long enought to get effective area into the free stream airflow.

"I'm at least 70% confident about whatever I say (90% of the time)"- college me

I'm also interested in the analytical effectiveness of fins on rockets like the Ventris or similar that have an expanded forward section.

The reducing-transition screws with the original boundary layer, but how much? Does it rejoin and match the forward thickness or do something funky?
If you try to go fast, then there's this weird balance between short for anti-flutter, but long enough to get effective area into the free stream airflow.

The flow layer will re-attach and get to its original thickness if it separates, but it's a question of how far down the rocket that happens. With a relatively smooth transition, the airflow will stay attached all the way down. If the Ventris was a boat, I would expect that the transition is smooth enough for the flow to stay attached to the skin all the way down. I also recognize that this is no help to you whatsoever. One way to get useful information may be to look at Re vs. drag coefficient plots for various gradual transitions. If you get big jumps in the drag coefficient in the laminar-turbulent transition area, you're probably seeing flow separation. The plot for a sphere is a classic version of that--you see a noticeable drop in drag coefficient when the flow becomes turbulent and the flow stays attached to the sphere for longer.

NAR L1 "Cheeto Dust", scratch 54mm, H54R (before it became a G54), Mansfield, WA
L2 "Arc Light", Madcow 2.6" Arcas, J285CL, Mansfield, WA, recovery by snowshoe

With a relatively smooth transition, the airflow will stay attached all the way down.

Thinking about diffusers (things, like ducts, that take an airflow and expand it to slow it down), IIRC to keep the flow laminar the divergence in the walls should not be above about 7-8 degrees or turbulence results (due to adverse pressure gradient I think). That's for a rectangular duct, but it would be a good starting point to think about round airframes. I suspect at any significant speeds the taper would need to be really gentle to keep the boundary layer stuck.

TRA 13430, Level 3

"Everybody's simulation model is guilty until proven innocent" (Thomas H. Lawrence 1994)

And thanks to that statement, I'm seriously considering slapping some tiny registry numbers on a fin when it comes time to paint mine. That comparison gave me a good laugh, so I thank you. Always thought the forward section had a bullet-like feel about it.

Here's a question though: Re has to have a uniform L-characteristic (Lc) in order to compare values for calculation, so for a geometry-change case like this, what's a decent to take as Lc? Choosing Lc has always been the annoying part for me when setting up stuff like this.
[And what does something like OR use when it develops its Re data? (Biggest diameter, major airframe length, etc...)]

Originally Posted by OverTheTop

Thinking about diffusers (things, like ducts, that take an airflow and expand it to slow it down), IIRC to keep the flow laminar the divergence in the walls should not be above about 7-8 degrees or turbulence results (due to adverse pressure gradient I think). That's for a rectangular duct, but it would be a good starting point to think about round airframes. I suspect at any significant speeds the taper would need to be really gentle to keep the boundary layer stuck.

That range sounds familiar to me as well. I think I linked to something similar in the Fin-airfoil thread. I checked the transition, and its ~5.7 degrees.

No. This calculates how far from the rocket body the laminar air flow extends. The fins must extend beyond that height to have an effect.

Is this flight intended to go beyond mach? I hadn't read on the effects of a shock wave on the boundary layer until reading through this thread.
Once again fluid shenanigans abound. The figure below demonstrates how the boundary layer rises, then decreases as flow reattaches to the surface.
I really need to revisit the end chapters of my Fluids textbook that weren't touched during undergrad.

One other thing to keep in mind is that you are almost exclusively* dealing with turbulent boundary layers at the speeds we're concerned about. The question is how thick the boundary layer is and (a corollary) how well it is attached to the surface of the rocket.

Here's a question though: Re has to have a uniform L-characteristic (Lc) in order to compare values for calculation, so for a geometry-change case like this, what's a decent to take as Lc? Choosing Lc has always been the annoying part for me when setting up stuff like this.
[And what does something like OR use when it develops its Re data? (Biggest diameter, major airframe length, etc...)]

I don't know how OR handles this, but you would typically be looking at diameters and/or fin chord for your characteristic length. These are usually all within a factor of three or so of each other. Since the transition between laminar and turbulent happens over a factor of 100 (10^4 < Re < 10^6), a factor of 3 is a relatively small error.

* This is a guesstimate on my part. I'm happy to be proven wrong, though.

NAR L1 "Cheeto Dust", scratch 54mm, H54R (before it became a G54), Mansfield, WA
L2 "Arc Light", Madcow 2.6" Arcas, J285CL, Mansfield, WA, recovery by snowshoe

Found this picture from NASA that shows turbulent flow down the length of a rocket, you can see how it will interfere with the airflow over the fins.

I have extracted data from OR for a flight I'm planning and have been playing around with a spread sheet but not getting the results I expected...

Unfortunately I am having trouble uploading to TRF so cannot attach anything at the moment.

The formula I am using is as follows (note that E10 contains the Reynolds number at the LE location and 623.5 is the length of the rocket at the
LE).
=IF(E10>5*10^5,(0.37/(E10^(1/5)))*623.5,0)

An example of the data is as follows, you'll notice that as the rocket speeds up the size of the boundary layer decreases
Also if I compare LE to TE I can see that the TE boundary layer is smaller, where is above picture shows that the boundary layer increases in size towards the end of the rocket. Anyone know what I am doing wrong?

Time

Velocity (Km/h)

Reynolds number

Turbulent Flow

0.068088

47.871

556513.8843

16.36601637

0.074856

54.1

627386.132

15.9783231

0.080845

59.633

690468.2783

15.67506808

0.086278

64.683

748111.6235

15.42570073

0.091287

69.364

801590.7347

15.21414826

QRS: 124 AMRS: 32 L2 RSO
Highest: 13,647 feet. Fastest: Mach 1.55.
Largest Motor: CTI 1115J530 IM
Current Projects: HPR X Wing

First up is the Reynolds number that are calculated based on the location of the leading and trailing edges.

For reference the velocity

And finally the the calculated turbulent flow

As previously mentioned it does not make sense to me, both the reducing size of the turbulent boundary layer as speed increases and also reducing along the length of the rocket.

QRS: 124 AMRS: 32 L2 RSO
Highest: 13,647 feet. Fastest: Mach 1.55.
Largest Motor: CTI 1115J530 IM
Current Projects: HPR X Wing

If you're using the turbulent boundary layer approximation, thickness is inversely proportional to a root of Re (which increases with speed). That decrease with speed doesn't sound too fishy. What may be interesting is a plot of the layer height across the length of the rocket at a handful of velocities. That sounds like a great Excel or MatLab excercise.

If you're using the turbulent boundary layer approximation, thickness is inversely proportional to a root of Re (which increases with speed). That decrease with speed doesn't sound too fishy. What may be interesting is a plot of the layer height across the length of the rocket at a handful of velocities. That sounds like a great Excel or MatLab excercise.

The Turbulent flow graph shows the boundary layer at the trailing edge is consistently about .5mm smaller than the boundary layer at the leading edge which is opposite of what I would expect. Will try graphing this at a few speeds to get a full picture.

QRS: 124 AMRS: 32 L2 RSO
Highest: 13,647 feet. Fastest: Mach 1.55.
Largest Motor: CTI 1115J530 IM
Current Projects: HPR X Wing

There is so much counter-intuitive in fluid dynamics. I have been reading a textbook for about six months on it. Half way through currently. My understanding is increasing but I am still not "in the groove" and thinking like a fluid dynamacist.

TRA 13430, Level 3

"Everybody's simulation model is guilty until proven innocent" (Thomas H. Lawrence 1994)

There is so much counter-intuitive in fluid dynamics. I have been reading a textbook for about six months on it. Half way through currently. My understanding is increasing but I am still not "in the groove" and thinking like a fluid dynamacist.

Join the club
Heck, I aced fluid mechanics when I took it 3 years ago and that just allowed me better define How Much I didn't know.
Thats why I love these technical inquiry threads. It gives me impetus to look back at things I don't use in my day job and apply it to something friggin awesome.

Found a mistake in my trialing edge calculations that was causing confusion. Will post an update graphs shortly and hopefully my spreadsheet. I'm a little shocked at the percentage of the fin that is within the turbulent boundary layer.

QRS: 124 AMRS: 32 L2 RSO
Highest: 13,647 feet. Fastest: Mach 1.55.
Largest Motor: CTI 1115J530 IM
Current Projects: HPR X Wing

I don't want to derail the thread but aren't fins less and less important as the ratio of engine size to weight increases.

For example:

Saturn V, tiny fins

Falcon 9, no fins

Chinese Bottle Rocket, no fins and in some cases of a bottle rocket I think the CP is actually ahead of the CG.

I think the engine size to weight ratio correlates to increase in speed, so again aren't fins less important the faster you are going? Not sure how that correlates to stability at apogee??

Maybe I'm all confused but very interested in this thread because I have two Space X Falcon 9 models that include the clear fins. I want to fly one without fins. I was trying to understand if this is possible with a large enough motor.

Bottom line I think we are both chasing the same rabbit or maybe I fell down the wrong rabbit hole.

I don't want to derail the thread but aren't fins less and less important as the ratio of engine size to weight increases.

For example:

Saturn V, tiny fins

Falcon 9, no fins

Chinese Bottle Rocket, no fins and in some cases of a bottle rocket I think the CP is actually ahead of the CG.

I think the engine size to weight ratio correlates to increase in speed, so again aren't fins less important the faster you are going? Not sure how that correlates to stability at apogee??

Maybe I'm all confused but very interested in this thread because I have two Space X Falcon 9 models that include the clear fins. I want to fly one without fins. I was trying to understand if this is possible with a large enough motor.

Bottom line I think we are both chasing the same rabbit or maybe I fell down the wrong rabbit hole.

Nope, the two bigguns do thrust vectoring. Bottle rockets have one huge stick-shaped fin that pulls the CP way back.

The fin doesn't care if the airflow is laminar or turbulent. The lift will be the same although the drag will not.

Also, the distance used in the Reynolds number calculation isn't necessarily the length of the rocket. Since the fins are only important at non-zero angles of attack you need to use the distance travelled at the angle of attack of interest.

At non-zero angles of attack you also get body-fin interference effects as the air flows off the body onto the fins. Lots of research on that.

The fin doesn't care if the airflow is laminar or turbulent. The lift will be the same although the drag will not.

Also, the distance used in the Reynolds number calculation isn't necessarily the length of the rocket. Since the fins are only important at non-zero angles of attack you need to use the distance travelled at the angle of attack of interest.

At non-zero angles of attack you also get body-fin interference effects as the air flows off the body onto the fins. Lots of research on that.

Since lift and drag are measured relative to the direction of airflow and with a non zero aoa the cp of the fin is not inline with the cg of the rocket relative to the direction of airflow an increase in drag will to a small degree contribute to an increase in the restoring toque on the rocket. Of course at any reasonable aoa this may be a negligible difference.