GSJames

I am working on doing some CFD modeling of various Q-500 wings and I am ready to run the computer models and see if there is any difference in induced drag that the program can detect. But to do so, I need to input an angle of attack to the CFD program. I know that we don't have angle of attack data, but if someone has ever logged any data on an "Eagle Tree" system or something like that and can tell me the load factors achieved, and at what speed they were, since I know the approximate weight, I can work the math backward to get an approximate angle of attack. In order for the results to be of any value, each computer wing model must generate the same amount of total lift, which will occur at differing angles of attack for different wings because of the differing cambers.

Anyone got any data that they'd like to share?

Thanks!

HighPlains

My Feedback: (1)

The math is simple. What speed and turning radius do you want to use?

djlyon

Measured on average 22Gs at 150-155 mph in the turn. Calculated on the claimed 50ft radius turn 30Gs. Assumptions: Stick fixed turn that produces a spiral turn. Couldn't decide if the spiral was converging or diverging so picked the special case, a circle. Other oversimplified calculations I did said a 66012 based Q500 wing will produce the necessary force (120lbs) for a 4 lb plane well short of stall and for that force structural design is not a problem.

Denis

tIANci

Wow ... 22Gs ... I thought maybe its like 10Gs or so. This is an eye opener!

Super Splatter

every time my little pilot climbs out of his plane after a heat, he's always a little flatter than the he was before..

daven

My Feedback: (1)

Gary, I would run the math on a 110 foot radius around 2/3 and a 75 radius on turn 1 and you should be in the ballpark.

424 speeds at 120 mph average in the straights, the quickest planes at 130
428 speeds at 170 mph average in the straights, the quickest planes at 180

So maybe a range over those speeds to see how the numbers run. If someone has better or differing data to those numbers, please chime in. Based on data up here using radar and Eagle Tree.

I would also run numbers on differing percentages of Camber at the root, and washout at the tips.

The math is something I stopped using years ago, so would really appreciated your findings.

HighPlains

My Feedback: (1)

Dave, you can go around turns two and three with a 71 foot radius and be right on the pins. So 50 feet is maybe slightly tight, but I'm sure some bang the stops on their transmitters.

What must also be included in the speed used is the wind, and worse yet, wind gusts. At a NMPRA championship in Texas many years ago, the gusts were so bad that half a dozen wings folded in the turns.

So if you assume a 60 foot radius and 190 mph (170 + 20 mph tailwind), then you could expect to see a load of at least 40 g's.

DonStegall

My Feedback: (1)

I have a handy dandy calculator here [link=http://www.rcpro.org/rccalc/GForce.aspx]G-Force Calculator[/link]

And a pitch speed calculator as well ... [link=http://www.rcpro.org/rccalc/PitchSpeed.aspx]Pitch Speed Calculator[/link]

If my calculations are off, somebody let me know.

GSJames

Thanks guys, that'll get me in the ballpark. I have to run the numbers over a range of angles of attack so I just needed to know if pylon racers are pulling right up to CLmax or are at much lower values. I'm looking to see if there is an advantage in reduced induced drag over a small range of Cl's as opposed to a "point design" I have no RC racing experience to draw on since I've never raced RC before. (first race is next Sunday :-) ) Though I've been an AMA member for 45 years, mostly I've flown CL Combat and racing. Last year I started flying RC Combat and it was a lot of fun and since there is a very active Club racing series here in North Texas, I decided to give racing a try and see if it is fun too. Thanks for the help.

djlyon

I would use something about 3/4 the way up the curve. Here's why. When I looked at it a couple of years ago I found that just short of cl max for a 66012 (about 1.0) a quickie 66012 wing could produce 240 lbs of force so 120lbs of force at cl .5 seemed reasonable. The data I used was actual wind tunnel data produced when I was a toddler. I'm now 70. A couple of big buts though. I did not correct the Reynolds number for speed and cord and I used the sectional data uncorrected for the wing. I was just trying to establish in my mind if 20 to 30 G turns was reasonable structurally and aerodynamically. It is.

Highplains suggestion of a 71 ft radius turn fits with measured data I've seen and mentioned in my first post.

In 428 when we pull turns much tighter than that (50 ft) we start to see the HOOK coming out of turn 1. I suspect that's approaching CL max at about 150 to 160 mph.

But there is a lot of intuitive guessing in my reasoning, always dangerous.

Denis

HighPlains

My Feedback: (1)

A ballpark angle of attack in the turns is about 6 degrees on the wing.

GSJames

I just made up a quick spreadsheet to take a look and you are correct. A wing CL range of about 0.5 to 0.8 or so looks to be about right. I plan to do the drag polars over that range and compare them at those CL's. The data will be corrected for Rn. Thanks! Of course all this is actually moot, since the skill of the pilot appears to be a much bigger factor than anything else. (That was the point of the original Q-500 rules, if I remember correctly...)

HighPlains

My Feedback: (1)

The problem of pylon design is that you are in the high g phase of flight for only 30% of the race. However because of induced drag, that is also when the airplane is slowing down the most. That is why an oval course is preferred in the faster racing classes. In the slower classes, flying a good course and compensating for the wind helps.

As you noted, flying the course is the first order of business. More time is lost in sloppy take-offs, and blowing the turns at two and three than any super design can compensate for. Most people worry about turning tight at pylon one, and that really doesn't gain you all that much.

daven

My Feedback: (1)

Thanks Bob for correcting me, I was thinking of a turn of 110 feet to get around 2/3, which would be a a 55 foot radius, so 60 is probably closer to where most people really fly.

Thanks,

djlyon

It'll be interesting to see what you come up with.

A couple of years ago I pushed numbers around trying to come with the shortest path around the pylons using 50 ft radius turns. It was 2.69 miles using 3 turns not 2 or in other words pretty close to Highplains oval.

Glancing at various airfoils that would be suitable for a quickie it looks like CL and CD don't change much at 6 Degs above the zero lift line whether the foil is symmetrical or not . But that's at an RN of 3000000. It might be different at a more realistic 1000000 when CL max comes down and the angle of attack has to go up. But as I said that's just a glance.

Denis

HighPlains

My Feedback: (1)

Reynolds on a 10 inch chord at 170 is around 1.3 million at standard conditions.

The radius of the turn at pylons 2 & 3 depend where you start the turn. If you start the turn when the airplane is straight inline with the two pylons, then 50+ feet would be the minimum radius. However a shorter course distance can be achieved by starting the turn 50 feet earlier with a radius of 71+ feet. This shaves off 34 feet per lap, or 340 feet for the race. That alone knocks off about 1.3 seconds. However since the airplane is flying at a lower g loading with the bigger radius, it also loses less speed.

The other thing is to not make a true 180 degree turn at each pylon (or pair of pylons). Instead only do about 165 degrees of the turn at each end of the course (pulling the high g loading), and do the rest of the turn in the straight section of the course. Which is why it ends up looking like an oval. On the long course with a 170 mph plane, the airplane should remain in a bank of around 45 degrees and be trimmed to fly level at that bank angle. This give a turn radius of about 1800 feet and a g loading of about 1.5 g's. It also adds a couple of feet per lap to the distance, but that is of no concern compared to the increase in average aircraft speed. Why does the average speed increase? Because with the 165 degree, 15 degree course, you end up removing almost two high g turns from a ten lap race. Since the wingspans are limited, avoiding time in high g loadings improves the times on your score sheet.

The 2.69 mile distance is pretty much a perfectly flown course.

djlyon

Just have to do 30 on-tops

Clark L

How about something actually measured?

HighPlains

My Feedback: (1)

It kind of tells me that you are not yet on the course.

GSJames

Thank you all for your input. I now have the range of data that I need to do the analysis and determine whether or not the CFD code can detect any difference in induced drag due to a modified spanwise lift distribution.

GSJames

Here are some "pretty pictures". That's all they are at this point, until I integrate the pressure distributions, but what they show is the pressure distribution at zero degrees angle of attack over 3 common Quickie 500 wings. The S8064 (Viper) , the MH 17 (HoleShot) and the N66-012 (Seeker). I ran the matrix of wing variations but I'll have to hand plot the results unless I can find a piece of freeware that allows the plotting of arbitrary data, MSExcel doesn't draw nice looking graphs.

Clark L

Startup, needle, ten laps, land, recover.... I don't have the exact time for the ten laps, but I can assure you it was "on the course"

Actually, I was doing a bit of testing when the LS came out....

[8D]

Scorpion Racing

GSJames, just out of curiosity, could you plot a pressure distribution for a 64-012 like you did the other airfoils in post #21? I would be interested in seeing how it compares to the 66-012.

Cool stuff!

Thanks,

djlyon

And a 66-212

GSJames

Sure, here is the 2-d section data for Q-25 and 428 speeds (424 will be somewhere in the middle) I'll do the 3D stuff later when I have some time.