scaling plans

[Al Williams]

I would like to know if there is a system to scale plans for different engine sizes.  I have a few 1/2A plans I would like to build but want to use a larger engine than a 0.049 engine therefore a larger plane. Engine size to be determined but between 0.050 to say a 0.15.
Thanks
Al Williams

[Steve Berry]

Very general rule of thumb is neighborhood of about 250 sq. in for 1/2A, with a max of 300 (assuming light construction and a strong 1/2A engine). For a .15, I'd shoot for 300-350 sq. in., with upper limit of about 400 (again, assuming light construction and a strong .15-size engine).

Every plane will likely be slightly different due to differences in airfoils, moments, and what the intended use is. You could probably fly a full-size Twister on a .15, if you're just looking for round-and-round training type flights. The SIG Akromaster is excellent for  a .15-size engine, as I've built several that way. If you switch out the kit wood with very light, yet strong, selections and build it light, a strong Norvel .061 can haul it around. It won't be a world-beater with that engine, but some mild stunts should be doable.

So, long story short...

1/2A = 250 squares or less
.10 to .15 = 200 to 350 squares
.20 to .35 = 300 to 450 squares
.35 to .46 = 400 to 550 squares
.51 to .76 = 530 to 700+ squares

It mostly will depend on how light you can build and how efficient the actual airframe is (airfoils, moments, etc).

Steve

[Dave_Trible]

Not really.  You can simply go to a Fed-Ex Kinkos and scale up or down to suit.  However when you scale a set of plans it is not necessarily the same airplane-just a different size.  To have desirable flying characteristics you may want to make some changes to suit heavier engine  and other things that become more an issue with scale.  This may mean lengthening moments-extending the rear fuselage and perhaps increasing the stab/elevator areas just as an example.  For what you describe,  a small airplane going just slightly larger I wouldn't worry about it.  There shouldn't be enough change to matter much.  Also these could not be higher performing airplanes in this size to be an issue.  Just hit the 'enlarge' button and have fun!

Dave

[Paul Wescott]

.10 to .15 = 200 to 350 squares
.20 to .35 = 300 to 450 squares
.35 to .46 = 400 to 550 squares
.51 to .76 = 530 to 700+ squares

Steve

1 - Pick a 1/2A design you want to scale up
2 - Pick an engine you want to use
3 - Refer to Steve’s chart to find the desired wing area for your chosen engine
5 - Divide the desired wing area by the 1/2A plan wing area and multiply by 100 for per-cent

Example: imagine 1/2A wing is 150 square inches, and you want to use a .35, the chart says 400 SqIn to 550 SqIn so let’s use 450 SqIn, 450 divided by 150 = 3.  That is a factor of 3.0 you need the plan enlarged 3x size, or multiply 3.0 by 100 to get 300%.

Good luck finding a kid in a copy shop who understands how to enlarge properly.

Next build the plane, mount the engine and fly it.  If it performs poorly build a smaller lighter airframe.  Rinse and repeat until the .35-powered model performs to your liking.

At least that’s how it was explained to me…

Paul W.

[Steve Berry]

5 - Divide the desired wing area by the 1/2A plan wing area and multiply by 100 for per-cent
Example: imagine 1/2A wing is 150 square inches, and you want to use a .35, the chart says 400 SqIn to 550 SqIn so let’s use 450 SqIn, 450 divided by 150 = 3.  That is a factor of 3.0 you need the plan enlarged 3x size, or multiply 3.0 by 100 to get 300%.

Paul W.

I thought it was (square root ((area desired) / (existing area))). That is, the √(450/150) = √3, which is approx. 1.732. Otherwise, assuming the plane you started from has a 24" wingspan (reasonable for a 1/2A), multiply by your factor of 300% (3) you would end up with a 72" wingspan, as opposed to a 41.5" wingspan resulting from the 1.732 scale factor, which sounds about right for 450 square inches.

Steve

[Brett Buck]

1 - Pick a 1/2A design you want to scale up
2 - Pick an engine you want to use
3 - Refer to Steve’s chart to find the desired wing area for your chosen engine
5 - Divide the desired wing area by the 1/2A plan wing area and multiply by 100 for per-cent

Example: imagine 1/2A wing is 150 square inches, and you want to use a .35, the chart says 400 SqIn to 550 SqIn so let’s use 450 SqIn, 450 divided by 150 = 3.  That is a factor of 3.0 you need the plan enlarged 3x size, or multiply 3.0 by 100 to get 300%.

  That is not correct - the linear scaling factor in this case is the square root of 3, or about 173%. All linear dimensions should be scaled up by this factor to go from 150 square inches to 450 square inches.

    I would add, while Steve's table is OK to first approximation, you can't really go strictly by displacement to figure out the desired wing area, because in any particular range of displacement, there is a huge variation in run quality and power. A Fox 15 can fly only very small airplanes, you could fly a 500 square inch airplanes with a 15FP, and you probably want around 350-400 square inches

       Brett

[Fred Cronenwett]

When you double the size of any plane the wing area goes up by a factor of 4. This changes the wing loading the model can handle. The larger models can handle more weight per square foot of wing area. If you added a servo a 1/2a model it might not even fly, but if you add the same servo to a 70" span version of the same plane it won't even notice the extra weight. this is why bigger models fly better.

My 96" span B-29 has a heavy wing loading of 38 ounces per square foot (or more), a smaller model would not handle that wing loading at all.

The structure for a 1/2a model is designed for a smaller model. I would suggest taking a larger model and scaling them down the size you want. either way you will have to adjust the structure to match the weight of the model.

[Gary Dowler]

  That is not correct - the linear scaling factor in this case is the square root of 3, or about 173%. All linear dimensions should be scaled up by this factor to go from 150 square inches to 450 square inches.

    I would add, while Steve's table is OK to first approximation, you can't really go strictly by displacement to figure out the desired wing area, because in any particular range of displacement, there is a huge variation in run quality and power. A Fox 15 can fly only very small airplanes, you could fly a 500 square inch airplanes with a 15FP, and you probably want around 350-400 square inches

       Brett

Very correct.  Assuming light build and perfect trim, amazing things can be done with small engines.  Paul told me that the Cobra he gave me (585 sq in according to the Brodak kit specs I just looked up) once flew the pattern with a 15FP on the nose. It was obviously trimmed to perfection. Its 40oz with a Fox 35, so perhaps 1-1.5oz less with the 15FP?  Conversely, I know a guy who has a Nobler with a 40FP on it. Said he had to go to that engine to get adequate line tension overhead. Should be a ton more power than it needs, but it is Obviously not trimmed right.

Id say that specific engine used and trim are equal partners in determining what is needed to fly a given size wing.

Gary

[Istvan Travnik]

My suggestion is to use 3rd root, if we want to scale up or down upon displacement of the engine.
Example: let us take equally "weak" engines: Enya .09  and Stalker .76.  Stalker is approximately 8 times bigger, 3rd root of 8 is right 2.
Hocus-pocus: the proper wingspan will be right the half of the Stalker, for Enya equipped model.
Try to make your own calculations of well known planes!   
Istvan

[GallopingGhostler]

Using cube root methodology for scaling factor seems to work as you have mentioned, Istvan Travnik. Not long ago I took Walt Musciano's C/L Fokker Triplane from his book on flying scale models at 12" wingspan for Cox .020 Pee Wee reed valve. For similar engine technology, in this case a .049 reed valve Babe Bee, came up with 16" wingspan.

Cube root of 0.020 is 0.2714
Cube root of 0.049 is 0.3659
0.3659/0.2714 is 1.348

12" x 1.348 is 16.18, the new wingspan for .049 engine, which seems typical for the designs of the times, for the sport type C/L aircraft being presented and kitted.

To make things easy, I round to 1.35 as the new scaling factor.

[M Spencer]

If you look at ancient 60's !  . 35 things ' around 48 inch '  , as in the spars likely a 4 foot asheet , or the trailing edge . Plus some wing tips .

A LIGHT .15 thing , 36 inch instead . Particularly if its a SCHNEURLE  . Maybe a old loop scavenged motor , 30 plus tips . Or If its for ' all weather ' use .

That gets you maybe 30 + 3 + 3 for 36 inch . Or 36 + 3 + 3 for 42 inch , maybe .

Some of the 1/2A T D jobs WERE 36 inch , so with a .10 Cu in , lower rever , as the TD is .089 or whatever anyway , should work good , with a shorter nose for the heavier engine .

Like wot Thisis .





foran example .

Look what happens when you put " Dee rice Oriental " in google pictures . 

REMEMBERING ; If its OVERPOWERED you can put a smaller intake in it , & it should run even better . Whereas if its underpowered , youve got problems . But power to weight is the problem . NOT power to size .
In calm air , anyway .
Say a 8 x 4 is about the biggest prop youd put on a Pinto . So a Old OS max .15 or a newer SF / FP .10 are about The Same WEIGHT . And'll swing the same prop . 35 foot lines , maybe 50 in flat calm , for lotsa room .

[M Spencer]

They were building 1/2A ships about HALF the span of their .40 ships , so 1/4 the area . Very Light control lines , .008 ??



Sorta Typical of the genre .

[Istvan Travnik]

It is suggested with good heart to memorize some other scaling equations:
-Scaling a prop when diameter is altered: the power used up is commensurable with the 5th power of diameter. (Eg. a double big prop with the same width in %  , revvig the same rpm, eats up 2^5 = 32 times more power.  When I put 5% bigger prop that means 1.05^5 = 1.276 more power! Not a joke...  When I know my engine is only 80% strong of the previous, 0.8^(-5) *  = 0.956 will be the proper scaling of the original.
What to do when I want to make 3 blade prop of 2 blades??  3/2 = 1.5.    5th root of 1.5 = 1,084, with this number must be shortened the blades. For example: 13" dia 2 blades equals 12" dia, 3 blades.  (It is proven, and works).
-What about the RPM? The power is commensurable with the 3rd power of rpm. When you double the rpm, that means 2^3 = 8 ! Not a joke, again...   When you can achieve 10% more rpm that means 1.1^3 = 1.331 more power.  (In the practice, you can change your prop's pitch from 5.5 down to 5.0"...           
Istvan
* in other words: 0.8^(-5)  means "fifth root of 0.8"

[Steve Helmick]

Personally, IMO power/weight ratio is more critical than power/area ratio. A very good...but old...rule of thumb is .049/.051 should weigh about 5 ounces, .15 = 15 ounces, .46 = 46 ounces, .60 = 60 ounces. Works for both FF & CL.

Schneurle port engines have changed these "ideal weights" by about 20% or so, but I would also point out that more experienced stunt fliers will be happy with less thrust , and lower skilled stunt fliers will prefer more thrust. I'm avoiding the word "power", because that is commonly interpreted as "horsepower", which isn't what we want or need.  Steve