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Author Topic: Shrinking larger models  (Read 1691 times)

Offline Glenn (Gravitywell) Reach

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Shrinking larger models
« on: March 24, 2010, 03:19:36 PM »
I was looking at that beautiful Dreadnaught that is being built in another thread and wondered (as I often do), "What a beautiful model, I wished it wasn't so big!".  So my question to all you stunt gurus......can a larger, say 60" model simply be scaled down to 48" without any problems?  I have often wondered if the airfoil would have to be changed, etc to allow for the smaller ship.  What say you all?
Glenn Reach
Westlock, Alberta
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Offline John Miller

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Re: Shrinking larger models
« Reply #1 on: March 24, 2010, 03:48:37 PM »
Yes, a simple scale down can be done rather easily, but there are a few items to keep track off, and pay attention to.

Simply scaling down to 48" span is basically linear scaling. Divide the wing span you want, by the wing span you got, (48/60=.8) The .8 is your lineal scale factor. Your copy place can scale to 80% (.8) and you're done, almost.

There are other considerations to work with when scaling a design. I prefer to consider the wing area when scaling. I believe it gives me a better size when I'm considering the engine I want to power the scaled plan with. Use the same idea as before, divide the area you want by the area you've got. Now, this is important. since we're dealing with squared numbers, our answer is also squared, so we must take the square root of our answer, to get a true scale factor.

For instance, I have a 700 sq inch, .60 and up, design that I want to shrink, and power with an OS .40LA.  The LA should be fine with a design having 620 sq. inches. this gives me enough area to keep the wing loading low. I'll have to do a few things to accomplish this, like remove weight. I can extend the scaled rib spacing back to the original spacing and get rid of a few ribs that won't be needed in the smaller sized wing. Removing weight by altering the structure is a good practise, though not often done, when scaling a design.

Here's an example using the above numbers.

620/700=.8857

The sq root =.94 as a scale factor.

This is a larger plane than it would be with simply scaling from 60 inch span to 48 inch span, but it fits the desired engine better.


Getting a line on life. AMA 1601

Offline Glenn (Gravitywell) Reach

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Re: Shrinking larger models
« Reply #2 on: March 25, 2010, 11:38:29 AM »
Thank you John, thats exactly the info I was wanting.  Now I can shrink that Rabe Bearcat I have to a size I can handle! LOL H^^
Glenn Reach
Westlock, Alberta
gravitywell2011 @ gmail . com

Offline L0U CRANE

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Re: Shrinking larger models
« Reply #3 on: March 27, 2010, 10:09:01 AM »
... and using John's formula, if you scale a 60" wing to 48", that's 80% in span. Square root of 0.8 is 0.894, or 89.4%.

If you started with a 700 sq in area, this percentage would go to 602 sq in, in stead of John's 620 for his numbers... This is seemingly close, but it is better to work from wing areas. Check what works on the engine you want to use, and shoot for that area.

Enjoy! Stuff like this can get habit-forming, especially when it works well and you understand why.

\BEST\LOU

Offline Serge_Krauss

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Re: Shrinking larger models
« Reply #4 on: March 27, 2010, 11:09:21 AM »
Lou-

I'm not sure we're talking about the same ideas here, but written language sometimes fails us in these things. So forgive me if I misunderstood your intention. If one scales a 60"-span wing back to 48", then the areas change in the ratio of (48/60) squared or .64. That is, the new area will be only 64% of the 60"-wing's area.

If you want a 620 in^2 area, you would scale the the span back in the ratio of (620/700)^.5 or the square root of (620/700) = square root of approximately .8857 = approximately .9411. That is, reduce the span of your 700 in^2 wing to 94.1% of its original value, and you will have an area of about 620 in^2.

Essentially, this is A2/A1 = (span2/span1)squared = (b2/b1)^2, which is the same as saying that b2/b1 = the square root of A2/A1.

or

A2 = A1 x (span2/span1)^2 and span2 = span1 x the square root of (A2/A1), or b2 = b1 x square root of (A2/A1)

Now my apologies to our host for using this notation, but the editing functions on these forums do not accomodate mathematical notation, and we have been substituting this notation for them for at least the ten years I've been posting. If anyone wants to substitute English language sentences for mathematical symbolism and then claim that the resultant convolutions show that math is hard, have at it. But mathematics is the language of science and engineering for good reason, and this is how you scale aircraft.

SK

Edit: I discovered when my lower-case "b" followed by a "1" gave me a smiley face, rather than a "BEE-ONE") that this forum's editing functions allowed for subscripts. So I corrected for the smiley face. I won't do it here now, but I could have then used the superscript two, as in 22 = 4. Now, if only there were a square-root symbol...
« Last Edit: March 27, 2010, 06:34:43 PM by Serge_Krauss »

Offline Glenn (Gravitywell) Reach

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Re: Shrinking larger models
« Reply #5 on: March 27, 2010, 11:42:05 AM »
Well I hate to say this but I was a straight "D" student in math.  That means my teachers passed me because they felt sorry for me, not because I understood what they were talking about! LOL

I think I understand what you guys are saying though.  Take the wing area I want, divide it by the large wing area and that gives me a percentage number to use to shrink the original large plan.  And if the large model was a good flier, the reduced model should also be a good flier.  Is this basically correct?
Glenn Reach
Westlock, Alberta
gravitywell2011 @ gmail . com

Offline Serge_Krauss

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Re: Shrinking larger models
« Reply #6 on: March 27, 2010, 06:26:48 PM »
Take the wing area I want, divide it by the large wing area and that gives me a percentage number to use to shrink the original large plan.  And if the large model was a good flier, the reduced model should also be a good flier.  Is this basically correct?

Glenn-

Unfortunately not. If you want to shrink the plane from one area to another, you would only start that way. Yes, take the ratio of the new smaller area to the old larger area. But then you must take the square root of that ratio to get the scale dimensional change for your plans.

For instance, if you have that 720 square inch plane and want to reduce it to 540 square inches, you would take the ratio of 540/720 = .75. However, this is not your scale factor. That would give you a 405 square inch wing - too small. You need to take the square root of .75 to get the linear scale factor to give the copier people. The square root of .75 is approximately .866 (because .8660254... squared is .75). Tell the copy people to make a copy 86.6% of the original size. That new model will then have a wing area of (approximately) 540 square inches.

Fortunately, almost any pocket calculator will do this. In the above example, you'd just punch in "540", the divide symbol, "720", "=", and the square-root symbol, in that order, and the scale factor appears in the window.

Most copy places will miss slightly. So you should go in knowing what span you actually want (multiply .866 in the above example by the measured span of the 720 square inch model). Then if (when) the span comes out slightly wrong and you want it closer, you'll have to measure the new plan span, divide the desired span by that number, and ask them to correct it by that factor (or multiply that ratio by .866 and give them the answer as a corrected scale factor with which to re-start).

I know,...'sounds annoyingly hard, but depending on how close the copy comes out, that may be the only way to get accurate plans, unless the copy guy/gal can make the adjustment him/herself. Miracles do happen.

SK


Offline John Miller

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Re: Shrinking larger models
« Reply #7 on: March 27, 2010, 07:07:01 PM »
Well I hate to say this but I was a straight "D" student in math.  That means my teachers passed me because they felt sorry for me, not because I understood what they were talking about! LOL

I think I understand what you guys are saying though.  Take the wing area I want, divide it by the large wing area and that gives me a percentage number to use to shrink the original large plan.  And if the large model was a good flier, the reduced model should also be a good flier.  Is this basically correct?

Not quite.

When dealing with wing areas, you are dealing with squared numbers. 620 square inches divided by 700 square inches = .8857 squared. You have to remove the sq. from the answer, and you do that by using the sq. root of the number, which in this case is .9411239.

I have the availability to have my computer find the area of an object. Here's a practical example using this principle. I'll use at least 3 decimal space as the cad program is accurate to 17 decimal spaces, just to show how accurate it is.

Gordan's Tony has a wing area of 743.321 sq inches, and a span of 63.901 inches. It's designed for a .72 or larger engine.

I want to scale one for an OS .46LA. I've seen this engine doing a great job powering a plane with 640 sq inches.

So I want to scale the Tony down to 640 sq inches, I don't know what the span will be because this will not be a lineal scaling job.

I have to find the scaling factor that will give me the square inches I want, then I can measure the span.

640 sq inches divided by 743.321 sq inches = .861....inches squared.

I need to get rid of the inches squared portion of the answer, so I find the square root, which = .9279012. (I'm going to use all the decimal places here because I want the exact square inches after all is done.) This is my scale factor.

Now going back to my Cad program where I had the computer figure the wing area, I can use the scale command, and use .9279012 as the scale factor.

The computer now resizes the wing using this information.

I do another inquiry for area, and the computer measures the new wing at exactly 640 sq inches. The span now measures 59.294 inches. The new wing is now 4.607 inches shorter, but the area is a little over 103 sq inches less.

In my earlier example, I may not have been clear on one point. That 60 I referred to was not the wing span. it has no place in figuring using sq inches of area. it referred to the engine size, a .60 engine related to a .40 engine. They were only used to illustrate the size airplane normally utilizing the wing areas noted.

I use the example of the Tony showing that the math works, and the Cad program shows the figures to be correct.

So simply put.

Take the area you want, divide it by the area you have, find the square root of the answer, and you have your scale factor, using the area (Inches squared).

If all you are concerrned about is a lineal distance, such as wing span, then it's simpler, but may not reflect exactly what you think it does.

Take the span you want, and divide it by the span you've got. No squared numbers, so no square root to arrive at the scaling factor. A clean and simple linear scaling.  H^^

Getting a line on life. AMA 1601

Offline Glenn (Gravitywell) Reach

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Re: Shrinking larger models
« Reply #8 on: March 27, 2010, 11:02:26 PM »
Serge and John....thank you both so very much for this information.....and you did it in a way that I can actually comprehend.  No small feat I assure you!  I have had many math teachers over the years that slowly fell into dispare and maddness trying to teach me the rudiments!  Thank you again and I know I'm not the only one out there that has wondered how to do this correctly. H^^
Glenn Reach
Westlock, Alberta
gravitywell2011 @ gmail . com

Offline L0U CRANE

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Re: Shrinking larger models
« Reply #9 on: March 28, 2010, 03:37:19 PM »
Glenn, Serge and John...

My bad. I was in a bit of a rush when I wrote that and put it down backwards.

Of course 0.8 squared is 0.64.... I knew that, but wife standing in door waiting to go somewhere made me a bit careless...

(I don't do it the way I wrote, way above, when I'm doing it for real.)
\BEST\LOU

Offline Serge_Krauss

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Re: Shrinking larger models
« Reply #10 on: March 28, 2010, 11:48:19 PM »
I know what you mean, Lou. My worst posts are when I have no time. I almost always wish I'd waited until life stopped "gettin' in the way."

SK

Offline Pinecone

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Re: Shrinking larger models
« Reply #11 on: March 29, 2010, 08:27:04 AM »
Fortunately, almost any pocket calculator will do this. In the above example, you'd just punch in "540", the divide symbol, "720", "=", and the square-root symbol, in that order, and the scale factor appears in the window.

Not my calculator. :)

I hit 540 - Enter - 720 - Divide - Square Root. :)

RPN rules.
Terry Carraway
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Offline Peter Hess

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Re: Shrinking larger models
« Reply #12 on: March 29, 2010, 09:36:22 AM »
Glenn:
Thank you for starting this helpful topic.  Thanks also to all of you who contributed your expertise.  However, speaking only about the Dreadnaught, the easiest way would be to buy a set of plans for the Dreadnaught 40 from Randy Smith.  The description of the Dreadnaught 40 on Randy's website says that it is the same size as the Vector 40.
Peter Hess
Canton, CT
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Offline EddyR

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Re: Shrinking larger models
« Reply #13 on: March 29, 2010, 09:59:41 AM »
Randy Smith has plans for a smaller Dreadnaudht .35-40 size
Ed
Locust NC 40 miles from the Huntersville field

Offline Glenn (Gravitywell) Reach

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Re: Shrinking larger models
« Reply #14 on: March 29, 2010, 12:44:00 PM »
I have looked at the smaller Dreadnought, and its a beauty, but I was actually thinking of my plans for Al Rabe's Bearcat.  The Dreadnought article made me think to ask the question! LOL  What a beauty that ship is though! H^^
Glenn Reach
Westlock, Alberta
gravitywell2011 @ gmail . com

Offline Jim Pollock

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Re: Shrinking larger models
« Reply #15 on: March 29, 2010, 03:18:14 PM »
Hmmmm,

Didn't we just go over this about a month ago in Scale Down?

I guess we didn't get it definitive enough huh?

Jim Pollock   HB~> HB~> HB~>

Offline Ward Van Duzer

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Re: Shrinking larger models
« Reply #16 on: March 30, 2010, 08:09:45 AM »
K.I.S.S.

I wanted a Bob Hunt Saturn for a piped .36. Checked all the reduced drawings and discovered that John Simpsons's Cavalier wing was (with wing tip adjustments) was 92.5 percent of the Saturn. The airfoils were no more that a 32nd of a inch off any where in the plot. (Wing avaliable in Aniston, Alabama.!) Use moments available from Modern 40 sized stunters to get the balance right. Order wing (or build from plan). Start building!

Ward-O  >:D

 
I hate spelling errors, you mess up 2 letters and you are urined!

Don't hesitate to ask dumb questions.
They are easier to handle than dumb mistakes!  Ward-O AMA 6022

Online Howard Rush

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Re: Shrinking larger models
« Reply #17 on: March 30, 2010, 12:01:12 PM »
My concern would have been that the metallic brown wouldn't scale well, but it seems to have come out just fine.
The Jive Combat Team
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Offline Steve Fitton

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Re: Shrinking larger models
« Reply #18 on: March 30, 2010, 12:25:01 PM »
Ward, Isn't a downsized Saturn a Europa?  Robbie's plane from before the (upscaled) Saturn.  Which .36 did you use, by the way?
Steve

Offline RandySmith

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Re: Shrinking larger models
« Reply #19 on: March 30, 2010, 01:39:31 PM »
Or you could just buy a Dreadnought 40  or  Tempest 40 plan  and build it as is     <=


Randy


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