Determine the Center of Gravity for a Swept Wing RC Model

[@davidquerry8869]

I enjoy how your videos promote thinking/learning about what is going on.

[@TimMcKay56]

David, Appreciate your feedback! Tim

[@Rabsian]

Thank you Tim for addressing swept back wings. I was looking for a simple, clear explanation to apply to a dlg wing. Your video hit the mark perfectly.

[@TimMcKay56]

Rabsian: Glad the video was helpful, and thanks for checking in! Tim

[@robertwoolley8632]

Thanks Tim. I wanted to check my ASK-11 tapered wing scale glider plan position for Cog was about right. I watched your straight forward approach. I also saw a more complex formula that resulted in an answer over 2cm difference to your mean chord. Then I remembered, I have used eCalc for prop/motor sizing and that a CoG calculator, and that provided a result within 2mm of your method for a root chord of 34cm. Your straight forward method wins the day again.

[@TimMcKay56]

Robert: Great news, thanks for checking in! Tim

[@planesstevee]

Great video mate. I have just scratch built an avro arrow and did not know how to find CG of it. Thanks!!

[@TimMcKay56]

PS: Glad it worked out! Tim

[@QuiDocetDiscit]

Thank you for a clear, intelligent explanation with graphics. You are a BIG help.

[@TimMcKay56]

Thanks for checking in! Tim

[@dradden1556]

Thank you for the instruction. You make it easy to understand.

[@TimMcKay56]

Glad it was helpful!

[@supaoranges1086]

Thank you for making this video! I’m currently scratch building an F-5 Tiger II, and right now I’m just working out all the dimensions and weight. This video is super helpful!

[@TimMcKay56]

Glad this was helpful! Tim

[@robertroe1956]

Great explanation of CG with a swept wing. At 3.10 you show the cord not including the ailerons. I am not sure about that. With the drawings the ailerons are recessed, and included.

[@TimMcKay56]

Good point!

[@TheGraemeEvans]

What you have calculated at 25% MAC is the aerodynamic centre (ac), the CG should ordinarily be behind this but infront of the neutral point. If you want to skip doing the math why not use an online tool like cgCalc. Check the Wikipedia artical on "Longitudinal static stability" its not as overly mathematical as some of the other pages. NASA GRC discussing AC: It has been found both experimentally and theoretically that, if the aerodynamic force is applied at a location 1/4 chord back from the leading edge on most low speed airfoils, the magnitude of the aerodynamic moment remains nearly constant with angle of attack. Engineers call the location where the aerodynamic moment remains constant the aerodynamic center (ac) of the airfoil.

[@TimMcKay56]

Graeme: Great update, thanks! Tim

[@robertwoolley8632]

Thanks for the video Tim. I shall be using your graphical method for my scratch built Beech C45 that has a tapered wing. I have adapted my plan from a similar twin fin aircraft plan by Ivan Pettigrew. Just to check with you: The final line you draw at position 9:56 on your video; is that at right angles to the fuselage? I notice that the tailplane cord, and its distance from the main wing do not feature in finding the COG?

[@TimMcKay56]

Yes, the final line should be at 90 degrees, my error. The tailplane chord or distance from the main wing really does not matter, rather the down force of the stabilizer and elevator is the important item.

[@robertwoolley8632]

@@TimMcKay56 Thank you Tim for your prompt response

[@randywetzler5976]

Hi Tim - Thanks for all the work you do to help all us beginners. With regard to the wing mean aerodynamic chord 25% rule and center of gravity is this the locator point to best optimize the COG for the model as a whole and/or is it also the COG for the wing itself? Trying to wrap my head around how the magic of the 25% wing rule works with it's relationship to the rest of the aircraft. I recently completed my first foam board model and took it out for a test and of course immediately crashed it on take-off, later determined it was way to tail heavy so that's why I'm really trying to get a handle on COG. Out of curiosity I've modeled a wing in my Autodesk Fusion CAD program and it shows the COG much further back so that only adds to the confusion.

[@TimMcKay56]

Randy: An improper center of gravity is probably the #1 reason for model airplane crashes. 20% distance back from the wing leading edge (straight wing) is a bit conservation (i.e. nose heavy), I'd start there. There is a bunch of aerodynamic research behind this location . . . do not worry about that, just go with the figure of 20% back from the wing's leading edge. If the width (chord) of the wing is 8", CG is 2" back from the leading edge. This balance point is for the entire model, everything that will fly . . . such as having the battery installed, etc. Just pretend you are hanging the plane as an Xmas tree ornament, with the string attached at the CG point. The plane must balance level. Tim

[@patrickhoyle8357]

Hi, thanks for sharing your time but I don't understand, at approx 10.00 min. you place ruler on from the MAC point 25% back from the leading edge and draw a line back to the root and say this is where you put your fingers to balance. The ruler is haphazardly placed at a strange angle neither parallel with the leading nor trailing edge , is this just 90 degree to the fuselage or am I missing something?

[@TimMcKay56]

Patrick: You are correct, I did a bad job with ruler placement. Missed on the final edit. Should be 90 degrees to the fuselage, the point 25% back on the MAC. Apologies! Tim

[@patrickhoyle8357]

@@TimMcKay56 Thanks for your reply. You might be interested in this, recently bought an Arrows Edge540 with a book CofG 90mm. Maiden went horrendous. Added 100g upfront and now flies ok. Now have an Eflite Extra300 (similar to Edge but wing shape bit different) with a book CofG also 90mm, thinking similar planes I added 100g on its nose before its successful maiden. Today, I used your way of determining C of G and guess what? your way puts both planes C of G at 70mm, which is how they both balance. You'd think the manufactures selling loads of them would get it right, after all, they know don't they? they're the experts? So thanks for sharing your knowledge Tim

[@gordonmckay4523]

@@patrickhoyle8357 Thanks for your kind words and glad all worked out OK. Few items more important than the correct center of gravity for any aircraft. Tim

[@nigelhutt9921]

Brilliant, learnt do much. I was with you till he very last moment then didn't follow. You went back 25% of the mean chord but when you drew you COG to the root what dictated where you drew to?

[@TimMcKay56]

Should be 90 degrees line to the fuselage for the CG location. Tim

[@nigelhutt9921]

Many thanks, I think I was caught out by the angle of the wing in relation to he camera angle which to my uninitiated eye looked at an angle. Msny thanks for clarifying

[@fpvcruzer]

Great video, at 9:45 - 10:00 I got a bit lost. When you say the "line to locate CG is 90 degrees to the wing root". I think you are indicating that the line drawn from 25% of the mean aerodyamic cord to the wing root meets the wing root at 90 degrees to the wing root?

[@TimMcKay56]

Yes, you are correct. I missed that portion after video out.

[@fpvcruzer]

@@TimMcKay56 Thank you, I really appreciate your in depth videos and in this one the graphical representation of how to find CG was really helpful. I am in the process of balancing a flying wing and wanted to check the if the location of the CG point was correct.

[@Claymore_83]

Great video. I noticed at the end when you drew from the mean aerodynamic chord to the fuselage, this wasn't to 90 degrees of the fuse (and mean aerodynamic chord line). Is CG 25% aft of the mean aerodynamic chord line? Also does this approach work for forward swept and trapezoidal wings?

[@TimMcKay56]

GMAC: Good catch . . . line should have been 90 degrees to the fuselage. CG still 25% aft. Good question on forward swept and trapezoidal wings. Not sure, really, but I think that would be a good start. Chris Golds, the famed UK RC model designer, used to make small profile "chuck gliders" to check out the CG of unusual models before building the larger model. This worked for his XB-70 design, for example. Tim

[@Claymore_83]

@@TimMcKay56 Exactly the reason I asked. I am a UK wing designer also about to bring out a trapezoidal wing. In the process of building a chuck glider version. I'll use your method to get a ballpark of the CG. Thanks.

[@ebikefe]

Thanks for this video. Do these calculations change if the model plane has canards up front?

[@TimMcKay56]

Great question, but I have no idea on CG and canards. Tim

[@carlosfern1708]

Thank you for your tips it was very helpful

[@TimMcKay56]

Carlos: Glad it was of use! Tim

[@adrianbassett9190]

Thanks for your help, are the calculations for a delta the same as your swept wing?

[@TimMcKay56]

Good question, I think they should be. Tim

[@fonwoolridge]

Nice trick.. I'll use it... thanks!

[@TimMcKay56]

😊👍🏻 Tim

[@igorbrito2695]

Does that also work for kinked wing designs like those seen on airliners?

[@TimMcKay56]

Pretty much. Tim

[@Eectechs]

Your are genius sir

[@TimMcKay56]

ET: Many thanks! Tim

[@jfamtd2770]

When calculating the chord, if you have a full length Alieron, do you include that in the measurment? When drawing the line from MAC, should it be a 90 degree angle (it lookd like your line to the root was at an angle?

[@TimMcKay56]

JF: The (strip) ailerons count as part of the total wing area for CG calculations. Good catch on the line from MAC, it should be 90 degrees from the chord line to the fuselage. Tim

[@fyzobaba]

Great video..but after following ur information I crashed 3 of my swept wing planes. I think u need to revaluate ur information for swept wing planes specifically the ones with fuselage with or without tail. After various researches and trials of mine, even with planes like v900, kraftei me 163, funjet and my scratch build swept wing planes what I found that was perfectly correct is for flying wings with fuselage without tail for example kraftei me 163 and funjet .I used ur method, additionally I consider the part of the fuselage that joins the 2halfs of wing to be a continuation of the wing as second panel, with assumption that it is rectangular having a uniform MAC which is same as the root cord of the wing. So for funjet it has 3 panels and kraftei me has 2 panels.for those wings with fuselage without a tail use 20% MAC, making sure it is perpendicular to the fuselage, mark the 2 spots along the fuselage and find the average.That is ur perfect CG. For wings with fuselage and a tail like v900 use above method but 25% of MAC , consider it has having 2 panels , find the average and u will be happy. Good luck!

[@TimMcKay56]

Hafis: Appreciate this feedback! Tim

[@johnwalsh1879]

Could you please show us how it is done on a mig 29 or f18. As the wing doesn't run straight along the fuselage

[@TimMcKay56]

I'd draw an estimate of the wing shape and sweep angles and adapt/sketch to straight lines, should be close enough. Also wise if in doubt to estimate a bit nose heavy as opposed to tail heavy.

[@johnwalsh1879]

@TimMcKay56 thank you for your quick response p.s your videos are brilliant very useful

[@TimMcKay56]

@@johnwalsh1879 Thanks!

[@johnwalsh1879]

Sorry for texting you again but I have left you a link to my KZbin page where I have a top view of sa mig 29 I have the 2 parts I'm confused about marked in blue and green Could you in your professional opinion tell me which is the mac cord. @johnwalsh 1879

[@wernervandenbosch1609]

Hello Tim, How would you calculate the center of gravity for a canard plane with delta wing or with normal wing ? Thanks 🙏

[@TimMcKay56]

Werner: Not sure, will add to video list! Tim

[@diablito2013]

Very good explaination! Been searching all over for this simple version. But the root - tip cord cross on a swept wing dont work for me, my cross ends up on a slight "longer" mac than the calculated line🤔 4 random drawn wings now, and all same result.

[@TimMcKay56]

Kristian: Maybe take a look at my foam F-4 video. I compute the CG of that swept wing airplane with the same method. Tim

[@Chris-vc1dh]

Fantastic information How to find CG if the wing is straight no sweep but it has a cut in the middle so the wing is narrowed in the root ?

[@TimMcKay56]

Chris: If the wing is simply constant chord (same width), CG is usually 25% back from the leading edge. In a wind with an inner constant chord, but a narrower end section, 25% back for the CG should work as well. Tim

[@Chris-vc1dh]

Tim McKay but the root will be before cutting or after?

[@spottyjack614]

very informative vidd.

[@TimMcKay56]

Glad it was helpful! Tim

[@Gary_Aviator]

I got confused. You've marked 25% aft LE for Tapered wing on MAC but then joined the line to root chord at an angle which is at same distance of MAC. So is MAC the actual CG.. please clarify?

[@TimMcKay56]

I did not do a good job in the video clearly showing this procedure. Key item is locating the Mean Aerodynamic Chord. Once you have this mark 25% back from the LE. This is your CG. Draw a line at 90 degree angle to the MAC over to the fuselage centerline, this is your CG. Tim

[@Gary_Aviator]

@@TimMcKay56 Thanks Tim for clarifying the confusion

[@richardwarner1373]

could you do a similar demonstration for a fixed chord for a 1/3rd of the span with a 1/3rd for each dihedral wing tip which has swept back le and a not so swept forward te please. i am modelling a fairchild a 10 tank buster. which is the wing i would like to calculate the cg for. its going to be about 6ft span running 2 90mm edfs.

[@TimMcKay56]

Richard: I can certainly add your request to my video list. However, the A-10 subsonic wing is essentially a rectangle. I'd just set the CG 25-28% back from the wing leading edge, should work fine. Tim

[@richardwarner1373]

@@TimMcKay56 thanks

[@denismoran670]

No go - I've watched it a dozen times+ and still can't find a measurent for 'where your fingers go, and no mark shows on the root chord, either?

[@TimMcKay56]

Denis: Try watching it again. I just did. At 9+43, I show the Mean Aerodynamic Chord line. The Center of Gravity is 25% back from the wing leading edge (along the MAC). Transfer that point (25% back) directly to the wing root. I did not do a great job in the video, but the line back to the wing root needs to be 90 degrees to the MAC. Tim

[@truthboomertruthbomber5125]

In about 1998 I built a solid sheet balsa delta wing with a sp400 motor and an 8 cell nicad. I researched the CG location and put it a bit ahead of the calculated location. Well, guess what? It was at least 2” rearward on a plane that was only about 18” in length. I had to extend the fuse so I could move the battery forward. The nose was simply a block of styrofoam that I sanded to shape. Eventually I got it figured out and the plane was a great flyer. The first 50’ of every flight was an adventure leading me to name the plane “ The Hairy Delta “.

[@TimMcKay56]

Great report, thanks for sharing!

[@truthboomertruthbomber5125]

@@TimMcKay56 The first flights [ attempts anyway ] were at Sugar Hill Elementary :). I used to do a lot of flying behind the school.

[@gordonmckay4523]

@@truthboomertruthbomber5125 Great!

[@jacobedman9240]

How would you calculate the MAC on a plane with wing sweep capabilities like the f-14 or mig 23

[@TimMcKay56]

Jacob: I'd use a CG for the wings in their landing configuration, usually swept full forward. In the landing phase is where the CG will be critical. As you fly faster and sweep the wings, the CG should be such that the model flies well. Most planes have a fore and aft range of CG anyhow, so flying at cruise speed should work out. Tim

[@jacobedman9240]

@@TimMcKay56 thank you, this helps a lot!

[@smitdmello1634]

70% of my wings cord is aerodynamic and rest 30% is flat extension for alirones. should i mark cg considering only 70% cord length or whole cord length

[@TimMcKay56]

Smit: Include the full wing chord (wing airfoil plus flat ailerons) for CG determination. Tim

[@smitdmello1634]

My aircraft wings are flat insted of aerodynamic wing (they don't produce lift) can same cg consept work on it

[@TimMcKay56]

Yes, CG location does not depend on airfoil. Tim

[@user-rs1fo2dd9b]

@@TimMcKay56 thats very interesting, why does the calculation not need to take airfoil into consideration? i thought a thicker wing with a more pronounced airfoil, would have more distance to take into account than a very thin wing, with negligible airfoil

[@bradhoult9921]

Great explanation! Does this work for flying wing designs?

[@TimMcKay56]

Brad: Yes, this CG approach will work for flying wing designs. Good luck! Tim

[@scootergeorge7089]

If possible, look at an actual aircraft. The mid point of underwing stores. If dropping bombs or fuel tanks the CG does not shift.

[@TimMcKay56]

Thanks! Tim

[@drmrjon]

Ok, now how do you figure that out without using paper when the plane is already built??

[@TimMcKay56]

Google "Swept wing center of gravity" there are various calculators out there.