For a swept and tapered wing, the CG should be back from the leading edge a distance between 25% and 33% of the mean aerodynamic chord (MAC) at the location of the MAC. I do not understand how your method of drawing the box and dividing it into thirds accomplishes this. More specifically, how does your method locate the MAC, and how does the 25/33% enter into the picture? I've watched Tim McKay's video: Determine the Center of Gravity for a Swept Wing RC Model and his method is quite different. Can you help me understand how your box method and Tim's more complicated method are equivalent, assuming that the two methods give the same result? Thanks!
Well the 3 boxes he did is equivalent to 33%. I'm not sure if that will work for swept wing. I will only believe if he flies 3 different wings with that set up
If this worked, why would you even need to make a box. Just use the square from tip of root to bottom of wingtip and divide by three.. Of course this is wrong and the CG would be nearer the center of the root chord.
@@donolson6335 So if the wing is swept the center of aerodynamic forces shifts proportionally backward so the projection on the root chord is not 33% anymore, it's more that - is that correct?
thanks that is a very clear illustration..
Simple, easy to follow. Thanks
For a swept and tapered wing, the CG should be back from the leading edge a distance between 25% and 33% of the mean aerodynamic chord (MAC) at the location of the MAC. I do not understand how your method of drawing the box and dividing it into thirds accomplishes this.
More specifically, how does your method locate the MAC, and how does the 25/33% enter into the picture?
I've watched Tim McKay's video: Determine the Center of Gravity for a Swept Wing RC Model and his method is quite different. Can you help me understand how your box method and Tim's more complicated method are equivalent, assuming that the two methods give the same result? Thanks!
Well the 3 boxes he did is equivalent to 33%. I'm not sure if that will work for swept wing. I will only believe if he flies 3 different wings with that set up
Totally agree , there's no way that this method is correct. CG will be further back than that.
If this worked, why would you even need to make a box. Just use the square from tip of root to bottom of wingtip and divide by three.. Of course this is wrong and the CG would be nearer the center of the root chord.
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Sir give a video of how to find cg of f-22 raptor Rc plane ,plz reply
I'm still waiting for your reply
I'm not super sure that's correct....
I assure you this is correct. Used it many times. It is a way to do the math by drawing it out. Makes it simple.
@@donolson6335 So if the wing is swept the center of aerodynamic forces shifts proportionally backward so the projection on the root chord is not 33% anymore, it's more that - is that correct?
It is not correct.