Now some considerations about the theory of the equilibrium of the quadcopter.

I’m curious to see if they will fit with the practice.

Vertical equilibrium


The force any prop can give is proportional with the square of the angolar speed: Fi=Kw^2


if we call ‘we’ the angular speed that keep the qpi in equilibrium  , we get:


Let’s Consider now to increase the w (for a ‘wd’ quantity) for all the motors .What will happen?

F0+F1+F2+F3=mg+ma =4kw^2=4mg(we+wd)^2/4we^2

mg+ma=mg (we^2+wewd+wd^2/we^2



>>wd=[-2gwe+_ sqr(4g^2we^2+4gawe^2)]/2g

that meas:

>>>Delta %=wd/we=sqr(1+a/g)-1


So if I  want to move up 1m the qpi in 1 second , it means a=2m/s^2. and g=9m/s^2

We get Delta%=sqr(1-2/9)-1  = 0,105.

it is necessary to increase the w by 10% to move up the qpi 1m in un second.

Vertical Equilibrium adding  roll or pitch

Let’s assume we  want to keep  the qpi at a given height but we want to move it horizontally.For exemple we want to roll the qpi.




k cos a (we^2+2wewd+wd^2+we^2-2wewd+2wd^2+2we^2)=mg

cos a (4we^2+2wd^2)/we^2=4

>>>Delta %= wd/we= sqr(2(1-cos a)/cos a)

If I want an inclination of 5′   Delta % = 8% ,so it is necessary to increase by 8 % the w for motor0 and decrease by 8% for motor2

Horizontal equilibrium with roll or pitch

k sen a (4we^2+2wd^2)=ma

mg sen a (4we^2+2wd^2)/we^2=ma

a=g sen a( 1+D^2/2)

with  angle a=10′ e D=0.17  the acceleration  a=1,58 m/s^2

s=1/2at^2  in 1  second qpi can move about  0,75m

Equilibrio di yaw


Mi=l*Fi=bk we^2

To obtain a yaw rotation it is necessary to increase the w for m0 and m2 and  reduce it for m1 and m3,  in order to keep the equilibrium on the other directions

Let’s say b the distance between the center of the qpi and each motor,and a the rotational acceleration

2bk(we^2+2wewd+wd^2-we^2+2wewd-wd^2)=Iz a

8bk we wd  =Iz a

with k=mg/4we^2

2bmgwewd/we^2=Iz a

Delta%=Iz a / (2mgb)

Let’s assume the qpi as 2 crossed bars  long 2b (b=200mm) , then

Iz= 2* m/2 * (2b)  ^2/12

>>Delta%=b a / 6 g

If we want a rotation on 1 round in 1 sec  the angle is

r=1/2 a * t^2    r=2pi , t=1   a =4 pi

Delta = b *4 pi /6 g=2 b pi/ 3 g = 2*200 *3.14 /3 * 9800= 4,2 %


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