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Vector Summation Principles

Creating Motion from Stillness

The magic of an AC motor isn't in any single component, but in how stationary parts work together to create something that moves. The key is combining two different kinds of shifts: a physical shift in space and a timing shift in the electrical supply.

Imagine three separate coils of wire, arranged in a circle inside the motor's housing. They aren't right next to each other. Instead, they are physically displaced, set 120 degrees apart. This is the spatial phase shift. Each coil is fixed in place.

Now, we power these coils with a three-phase AC supply. The currents flowing into each coil are identical in magnitude and frequency, but they peak at different times. Each current waveform is offset from the next by one-third of a cycle, or 120 degrees. This is the temporal phase shift.

When current flows through a coil, it generates a magnetic flux. Because the current is AC, this flux is a pulsating vector. It grows to a maximum, shrinks to zero, reverses direction, and repeats. Our goal is to see what happens when we add these three pulsating, spatially separated magnetic fluxes together.

Each phase produces a magnetic flux that varies sinusoidally with time. We can represent these fluxes as vectors. The direction of each vector is fixed along the axis of its coil, but its magnitude changes. Let's write them down, where ΦmΦ_m is the maximum flux from any single phase, and ωω is the angular frequency of the AC supply.

ΦA=Φmcos(ωt)0ΦB=Φmcos(ωt120)120ΦC=Φmcos(ωt240)240\vec{\Phi}_A = \Phi_m \cos(\omega t) \angle 0^\circ \\ \vec{\Phi}_B = \Phi_m \cos(\omega t - 120^\circ) \angle 120^\circ \\ \vec{\Phi}_C = \Phi_m \cos(\omega t - 240^\circ) \angle 240^\circ

The total, or resultant, flux ΦrΦ_r is the vector sum of these three individual fluxes. To find this sum, we resolve each vector into its horizontal (x) and vertical (y) components and add them up. The horizontal component of a vector is its magnitude times the cosine of its angle, and the vertical component is its magnitude times the sine of its angle.

The Mathematical Proof

Let's calculate the total horizontal component, ΦxΦ_x.

Φx=ΦAcos(0)+ΦBcos(120)+ΦCcos(240)=[Φmcos(ωt)](1)+[Φmcos(ωt120)](0.5)+[Φmcos(ωt240)](0.5)\begin{aligned} \\ \Phi_x &= \Phi_A \cos(0^\circ) + \Phi_B \cos(120^\circ) + \Phi_C \cos(240^\circ) \\ &= [\Phi_m \cos(\omega t)](1) + [\Phi_m \cos(\omega t - 120^\circ)](-0.5) \\ &\quad + [\Phi_m \cos(\omega t - 240^\circ)](-0.5) \\ \end{aligned}

Using the trigonometric identity cos(X)+cos(Y)=2cos(X+Y2)cos(XY2)\cos(X) + \cos(Y) = 2 \cos(\frac{X+Y}{2}) \cos(\frac{X-Y}{2}), the terms cos(ωt120)+cos(ωt240)\cos(\omega t - 120^\circ) + \cos(\omega t - 240^\circ) simplify to cos(ωt)-\cos(\omega t).

Substituting this back in gives us:

Φx=Φm[cos(ωt)0.5(cos(ωt))]=Φm[cos(ωt)+0.5cos(ωt)]=1.5Φmcos(ωt)\begin{aligned} \\ \Phi_x &= \Phi_m [\cos(\omega t) - 0.5(-\cos(\omega t))] \\ &= \Phi_m [\cos(\omega t) + 0.5\cos(\omega t)] \\ &= 1.5 \Phi_m \cos(\omega t) \\ \end{aligned}

Now for the total vertical component, ΦyΦ_y.

Φy=ΦAsin(0)+ΦBsin(120)+ΦCsin(240)=[Φmcos(ωt)](0)+[Φmcos(ωt120)](0.866)+[Φmcos(ωt240)](0.866)\begin{aligned} \\ \Phi_y &= \Phi_A \sin(0^\circ) + \Phi_B \sin(120^\circ) + \Phi_C \sin(240^\circ) \\ &= [\Phi_m \cos(\omega t)](0) + [\Phi_m \cos(\omega t - 120^\circ)](0.866) \\ &\quad + [\Phi_m \cos(\omega t - 240^\circ)](-0.866) \\ \end{aligned}

Using the identity cos(X)cos(Y)=2sin(X+Y2)sin(XY2)\cos(X) - \cos(Y) = -2 \sin(\frac{X+Y}{2}) \sin(\frac{X-Y}{2}), the vertical component simplifies dramatically.

Φy=0.866Φm[cos(ωt120)cos(ωt240)]=0.866Φm[2sin(ωt180)sin(60)]=0.866Φm[2(sin(ωt))(0.866)]=1.5Φmsin(ωt)\begin{aligned} \\ \Phi_y &= 0.866 \Phi_m [\cos(\omega t - 120^\circ) - \cos(\omega t - 240^\circ)] \\ &= 0.866 \Phi_m [2 \sin(\omega t - 180^\circ) \sin(60^\circ)] \\ &= 0.866 \Phi_m [2(-\sin(\omega t))(0.866)] \\ &= -1.5 \Phi_m \sin(\omega t) \\ \end{aligned}

We now have the components of our resultant flux vector. To find its total magnitude, ΦrΦ_r, we use the Pythagorean theorem: Φr2=Φx2+Φy2Φ_r^2 = Φ_x^2 + Φ_y^2.

Φr2=(1.5Φmcos(ωt))2+(1.5Φmsin(ωt))2=(1.5Φm)2(cos2(ωt)+sin2(ωt))=(1.5Φm)2(1)Φr=1.5Φm\begin{aligned} \\ \Phi_r^2 &= (1.5 \Phi_m \cos(\omega t))^2 + (-1.5 \Phi_m \sin(\omega t))^2 \\ &= (1.5 \Phi_m)^2 (\cos^2(\omega t) + \sin^2(\omega t)) \\ &= (1.5 \Phi_m)^2 (1) \\ \Phi_r &= 1.5 \Phi_m \\ \end{aligned}

The angle of this resultant vector, θθ, is given by arctan(Φy/Φx)\arctan(Φ_y / Φ_x), which simplifies to arctan(tan(ωt))=ωt\arctan(-\tan(\omega t)) = -\omega t. The negative sign indicates a clockwise rotation at a constant angular velocity ωω. Stationary coils and time-shifted currents have produced a magnetic field of constant magnitude, rotating smoothly in space.

Two-Phase vs. Three-Phase

What if we only had two phases? In a two-phase system, the coils are spatially separated by 90 degrees, and the currents are temporally shifted by 90 degrees.

ΦA=Φmcos(ωt)0ΦB=Φmcos(ωt90)90\vec{\Phi}_A = \Phi_m \cos(\omega t) \angle 0^\circ \\ \vec{\Phi}_B = \Phi_m \cos(\omega t - 90^\circ) \angle 90^\circ

If you run through the same vector summation, you'll find the horizontal component is Φx=Φmcos(ωt)Φ_x = Φ_m \cos(\omega t) and the vertical component is Φy=Φmsin(ωt)Φ_y = Φ_m \sin(\omega t).

The magnitude of the resultant flux is:

Φr=(Φmcos(ωt))2+(Φmsin(ωt))2=Φmcos2(ωt)+sin2(ωt)=Φm\begin{aligned} \\ \Phi_r &= \sqrt{(\Phi_m \cos(\omega t))^2 + (\Phi_m \sin(\omega t))^2} \\ &= \Phi_m \sqrt{\cos^2(\omega t) + \sin^2(\omega t)} \\ &= \Phi_m \\ \end{aligned}

A three-phase system produces a rotating magnetic field that is 50% stronger than a two-phase system using coils with the same peak flux. This contributes to the higher power density and smoother operation of three-phase motors.

Quiz Questions 1/5

What are the two key principles combined in an AC motor to create a rotating magnetic field from stationary components?

Quiz Questions 2/5

In a typical three-phase AC motor, the stationary coils are physically arranged how many degrees apart?

This mathematical superposition is the principle behind how we turn electrical energy into smooth, continuous rotational motion.