The 60 Degree Question:

[rattus]

This question arises from time to time:

"Why does the delta diagram include 60 degree angles when the phase separation is 120 degrees?"

Look at it this way:

Say you wire up 3 secondaries into a wye configuration, and to impress the Boss, you draw out the wye phasor diagram complete with arrows.

Now the Boss comes along and says, "Dummy, that is supposed to be a delta!" You sheepishly correct your mistake. Now how do you correct your phasor diagram?

You merely slide the arrows around until you have formed a triangle. You do not change the directions of the arrows! You now have the familiar delta phasor diagram which is an equilateral triangle with internal angles of 60 degrees. The phase separation however is still 120 degrees.

If you know a little trig, you can see that the voltage of any one side is,

V = 2 x Vphase x cos(60) = Vphase

Of course that is obvious since we have an equilateral triangle.

For the line to line voltage in a wye, we use the same approach but the angles are different, 120, 30, 30.

Then Vline-line = 2 x Vphase x cos(30) = 1.732 x Vphase

 

[charlie b]

Of course that is obvious since we have an equilateral triangle.

That is even more obvious, since all triangles are equilateral. I have proof! I just don't know how to draw a picture and post it on this site.

 

[rattus]

Huh?

Huh?

That is even more obvious, since all triangles are equilateral. I have proof! I just don't know how to draw a picture and post it on this site.

All triangles are equilateral? Prove it in words, CB, or try "Manage Attachments" below the message dialog box.

 

[charlie b]

OT: Proof that All Triangles are Equilateral.

OT: Proof that All Triangles are Equilateral.

PART ONE: Construction

1. Construct general triangle ABC. See the attached sketch.

NOTE: Without loss of generality, I will start with vertex A. I will show that side AB is congruent to side AC. I assert that had we started with vertex C, the same process would show that side AC is congruent to side BC. This will complete the proof that the triangle is equilateral.

2. Construct the bisector of vertex ?A.? You won?t need to draw it further than about halfway into the triangle, towards the other side.

3. Construct the perpendicular bisector of side BC. Here again, you won?t need to draw it further than about halfway into the triangle. Label the mid point of side BC as ?Point G.?

4. Label the point at which items 2 and 3 intersect as ?Point D.?

5. From Point D, draw a line segment to Point B, and draw another line segment to Point C.

6. From Point D, construct a line segment perpendicular to side AB. Label the point at which this line segment intersects the side as ?Point E.?

7. From Point D, construct a line segment perpendicular to side Ac. Label the point at which this line segment intersects the side as ?Point F.?


PART TWO: Proof

8. Consider triangles BDG and CDG. Side BG is congruent to side CG, since we started by bisecting side BC. Side DG is common to both triangles. Therefore, BDG is congruent to CDG by the rule of ?Hypotenuse-Leg of a right triangle. That gives us BD congruent to CD as corresponding parts of congruent triangles.

9. Consider triangles AED and AFD. Angle EAD is congruent to angle FAD, since we started by bisecting vertex A.. Side AD is common to both triangles. Therefore, AED is congruent to AFD by the rule of ?Hypotenuse-Angle of a right triangle. That gives us AE congruent to AF, and DE congruent to DF, as corresponding parts of congruent triangles.

10. Consider triangles BED and CFD. Side DE is congruent to side DF, from Step 9. Side BD is congruent to side CD, from step 8. Therefore, BED is congruent to CFD by the rule of ?Hypotenuse-Leg of a right triangle. That gives us BE congruent to FC as corresponding parts of congruent triangles.

11. Since AE is congruent to AF (from step 9), and since BE is congruent to FC (from step 10), we have AB congruent to AC, simply by adding the two pairs of line segments together.

12. This proves that the triangle is isosceles.

13. By starting with the bisection of vertex C, and using the same steps, it will be proven that side AC is congruent to side BC.

14. This proves that all three sides are congruent to each other, making the triangle an equilateral one.

QED.

 

[drbond24]

Do it with the triangle I attached.

 

[drbond24]

How about this for a proof:

Time = Money -- time is money
Money = sqrt(evil) -- money is the root of evil
Women = time * money -- women require time and money

time * money = money^2 -- since time is money

money^2 = evil -- square of the square root

Women = time * money = evil -- Women are evil!!

If you require further evidence, I'll introduce you to my wife.

 

[rattus]

I will guess at it Charlie and say that point D may not be attainable with anything other than an equilateral triangle. I am too old to grind it out for myself.

 

[bcorbin]

I will guess at it Charlie and say that point D may not be attainable with anything other than an equilateral triangle. I am too old to grind it out for myself.

You are correct. The line from point D to G is a complete fabrication. It can not exist, as defined. It is produced by stating you are bisecting the first angle, and then drawing a line which does not.

 

[drbond24]

http://www.mathematik.com/Isoscele/index.html

Check this out. I think this is what he was going for, although I maintain that it only seems to work when you begin with a triangle that is visually close to equilateral to begin with.

 

[charlie b]

I learned this one from my HS calculus teacher. I will admit that I was not sharp enough to see the fallacy in the proof.

As to the location of Point D in the image, this very “proof” can be used as a real, valid proof that the two lines drawn in steps 2 and 3 cannot meet at a unique point inside the triangle. Indeed, for an Isosceles or Equilateral Triangle, the two are actually the same line.

 

[bob]

That is even more obvious, since all triangles are equilateral. I have proof! I just don't know how to draw a picture and post it on this site.

An equilateral triangle is defined as one in which all three sides have the same length. Kinda like a delta with 60 degrees.

 

[charlie b]

An equilateral triangle is defined as one in which all three sides have the same length. Kinda like a delta with 60 degrees.

Thanks, but I think I knew that. :roll:

 

[Bob NH]

In an isoceles triangle the bisector of angle A is coincident with the the perpendicular bisector of side BC. Therefore, there is no intersection point D. It is "undefined", as though dividing by zero.

If the triangle is anything other than an isoceles triangle, then the intersection point exists but it is outside the boundary of the triangle. Therefore, the whole construction fails.

 

[charlie b]

If the triangle is anything other than an isoceles triangle, then the intersection point exists but it is outside the boundary of the triangle. Therefore, the whole construction fails.

True. Indeed, that is pretty much what I said in post #10.

 

[nakulak]

the answer is simple - you are perceiving the diagram inverted. The relative phase angles in the "delta" diagram are the angles on the outside of the triangle, not the inside. 180 - 60 = 120.

 

[Bob NH]

True. Indeed, that is pretty much what I said in post #10.

Yes. I missed that and was answering the original question. Sorry about that.

 

[rattus]

Good point:

Good point:

the answer is simple - you are perceiving the diagram inverted. The relative phase angles in the "delta" diagram are the angles on the outside of the triangle, not the inside. 180 - 60 = 120.

Supplementary angle is the geometric term for it. Some draw an equilateral triangle without the arrowheads and call it a phasor diagram. This leads them to think the phase separation is 60 degrees.

It is possible to have unbalanced delta voltages in which case the triangle would not be equilateral. Still the angles of separation would be the supplementary angles.

 

[Smart $]

Supplementary angle is the geometric term for it. Some draw an equilateral triangle without the arrowheads and call it a phasor diagram. This leads them to think the phase separation is 60 degrees.

It is possible to have unbalanced delta voltages in which case the triangle would not be equilateral. Still the angles of separation would be the supplementary angles.

Yes, but in truth, that just happens to be a geometric coincidence. That is, one must draw the vector before the supplementary angle exists. Whereas, the vector heading IS the phase angle. It's angle is measured relative to an assigned vector heading of 0?, which is typically that of Line A or Phase A... and as you wrote earlier, "You merely slide the arrows around... You do not change the directions of the arrows!"