# angle outside triangle

#### xl5899

[FONT=&quot]this is actually mohr's circle formula, forget about the theory,let's focus on the mathematics part. I couldnt understand why the tan( 2 θs1) = -(σx -σy) / 2τxy ?[/FONT]
[FONT=&quot]2θs1 is outside the triangle[/FONT]

[FONT=&quot]theorically, tan(180-α )= -tan( α ) , so tan( 2 θs1) = -(σx -σy) / 2τxy , the vertical staright line of triangle should represent -(σx -σy) , whereas horizontal axis represent 2τxy , am i right?[/FONT]

#### chiro

MHF Helper
Hey xl5899.

Do you know the relationship between tan(theta) and the gradient (i.e. (y2-y1)/(x2-x1))?

#### xl5899

Hey xl5899.

Do you know the relationship between tan(theta) and the gradient (i.e. (y2-y1)/(x2-x1))?

#### chiro

MHF Helper
It's not that - it's in terms of the gradient.

In a normal line (that's embedded in two-dimensional space) you have y - y0 = m(x-x0) where m = tan(theta) = (y2-y1)/(x2-x1).

The rate of change that is m in terms of tan(theta) is what I'm discussing above.

#### xl5899

It's not that - it's in terms of the gradient.

In a normal line (that's embedded in two-dimensional space) you have y - y0 = m(x-x0) where m = tan(theta) = (y2-y1)/(x2-x1).

The rate of change that is m in terms of tan(theta) is what I'm discussing above.
But , the problem is the angle is outside the triangle....

#### chiro

MHF Helper
We are referring to the lines made by the triangle that form the gradient.

You can apply them to any triangle.

#### xl5899

We are referring to the lines made by the triangle that form the gradient.

You can apply them to any triangle.
can you explain further?I'm still blurred...

#### chiro

MHF Helper
Let m = (y2-y1)/(x2-x1) and you have points on the line (x1,y1) and (x2,y2).

In the triangle you have two points - one in the top left hand side and one in the bottom right.

For the top left you have (x2,y2) = (txy, [sigma_x - sigma_y]/2) and (x1,y1) = (0,0)

Since m = tan(theta) you have

m
= tan(theta)
= [[sigma_x - sigma_y]/2] / txy
= [sigma_x - sigma_y] / {2*txy}

which is what the result is.

You can do the other triangle by using the bottom right vertex for (x2,y2) and get the result that way.

• 1 person

#### xl5899

Let m = (y2-y1)/(x2-x1) and you have points on the line (x1,y1) and (x2,y2).

In the triangle you have two points - one in the top left hand side and one in the bottom right.

For the top left you have (x2,y2) = (txy, [sigma_x - sigma_y]/2) and (x1,y1) = (0,0)

Since m = tan(theta) you have

m
= tan(theta)
= [[sigma_x - sigma_y]/2] / txy
= [sigma_x - sigma_y] / {2*txy}

which is what the result is.

You can do the other triangle by using the bottom right vertex for (x2,y2) and get the result that way.
it should be (x2,y2) = ([sigma_x - sigma_y]/2 , txy) , the y-axis(vertical axis) represent txy , right ?

#### chiro

MHF Helper
Yes - I'm sorry about that you are correct.

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