Calculating the Unequal Angle of a Triangle

180-[tan(y/(x/2)) *2]. ?

Regards JD.

since we deal with angles the "atan" should be present instead of "tan" :

180-2*[atan(y/(x/2))].

When computing it shlould be taken care that the "atan" function gives results in degrees and not in radian. EXCEL works usually with radians so that again take care for the units.

Sorry but it had to be corrected.

Nick

Hi Nick

Thank you, no problem, not the first time.

Regards JD.

Boy i must be dumb, i forgot to menton that, i dont think i can caculate sin, cosine or tangents that easily. With your forumla how do i calculate the sin, cosine or tangent? for variables such as when the operators change the die. I will check over the CPU tomorro to see if it has this ability on board.

Not sure by what you mean when you say atan?. but i do think that i get the rest of you forunla.

Thanks Mate

Best Regards

Joe

Do you have excel or openoffice spread sheet? if yes you have the function.

Atan = arctangent= angle which has as tangent value what is computed between brackets.

Most of scientific hand calculators for college have the functions as well.

Trigonometry is the part of mathematics which makes the link between angular and linear quantities thus the need to use it.

Nick

... but the tan (or atan) function is inappropriate, as we have one side (Y/2) and the hypotenuse (X).

Good Answer John, but I think the angle you solved for and show in your diagram (b) is actually half of the angle the OP is looking for. For the OP question is the unique angle of the isosceles triangle. You've calculated the bisecting angle that the right triangle creates. But the OP question is a little ambiguous as to whether the desired angle is b or 2b. (Alas poor Yorick. )

Also, don't forget that sheet metal tends to spring back a little after being initially bent. So in your break (brake?) you will have to over-bend a little to compensate for this. Unfortunately the over-bend will change depending on thickness and alloy type.

You are dead right, the angle i am looking for is 2b, so i could use John calc and multiply the result by 2.

As for the spring back, this will be acounted for by adding a correction factor to each and every bend, the operator will get the hang of this very quickly

Thanks All for your input.

The angle shown (and the formula given for finding it) in the diagram is indeed half the required angle - that's why the formula given in the spreadsheet example is 2 x the result shown in the diagram.

I didn't go into the practicalities of folding an exact angle (allowing for spring-back etc), as it seemed that Joe Sparky was well up on brake-press operation, & just needed a hand with the trig.

BTW - I expressed angle b (half the required angle) in terms of angle a so that the formula would accord with that given by jdretired. It would (imho) have been a little easier to go directly for b = atan ((y/2)/x), b still being half the required angle.

Thanks,

John.

JD formula gives the full angle! But it was not understood.

Agreed - but he must've got there by taking 2 x the half-angle - hence 180 - 2* [...]

Look ar #3

Exactly - that's what I did!

My point is that jdretired's formula could only have been obtained by calculating the half-angle, & doubling!

Hi JD yes we are looking a b, b x 2 as this will give us the absolute angle (in theroy that we are bending) the problem with pressbrakes is that they are designed to only work to the maximun limut, then because of the tonnage required to complete the bend, the result will vary depending on the material used and the lenght of the material being folded. When you are folding a long sheet the result is always less than what you had calculated on, this is due to the folder growing (streching) we have a factor that we can use to calculate this error, it is not perfect but it works. The calculation depends on materal used (steel type) material thickness and lenght of fold. Which ends up being a long equasion, this is why i asked the boffins how to sort out my Trig.

Again thanks to all contributers, your advise has been taken on board, watch this space, I will let you all know how the system developes, mind you that may take a few months as there is still a lot of ground work to be done, and i cant afford to spend all my time on R and D even though it is my favorite work. It dont pay the bills though.

Thanks All!!!!!!

http://en.wikipedia.org/wiki/Isosceles_triangle#Types_of_triangles

Here's an excel formula that works (giving the answer in degrees):

2*ASIN((B2/2)/B1)*180/PI()

Where B2 contains the length of the un-equal side (Y), and B1 the equal sides (X)

Good answer DJ.

And for the "Z" (or X axis of the press) travel, z = SQRT(X^2 - (Y^2)/4)

Hi JD

What is the most important thing is the perpindicular (Spelling wrong) so i can calculate the depth that the pressbrake must travel to make the fold. I think you had it before.

We only need 2 dimensions to calculate the third using opposite / adj = Tan, i think call you me dumb. But what i have to do is reverse engineer all calc's to get the required result, it is a bit of fun in a PLC.

What PLC do you have?

The PLC we are using is a Unitronics "Vision 350"

'Fraid I've never met one. The docs on the Unitronics website seem to only show +, -, x, ÷ and √ .

Good luck!

Hi John

When you look at the math functions, there is a function called forumla it then opens up the edit formula menu, you can use all sorts of functions in this including sin cosin and tangent.

Cheers.

John,

Just looking back, you used the Y and X Values, the other value is the Z value, I need to calculate the value using the Y and Z value's? Also how do you manage to insert diagrammes into this forum?

Cheers

Joe

Hi, Joe,

Getting myself a bit mixed up, here - are you referring to my post #9 or #11?

Maybe we'd better start again, with a reference drawing:

The triangle vertices are P, Q and R. S is the mid-point of the unequal side. A is the unequal angle at vertex Q, and B is one of the equal angles.

X, Y & Z obvious.

Which dimensions or angles do you have, and what do you want to calculate?

[BTW To insert an image, you have to have it on your PC. Not sure what formats work - .gif and .jpg are definitely ok.

I did this drawing in AutoCAD, then did a print preview and captured it from the screen (using Faststone Image Viewer, which you can get as a free download). I saved it as Triangle.jpg, then inserted it in this post using the "Insert/Edit Image" tool: ]

Hi John

I want to use Z and Y to calculate Angle A.

That Faststone Image Viewer is neat, thanks John.

Cheers

Joe

OK, it's more-or-less what I did before, but here it is - a bit clearer (I hope ):

You have to split the angle in half, and multiply the result by 2 to get the full angle, thus:

The corresponding Excel worksheet (for a right-angled triangle) would be:

[ Again, the factor 180/pi is used to convert from radians to degrees ].

Hope this helps,

John

Hi John

This is as far as I have got so far. I thought it fair to share it with you, and I can attach drawings (thanks Mate). The first calc is for the Atan.

Then we have to convert this to degrees.

Having got this information, we can now save tool dimensions.

The next challange that I have to work out, is how to calculate tool collision risk and preventing due to operator error and preventing it happening.

I have my work cutout, this project is no small one, but hopefully the benifits will be worth it, I will keep you informed.

Again John Thanks for Your Valued Input

Cheers

Joe

Hi Joe,

Thanks for the update - always good to know how a project's going.

Glad to have helped - keep up the good work !

John

Hi John

You must think that I am a dumb Bas@#d, but now I have to reverse engineer the whole thing, the reason for this is we have a folding tool (top tool) and a Die Tool.

These tools are not matched, they can be of different angles, it is also possible that the Top Tool has a greater angle than the Die Tool, and then if we tried to move the Top Tool into the Die Tool, this could be nasty (damaging the tools).

We know the angle of the Top Tool and the Die Tool We also know its X Dimension of the Die Tool. We also know the material thickness, so we must subtract the material thickness from the X dimension of the Die Tool to determine the Max travel that the Top Tool can travel before there is a collision risk.

Let me show you a diagramme.

We only can bend an angle to the (max angle of the Top Tool) but we must avoid collision with the Die Tool.

To calculate this, we know the Die Tool X, we must also add the material thickness to this, then we must calculate the Y Depth that the folding beam can travel, any idears

would be appreciated.

See Dwg.

Cheers

Joe

Hi Joe,

Just so we're both singing from the same songsheet ...

... this is what you need to calculate, yes?

[This may take a little while - got to pretend I'm working ]

OK, I haven't heard back - so I'll assume I'm going in the right direction.

First, let's simplify it. These sketches:

... should persuade you that (provided the top tool angle is greater than or equal to the die tool angle) the depth of the 'Vee' in the die tool doesn't matter; it's only the aperture width that is important.

That said, the next stage is to draw up the important bits. I've halved the picture (as before) because it's easier to deal with the right-angled triangles. To begin, we'll ignore the piece-part, just concentrating on the top and die tools:

In this diagram, we have:

TXh = half the top tool X dimension,

TY = top tool Y dimension,

DXh = half the die tool X dimension (ie half the die tool aperture),

DY = depth of entry of the top tool into the die tool (the unknown!).

This bit is actually very easy - we don't need any trigonometry, it's solved using the properties of similar triangles.

We have DY/TY = DXh/TXh,

so

DY = (DXh * TY) ÷ TXh

And since

DXh = 2 * DX (half die tool aperture width = 2 times die tool aperture width)

and

TXh = 2 * TX (half top tool width = 2 times top tool width),

we can cancel out the "2"s, so that:

DY = (DX * TY) ÷ TX, or

Depth_of_entry_of_top_tool = width_of_die_tool_aperture * height_of_top_tool ÷ width_of_top_tool

_{^{<DEEP BREATH>}} that was the easy bit. Adding in the effect of the material thickness is (fortunately ) another problem, which does however depend on the angle of the top tool 'Vee' and needs some trigonometry. I'll probably have to leave that one 'til tomorrow (unless anyone else fancies a go?).

Hi John,

You are right on track, I have to have a bit read to understand it but it looks good.

Its great to have someone to bounce ideas with

Thanks John

Best Regards

Joe

Hi John,

Yes it makes sense now that I have had a good read.

With regard to the material thickness, I thought that was the easy part, you have done the hard work already. If we subtract the material thickness from DXH within the formula, that should give us the true result for DY

Cheers

Joe

Unfortunately, you can't just subtract the material thickness, because the angle comes into it. Here's a view of the top tool in the die with the material sandwiched between:

Zooming in on the point where they meet:

... you can see that Tx is quite a bit bigger than the material thickness, T.

Zooming in on the bottom of the top tool:

... you can see that the depth must be decreased by the amount Ty. Ah is half the top tool angle.

To find Ty, consider this triangle:

sin(Ah) = T ÷ Ty, which can be re-arranged as:

Ty = T ÷ sin(Ah)

So the final entry depth is the depth as calculated in #51 minus Ty as calculated here. Remember Ah is half the top tool angle.

Hope this makes sense (and I haven't made any goofs ).

In fact you should take into consideration the true form of the tools this means the radii of knife and of the support sides. Depending of what you want to do the differences can be of importance. It is as well important to compute the changes in the depth as function of the angle opening due to spring back. This last value can be used for adaptive corrections and tolerances (angular) reduction.

You are dead right, depending on the angle of your tool will also have effect on the angle produced, again depending on the angle.

Good thinking Mate

Hi John,

I am going to have to think about that.

Cheers Mate

Joe

I've been out all day (up to London to get a visa for my daughter to go to China) - but I've thinking about it on-and-off.

I know that the angle matters (when the thickness of the material is introduced) - but I don't think I've made the correct assumptions in my calculaions (I think I said it wasn't the easy bit ).

So "I am going to have to think about that" too.

I'll be back!

Looks as if it works! Here are some constructions, done with the aid of (and dimensioned by) AutoCAD, with the calculations below:

Entry without material:

DY = (40 x 80) ÷ 277.13 = 11.55

Correction for material thickness:

Ty = 5 ÷ sin(60°) = 5 ÷ 0.87 = 5.77

Entry with material = 11.55 - 5.77 = 5.77

And:

Entry without material:

DY = (40 x 80) ÷ 92.38 = 34.64

Correction for material thickness:

Ty = 5 ÷ sin(30°) = 5 ÷ 0.5 = 10

Entry with material = 34.64 - 10 = 24.64

Hi John, That looks dam good, you know you trig.

I have to leave the New Zealand urgently, I will get back to you when I get back or I may reply while overseas if I get time, Thanks John.

Cheers

Joe

Simply draw your isosceles triangle then add a vertical line through the middle of your uneven angle to give you 2 right angle triangles. The uneven angle will be split exactly in 2 so you must remember to double your answer when you find it.

You know the height of the right angle triangle (X) an the base of the right angle triangle (0.5Y). The angle we want to find is (@).

Formula

SIN @ = 0.5Y / X

use this and then double the value of @ to give you your original uneven angle.

All the best

You have made it so simple, now all i have to do is see if my programme can handle sin.

Thanks

Guest's formula (using the sin() function) is OK assuming the X is measured along the side of the V (as per the diagram in #9). If you're using the depth of the V, you need the tan() function.

Also, to find the angle (from the sin or tan) you need the inverse function (asin or atan).

If your program can't handle trigonometric functions, you can always use the sum of the first seven or eight terms of the Taylor series expansion for the function, depending on your desired level of accuracy.

If your X measurement is the actual length of the equal sides (figure in #9), then the arcsin function should be used. The first seven terms of the Taylor series for arcsin are:

/ a = arcsin(w) = w + (1/6)w³ + (3/40)w⁵ + (5/12)w⁷ + (35/1152)w⁹ + (63/2816)w¹¹ + (231/13312)w¹³

where w = Y/2X from your original measurements and w<1.

This will calculate half of your desired angle in radians, so your final angle will be 2(arcsin(w)) or 2a. To convert radians to degrees, multiply by (180/pi). The accuracy of this series is within 0.5% for half-angles less than 65° (total 130°).

If your X measurement is the projected depth of the equal sides (figure in #11), then the arctan function should be used. The first eight terms of the Taylor series for arctan are:

/ b = arctan(w) = w – (1/3)w³ + (1/5)w⁵ – (1/7)w⁷ + (1/9)w⁹ – (1/11)w¹¹ + (1/13)w¹³ – (1/15)w¹⁵

where w = Y/2X from your original measurements and w<1.

This will also calculate half of your desired angle in radians, so your final angle will be 2(arctan(w)) or 2b. However, note that this series is only valid for w<1 which means the half-angle must be less than 45° (total 90°). The accuracy of this series is within 0.5% for half-angles less than 40° (total 80°).

Thanks Mate for your advise, But i have, after a bit of reasearch found that i can use the Tan, Cosine and Sine rules within my PLC. All the advise that I have received from this forum has been great.

I have also found out there is a calculation for the spring back in materials being folded, this does depend on a few factors, 1. Material Thickness 2. Material Type S/Steel v mild steel 3. Material Width. 4. Maximun Pressing Pressure.

There is a reasons for this problem, 1. the pressure that is pressed by the pressbarke is contant, on many pressbrakes they have stop nuts to limit the travel on the folding action this is also controled by hydralic pressure. I have also been told by a very experienced person that has manufactured pressbrakes that what happens is that if you use a high tonnage pressbrake to fold a thin material it should work every time if you are using the correct die.

But if you are trying to fold a thick material (towards the limit) on the pressbrake, what happens is that everything moves including the pressbrake, everything strecthes and then our calculations go out the window.

This is a charastic of pressbrakes, cant do much about this, but i can use all the information that i have received from this forum, and then include it to get reasonable angles.

Thanks All To Contributers

if you bisect the unequal angle, the resulting line will create two equal right triangles. just measure the isoceles side and the bisecting line, find their ratio and use trig to get half of the angle. multiply by two and you have the angle of the third corner of your isoceles triangle. Or you can measure the other two angles with a protractor and subtract that sum from 180. But, of course, if you have a protractor, you can simply measure the third

That's right thats is what i have to do, it dose still require the trig calc, and that is what i was having difficulity with.

Here is a link to a quality website where this guy has half a dozen nice programs that he wrote. Free and downloadable. The Geometry program is a comprehensive shape, area and volume calculator. Works nicely.

http://www.sjweeks.co.uk/

Sorry, i forgot to put the link in the last post.

U can try 2 solv this problm by juat incorporating a wheel in between the driving end cam and driven end connected at opp end of the tangent of the disk. The wheel should b rotating with center piviot to facilitate rotary as well as reciprocating motions symiltnsly. It will be something like excentrik wheel

I think I Know what you mean, but this requires another encoder, we dont want the extra compulaction, other theroy's will work just fine, thanks mate.