Tech Drawing (H)

Feral Mutant

Raphael wrote:

Them thingies. You just bend it to get the curve that fits your points.

Some guys in my class have those 'cept they call it a jackerflex.

Raphael

Funkstard wrote:

Yeh that's the same curve I & all my class got. I'm usually good at the theorem, but I just couldn't get it started today. How did you start it?

And about the spiral, I don't know which way I did it...I got the first and second length, swung up to a 30degree line, swung it back down and repeated it, each length getting bigger each time

Area's of figures? Call the point outside the circle x. You know that the distance from |xC|.|xA|=|xD|.|xD| , which let's you get A and C

Now, if you go off at 20 degrees from A and C you can get O (angle at the centre = twice angle at the circumfrence. Draw the circle.

D you can get by swinging an arc length 120 from x and getting where it hits the circle.

You know the area of ABC is half ADC. since the base is common, that means the height is half. So half the distance from D to AC, and then draw a light line that distance from AC on the other side. Where it hits the circle on the right is B

Feral Mutant wrote:

Some guys in my class have those 'cept they call it a jackerflex.

A......jackerflex? They're clearly idiots, you;re going to have to kill them for the good of the human race.

Funkstard

Right I'll have to get it out now and try it again. The rest of it seemed fine, I just couldn't get it started & already had 4 questions done so I just left it

"You know that the distance from |xC|.|xA|=|xD|.|xD| , which let's you get A and C"

How do you know that? It doesn't say it's a similar triangle or anything..

sticky9cd

Raphael wrote:

Yeah, you just draw an auxiliary view with DE as a point, then join from there perpindicularly to AB. Get the point where it hits AB in plan, and because it was a true length in the auxiliary then the line is parallell to the X1Y1

yes - got that

Raphael

It's a formula about circles. Like the one with crossing over chords.

Edit: "If a secant and a tangent are drawn to a circle from an external point, the product of the secant and its external segment equals the tangent squared"

sticky9cd

this is what i got for q3

Funkstard

Oh right. Never learnt that! Haven't seen it in a pass paper either

Raphael

...it was in last years paper

Edit: There's another solution for last years, but it's also in '02 and '00 with no obvvious alternative solution.

Funkstard

I must have done it a different way 'cos I didn't use it doing that question.

whassupp2

sticky9cd wrote:

this is what i got for q3

A doesnt overlap with the cone and you left out the other cylinder.

Nichololas

Can anyone give me a quick run-down of dihedral angles and getting auxiliary points in perspective? (I missed the 3 classes when we did them. )

eamoss

Im prob the best person to be expaining this but anyway I will give it a go. Also I have very very bad english so try & make out what I am saying

dihedral angles
how to find D.A

Red Line = Draw a line which touches both sides of the roof

Green Line = Fold down the hip rafter

Then when the green line hits the red line come off perpendicular to the top of the hip rafter(shown by the blue line).

Then with ur compass using your two points(both ends of the blue line) bring it down till it hits the bottom of the rafter(purple line).

Then with the two points from the red line join them up with the line you just took down(yellow line)

The gray line is the d.a

Auxiliary V.P

Raphael

Hang on a sec, the blue line should be where the red line hits the base, shouldn't it? By the base I mean the other green line that you flattened the hip rafter around

Nichololas

What does folding down the hip rafter entail? I can't figure it out from the diagram.

eamoss

Raphael wrote:

Hang on a sec, the blue line should be where the red line hits the base, shouldn't it? By the base I mean the other green line that you flattened the hip rafter around

Whoops, yep ur right.

Enlil_Nick wrote:

What does folding down the hip rafter entail? I can't figure it out from the diagram.

Its basically just looking along so that you can see the true height of the hip rafter.

Raphael

Just rebatting it around.

Imagine the hip rafter as being a trap door that's open, what you're doing is pushing it so it falls closed

eamoss: oh good, was afraid I'd been doing it wrong ^^;

Nichololas

Oops, I get it now. I don't usually do it superimposed onto the plan also.

whassupp2

Cheers fo the info. I'll print that and learn it. By the way, did u draw them in MS Paint and how did you get them on screen without using an attatchment???

Raphael

www.photobucket.com, or somesuch

sticky9cd

whassupp2 wrote:

A doesnt overlap with the cone and you left out the other cylinder.

ya i drew it wrong but at least i know you realised what i had wrong. i done it right in the exam(i hope):D

eamoss

whassupp2 wrote:

Cheers fo the info. I'll print that and learn it. By the way, did u draw them in MS Paint and how did you get them on screen without using an attatchment???

Yeah I did draw that on MS Paint. When it comes to drawing lines I wouldnt use PSP or PS.

All you need a sever to host ur photos/images I have my own sever but I use other severs to host images(like www.imageshack.us ) for space & its just easier(means I dont have to CHMOD the images) there is a few out there

Also remember I made a mistake in drawin the d.a so do wat Raphael said.