WEBVTT

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>> In the last part
of the lecture

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on descriptive geometry
it will cover the topics

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in Homework Number 16.

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The first part is on
angle between two planes.

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So you're given two angles
in space and you want a view

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that will show the angle
between the two planes

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and such a view would be a view

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that shows the two planes
simultaneously as edges.

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So knowing that in order to
find an edge of a plane we need

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to find the point view in
the plane finding a view

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that shows two planes
simultaneously

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as edge views involves
finding a line

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that is containing both planes
and finding the point view

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of that line and of course
that line that's common

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to both planes is the
line of intersection.

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So the steps would be first
find the line of intersection

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and we've talked about this

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in previous descriptive
geometry lectures.

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Once you find the line
of intersection you want

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to find the point
view of that line

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but before you find the point
view of any line you need

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to first find the true
length of that line,

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so in this case the
line of intersection.

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How do you do that?

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You take a reference plane
parallel to the line in any view

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and once you have
the true length

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of the line you take the
point view of the line

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by taking a references
plane perpendicular

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to the true length of the line.

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The point view of the line
of intersection will be shown

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and consequently the
two planes will be shown

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as their edge views.

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After you have those
planes and the edge views

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and you can see the angle
between the edge views you need

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to show the correct
visibility in all the planes

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because some part of one plane
may be able to cover the other,

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the other parts of the
plane, the other plane.

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The last topic is
on skewed lines.

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Skewed lines are lines that are
not parallel, not intersecting.

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So we know that when two lines
are parallel those two parallel

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lines define a single plane

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and when two lines are also
intersecting those two lines

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also define a single plane.

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So any pair of nonparallel,
nonintersecting lines are lines

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that are not in the same plane

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so a skewed length can
be defined as lines

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that are not co-planar
or don't lie

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in the same plane,
so that's a typo.

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The lines do not lie
in the same plane.

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Topics that will be covered here
- finding the shortest distance

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between the two skewed lines,

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so when two lines are
intersecting sometimes we're

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interested in the
shortest distance

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between them also shortest
horizontal distance

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between the lines and also
shortest distance along a

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given slope.

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So first part, finding
the shortest

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between skewed lines
there's two methods.

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The first one is
called the line method

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and it involves finding the
point view of one of the lines.

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So the shortest distance between
the two skewed lines can be

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showing the view wherein one

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of the lines is shown
as a point view.

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So what you need to do is
find the point view of one

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of the lines and the way
to do that is first try

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to find the true
length of that line

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by taking a reference plane
that's parallel to the line

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as a result of that
reference plane parallel

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to the line you will get the
true length of the line and then

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from the view that's
showing the tree length

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of that line you take the
reference plane perpendicular

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to that true length and
the line becomes point view

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and the shortest distance

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between the lines is the
perpendicular distance

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from the point view of the
line to the other line.

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And then once you find that
shortest distance you need

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to project those back to all

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of the views including
the original given views.

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That's called the line method

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because we're finding
the shortest distance

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between the skewed lines

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by finding the point
view of one of the lines.

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On the other hand,
there's another method

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of finding the shortest
distance between two lines

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by using what's called
the plane method.

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The idea behind this is
that the shortest distance

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between two skewed lines
can be seen in a view

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that shows the two of them
being parallel to each other,

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so any pair of skewed
lines or any pair

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of skewed lines it's
also possible to find

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at least one view where
the two lines appear

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to be apparently parallel and
that's what we're going to do.

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This view were in
the view parallel

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to each other can be
constructing a plane containing

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one of the lines
parallel to the other

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and then finding the
edge view of the plane.

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So the key here is we
create a fake plane,

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a plane that contains one of
the lines and then parallel

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to each other and once we have
that plane we find the edge view

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of that plane and the resulting
view will show the two lines

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appearing apparently
parallel to each other.

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So the steps are number one
construct a plane containing

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let's assume that the two lines
given are line A, B and line C,

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D. What we do is we construct
a plane containing line A,

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B and at the same time
parallel to line C,

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D okay and then once we have
that we find a horizontal.

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We do that by making sure that
the line is parallel to one

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of the edges of the plane that
we're constructing is parallel

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to the line C, D in both of
the views that are given.

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The length of Ax is
arbitrary but x in the top

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and front views should align.

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Here's an illustration of that.

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Okay, so here's the two
lines that are given here,

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top view and front views.

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A, B; C, D in the front; A, B;
C, D in the top so according

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to this we need to construct
a plane containing line A,

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B and parallel to C,
D and the way we do

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that is we construct line Ax
that is parallel to line C,

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D. So we construct this
line Ax parallel to C,

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D in the front view and the
corresponding Ax parallel

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to the same line C, D in the
top view so Ax is parallel to C,

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D in both views as
it does previously.

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The length of Ax is
arbitrary but the x in the top

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and front views should align.

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So the length of this Ax
that is parallel to C,

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D I can end that
anywhere I want.

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The x could have been
somewhere here or here

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or I selected it here but once I
defined the horizontal position

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of x okay, the length
of Ax in the front view

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that is also parallel to C,D in
the front view should correspond

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in the same vertical position
as horizontal position

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as the x in the top here.

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So this view here now shows a
plane, A, Bx that contains one

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of the original lines A, B and
parallel to the other line C,

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D. We know that the plane A, Bx
is parallel to C, D because one

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of its edges, namely
Ax is parallel to C,

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D both in the front
and the top views.

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It's very important that Ax
be parallel to same line Ax,

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B parallel to C, D.
Once we have that okay,

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we focus on the plane here.

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We find the edge
view of the plane A,

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Bx and how do you
find the edge view?

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We take a horizontal line here
let's call this point point y.

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Wherever y intersects Bx
okay we project that up

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and wherever it intersects
Bx that will be the y

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in the top view and this Ay

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in the top view becomes
true length.

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Once you have the true length

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of Ay we take a reference plane
here called ARP on this side

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and H on this side
that is perpendicular

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to that true length of Ay

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and then we project
all the points A, B; C,

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D including x. We want x. Okay,
but if we project x, A, B and A

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and B and we transfer the
measurements of height

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from the front view
so you have the height

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of C along the projection of C

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that defines the C. Here's
D. Measure this height of D

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from H will be the same as
this height of D from this H

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and then do the same thing
from A. Along the projection

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of A this distance here
is the same as this height

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of A below the horizontal
plane and then B

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as well here is the
projection, that distance

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of B along the projection
of B this side of the ARP.

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Anyway, that did
correctly number 1 plane A,

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B and x I didn't plot x here

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but x will be somewhere here
along this projection, okay.

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This distance here.

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So x will be here if I did
everything correctly A,

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B and x will line up.

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Now in this view here I'm
focusing on A, B and C,

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D. Now if I did everything
correctly lines A, B and B,

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C would be parallel
to each other.

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Okay, so these two lines now

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which are really
not truly parallel

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in reality we've managed to
find a view showing the two

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of them parallel to each
other because the plane A,

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Bx we use to find the edge
view of A, Bx is parallel to C,

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D therefore the edge view of
the plane which contains A,

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B as well will be parallel to C,

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D. And from here this
perpendicular distance

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between these two parallel lines
will be the shortest distance

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but of all of those possible
shortest distances between C,

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D and A, B, if you want to know
exactly where it is it's the one

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that was true length in this
view and the way we find

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which one it is we take an
auxiliary reference plane

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parallel to the two parallel
lines and hence perpendicular

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to the true length

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of the shortest distance
we're looking for.

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So we take a reference,
it's not shown here.

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We take a references plane here,
this way parallel to A, B and C,

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D and then we project both lines
A, B and C, D perpendicular

00:11:00.936 --> 00:11:04.416 A:middle
to that secondary
auxiliary reference plane

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and we construct A, B and C,
D in the second auxiliary view

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by transferring this distance
away from ARP or ARP 1 to find

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where they are in the
secondary auxiliary

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and then we should
see A, B and C,

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D in the secondary auxiliary
view intersecting each other

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and the shortest
distance between A, B; C,

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D will be the apparent
line of intersection

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in that auxiliary reference
plane, secondary auxiliary view.

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I will not show it
here because I will

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on the next page I will do the
problem for Homework Number 16

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which is finding the horizontal
distance between skewed lines.

00:11:53.426 --> 00:11:56.626 A:middle
So the steps in finding,
the first few steps

00:11:56.626 --> 00:11:59.506 A:middle
for finding the shortest
horizontal distance

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between skewed lines is
similar to the first few steps

00:12:02.836 --> 00:12:06.076 A:middle
of finding the distance,
shortest distance

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between the two lines.

00:12:08.276 --> 00:12:14.846 A:middle
So for the first problem back
here is the shortest distance is

00:12:15.156 --> 00:12:16.236 A:middle
along this direction,

00:12:16.236 --> 00:12:17.996 A:middle
the perpendicular
distance to the parallel.

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On the other hand,
if we're looking

00:12:19.356 --> 00:12:22.016 A:middle
for the shortest
horizontal distance it has

00:12:22.066 --> 00:12:26.686 A:middle
to be a line that's parallel to
the horizontal reference plane.

00:12:26.686 --> 00:12:27.496 A:middle
So the true length

00:12:27.496 --> 00:12:31.346 A:middle
of the shortest horizontal line
is going to be going this way

00:12:31.576 --> 00:12:35.666 A:middle
between A, B and C, D parallel
to x. We don't know which one

00:12:35.666 --> 00:12:39.076 A:middle
of those infinitely horizontal
lines will be the shortest,

00:12:39.366 --> 00:12:41.376 A:middle
so the way we will find
it is we are going to take

00:12:41.856 --> 00:12:47.636 A:middle
as our second auxiliary
reference plane 1 and ARP 2

00:12:47.636 --> 00:12:50.836 A:middle
that is perpendicular
to ARP 1 as opposed

00:12:50.836 --> 00:12:54.576 A:middle
to this problem here finding the
shortest distance our ARP 2 is

00:12:54.656 --> 00:12:58.146 A:middle
parallel to A, B
okay for the problem

00:12:58.146 --> 00:13:02.186 A:middle
of finding the shortest
horizontal distance our ARP 2 is

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perpendicular to initial ARP 1.

00:13:05.186 --> 00:13:08.716 A:middle
It can be perpendicular over
here or perpendicular over here.

00:13:08.716 --> 00:13:13.326 A:middle
It doesn't matter but we want to
find lines that are horizontal

00:13:13.326 --> 00:13:15.856 A:middle
and the horizontal lines
will add true lengths

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that are parallel to H here.

00:13:19.196 --> 00:13:21.406 A:middle
So the steps are the same
for the plane method.

00:13:21.406 --> 00:13:24.666 A:middle
Steps are similar to those

00:13:24.666 --> 00:13:26.776 A:middle
for shortest distance
except for step four.

00:13:26.836 --> 00:13:31.166 A:middle
So after step three here okay?

00:13:31.396 --> 00:13:34.086 A:middle
We will see that A, B and
C, D parallel to each other

00:13:34.086 --> 00:13:37.346 A:middle
so after step three this
will be exactly the same.

00:13:38.206 --> 00:13:41.406 A:middle
But as I said instead of
finding the shortest distance,

00:13:41.406 --> 00:13:44.006 A:middle
we're finding the shortest
horizontal distance.

00:13:44.006 --> 00:13:47.136 A:middle
So for step four to
find a point view

00:13:47.136 --> 00:13:51.356 A:middle
of the shortest horizontal
distance the next step is this.

00:13:51.456 --> 00:13:56.006 A:middle
We take an ARP which will
be our ARP 2 perpendicular

00:13:56.086 --> 00:14:02.286 A:middle
to the horizontal line
which is the H perpendicular

00:14:03.546 --> 00:14:06.956 A:middle
to the previous of
the reference plane.

00:14:06.956 --> 00:14:10.306 A:middle
The shortest horizontal distance
will appear true length the

00:14:10.306 --> 00:14:13.966 A:middle
resulting you and hence is
the apparent intersection

00:14:13.966 --> 00:14:16.616 A:middle
of lines A, B and C,
D. What does that mean?

00:14:16.766 --> 00:14:21.256 A:middle
So here's the continuation
for that problem.

00:14:21.256 --> 00:14:24.426 A:middle
So Homework Number 6 is this.

00:14:24.756 --> 00:14:29.986 A:middle
We are given 1, 2 and
A, B. Number 16 rather.

00:14:29.986 --> 00:14:33.896 A:middle
We're given two skewed lines
1, 2 and A, B as the top view.

00:14:34.306 --> 00:14:38.106 A:middle
1, 2; A, B, there's 1,
2, A, B in the front view

00:14:38.106 --> 00:14:40.956 A:middle
and we're asked to find the
shortest horizontal connection

00:14:41.376 --> 00:14:42.336 A:middle
between the two lines.

00:14:42.336 --> 00:14:48.726 A:middle
I think in the problem itself
the two lines are center lines

00:14:48.726 --> 00:14:50.006 A:middle
of freeways and we're trying

00:14:50.046 --> 00:14:53.556 A:middle
to find the shortest
horizontal on-ramp

00:14:53.686 --> 00:14:56.226 A:middle
that would connect
the two freeways.

00:14:57.076 --> 00:14:59.566 A:middle
So just like before the
first thing you need

00:14:59.596 --> 00:15:03.616 A:middle
to do is construct a
plane containing one

00:15:03.616 --> 00:15:05.416 A:middle
of the lines parallel
to the other.

00:15:05.586 --> 00:15:10.776 A:middle
So in this case we're
constructing the plane A,

00:15:10.946 --> 00:15:14.296 A:middle
Bx is parallel to 1, 2.

00:15:14.556 --> 00:15:19.346 A:middle
Why? Because we, SO the first
step is draw a horizontal,

00:15:19.646 --> 00:15:23.956 A:middle
draw a line parallel to 1,
2 passing through point A

00:15:24.096 --> 00:15:29.176 A:middle
so I draw this line here,
Ax parallel to 1, 2.

00:15:29.336 --> 00:15:30.756 A:middle
The length is arbitrary

00:15:31.326 --> 00:15:33.226 A:middle
and in the top view
I do the same thing.

00:15:33.226 --> 00:15:37.606 A:middle
I draw one, an Ax
parallel to 1, 2, as well.

00:15:38.066 --> 00:15:39.236 A:middle
So the starting point,

00:15:39.286 --> 00:15:44.806 A:middle
the starting lines are draw one
Ax parallel to 1, 2 in the front

00:15:45.126 --> 00:15:48.006 A:middle
and Ax parallel to
1, 2 in the top.

00:15:48.546 --> 00:15:53.326 A:middle
As I said earlier the actual
position of x doesn't matter

00:15:53.326 --> 00:15:56.736 A:middle
as long as it's along this
line but once I've selected x

00:15:56.736 --> 00:15:59.616 A:middle
in the top or the front
they should correspond

00:16:01.076 --> 00:16:02.476 A:middle
to each other, okay?

00:16:02.476 --> 00:16:05.876 A:middle
So here's the length that I
arbitrarily selected for Ax

00:16:05.876 --> 00:16:08.516 A:middle
so that x must be around here.

00:16:08.606 --> 00:16:12.406 A:middle
Then I will construct
my A, Bx, A,

00:16:12.406 --> 00:16:14.816 A:middle
Bx both in the front
and the top.

00:16:15.656 --> 00:16:19.736 A:middle
Next thing I do is I will try
to find edge view of that A, Bx.

00:16:19.736 --> 00:16:21.916 A:middle
So I'm focusing on A, Bx.

00:16:22.406 --> 00:16:23.886 A:middle
I temporarily ignore 1, 2.

00:16:24.476 --> 00:16:28.126 A:middle
To find the edge view of A,
Bx I look for it on the line.

00:16:29.146 --> 00:16:30.136 A:middle
Here's the horizontal line.

00:16:30.136 --> 00:16:34.506 A:middle
Let's call it Ay so
point y is on Bx.

00:16:34.556 --> 00:16:39.286 A:middle
I project that to Bx in the top
so this point here would be y

00:16:39.586 --> 00:16:42.576 A:middle
so this line A2y
will be true length

00:16:43.406 --> 00:16:46.586 A:middle
and what I do next is
perpendicular to its true length

00:16:46.586 --> 00:16:51.466 A:middle
and is my first projection
plane.

00:16:51.466 --> 00:16:55.036 A:middle
So this projection plane
here or reference plane is H

00:16:55.036 --> 00:16:58.366 A:middle
on this side for the top
view and ARP 1 on this side.

00:16:59.316 --> 00:17:07.066 A:middle
Then I project all the points
including A, B and the height

00:17:07.066 --> 00:17:11.026 A:middle
of A here I borrow from the
front view is the same height

00:17:11.076 --> 00:17:14.426 A:middle
here which is distance
from H here is the same

00:17:14.426 --> 00:17:19.506 A:middle
as this perpendicular distance
of A from H. Do the same thing

00:17:19.506 --> 00:17:23.196 A:middle
for B. This distance here will
be the same as this distance

00:17:23.566 --> 00:17:25.196 A:middle
and that will give me A,

00:17:25.196 --> 00:17:29.226 A:middle
B and then I will do the
same thing for line 1, 2.

00:17:29.226 --> 00:17:32.866 A:middle
Here are the projections
and the actual positions

00:17:32.866 --> 00:17:36.526 A:middle
of 1 will be defined by this
height here equal to this height

00:17:37.386 --> 00:17:40.466 A:middle
and then 2, this height
here will be equal

00:17:40.556 --> 00:17:41.916 A:middle
to the height here.

00:17:42.756 --> 00:17:46.456 A:middle
Now when I connect 1 and
2 and connect A and B

00:17:46.666 --> 00:17:50.686 A:middle
if I did everything correctly
those two lines 1, 2 and A,

00:17:50.686 --> 00:17:54.746 A:middle
B in the primary auxiliary
view should be parallel

00:17:54.746 --> 00:17:55.206 A:middle
to each other.

00:17:56.656 --> 00:18:02.666 A:middle
Okay? And as said here in
Step 4 the next step is

00:18:02.746 --> 00:18:05.996 A:middle
to take an auxiliary
reference plane perpendicular

00:18:06.076 --> 00:18:08.276 A:middle
to a horizontal line.

00:18:09.456 --> 00:18:14.766 A:middle
So take note that this line
here is the horizontal reference

00:18:14.806 --> 00:18:15.976 A:middle
plane, H on this side.

00:18:15.976 --> 00:18:20.646 A:middle
So on the first auxiliary
view ARP 1 any length parallel

00:18:20.646 --> 00:18:21.856 A:middle
to this will be horizontal,

00:18:21.956 --> 00:18:24.956 A:middle
so any length going
this way is horizontal.

00:18:25.976 --> 00:18:29.606 A:middle
So I'm trying to-- there are
infinitely many horizontal lines

00:18:30.056 --> 00:18:34.486 A:middle
connecting 1, 2 and A, B. I'm
trying to find which of all

00:18:34.526 --> 00:18:37.696 A:middle
of those is the shortest.

00:18:37.696 --> 00:18:40.086 A:middle
Now the shortest of all
of those lines is the one

00:18:40.086 --> 00:18:41.336 A:middle
that is already true
length here.

00:18:42.116 --> 00:18:43.566 A:middle
I don't know where
it is at this point.

00:18:43.566 --> 00:18:47.096 A:middle
O, P is not here initially
so what I do is I try to find

00:18:47.096 --> 00:18:50.646 A:middle
that both O by trying-- I know
the direction is going this way,

00:18:51.326 --> 00:18:52.496 A:middle
parallel to this horizontal.

00:18:52.496 --> 00:18:58.036 A:middle
I want to make that O, P become
a point view and then edge view

00:18:58.036 --> 00:19:00.996 A:middle
by taking a reference plane
perpendicular to the horizontal

00:19:00.996 --> 00:19:06.596 A:middle
so this reference plane here
ARP 2 is perfectly perpendicular

00:19:06.596 --> 00:19:07.546 A:middle
to ARP 1.

00:19:08.786 --> 00:19:11.716 A:middle
So the next step is take
this reference plane ARP 2

00:19:11.976 --> 00:19:16.056 A:middle
 perpendicular to ARP 1 then
project all the points 1 here,

00:19:16.996 --> 00:19:20.906 A:middle
2 here, A here and B here
all the projection lines

00:19:20.906 --> 00:19:22.776 A:middle
or construction lines
are perpendicular

00:19:22.776 --> 00:19:23.646 A:middle
to the reference plane.

00:19:23.646 --> 00:19:26.576 A:middle
By the way, there's
ARP 1 on this side

00:19:26.576 --> 00:19:27.836 A:middle
and ARP 2 on this side.

00:19:27.836 --> 00:19:30.956 A:middle
This whole view is
auxiliary view number 1.

00:19:31.336 --> 00:19:34.016 A:middle
So how do we find
say the location

00:19:34.016 --> 00:19:37.506 A:middle
of point 1 away from the ARP 1?

00:19:37.666 --> 00:19:39.516 A:middle
Well we borrow that
from this ARP.

00:19:39.546 --> 00:19:43.226 A:middle
We said this ARP 1
corresponds with to ARP 1

00:19:43.226 --> 00:19:47.906 A:middle
so the distance away from
ARP 1 were the top view,

00:19:47.906 --> 00:19:52.506 A:middle
this distance here is
the distance of point 1

00:19:52.506 --> 00:19:56.596 A:middle
from ARP 1 should be exactly
the same as this distance

00:19:57.026 --> 00:20:00.346 A:middle
which is again the
distance of 1 from ARP 1.

00:20:00.346 --> 00:20:05.456 A:middle
So I transfer this distance
from ARP 1 along the projection

00:20:05.456 --> 00:20:07.676 A:middle
of 1 for that distance.

00:20:07.886 --> 00:20:08.886 A:middle
I do the same thing.

00:20:08.886 --> 00:20:11.326 A:middle
Here is the distance
of 2 A from ARP 1.

00:20:11.326 --> 00:20:13.376 A:middle
I transfer that along
the projection of 2

00:20:13.376 --> 00:20:14.526 A:middle
so that's this distance.

00:20:14.956 --> 00:20:19.816 A:middle
Same thing, here is the
projection of A is the distance

00:20:19.816 --> 00:20:22.446 A:middle
of A from ARP 1, so along
the projection of A,

00:20:22.866 --> 00:20:26.116 A:middle
I measure this perpendicular
distance are from ARP 1.

00:20:26.116 --> 00:20:28.246 A:middle
That will define
A. Same thing here,

00:20:28.246 --> 00:20:30.586 A:middle
this distance here
will be the same

00:20:30.586 --> 00:20:34.016 A:middle
as this distance here
along the construction line

00:20:34.016 --> 00:20:39.506 A:middle
for B. Then I connect A,
B. I also connect 1 to 2.

00:20:39.626 --> 00:20:41.426 A:middle
Now the apparent
intersection here

00:20:41.936 --> 00:20:45.846 A:middle
if the shortest horizontal
distance are connecting lines

00:20:45.846 --> 00:20:46.576 A:middle
that we're looking for.

00:20:47.006 --> 00:20:50.066 A:middle
So I'm going to call that
O, P and this apparent,

00:20:50.066 --> 00:20:52.216 A:middle
this section here I
project that here.

00:20:52.756 --> 00:20:56.246 A:middle
That will be the true length of
the shortest horizontal distance

00:20:56.706 --> 00:21:01.416 A:middle
from line A, B to line 1, 2
and then if I want to find what

00:21:01.566 --> 00:21:09.016 A:middle
where O, P is in those front
view I just project P into 1,

00:21:09.096 --> 00:21:15.736 A:middle
2 and O into A, B and connect
them here and find that O,

00:21:15.836 --> 00:21:17.666 A:middle
P in the front view
I'd do the same thing.

00:21:18.216 --> 00:21:20.476 A:middle
The O here we'll be
projected it to A,

00:21:20.476 --> 00:21:25.216 A:middle
B and the P here will
be projected to 1, 2.

00:21:26.096 --> 00:21:28.196 A:middle
Okay, and then the two lengths

00:21:28.196 --> 00:21:34.936 A:middle
as I said will be this while
I think the problem also asks

00:21:34.936 --> 00:21:36.886 A:middle
for the compass bearing, so
whatever the line is here

00:21:37.796 --> 00:21:38.776 A:middle
in the front view you need

00:21:38.856 --> 00:21:43.286 A:middle
to find its direction the way
you find directions of any line,

00:21:43.846 --> 00:21:45.016 A:middle
always from the top view.

00:21:45.326 --> 00:21:50.036 A:middle
All the directions come from
the top view so if P or O,

00:21:50.036 --> 00:21:54.176 A:middle
P appears something like this
between this measure it's angle

00:21:54.176 --> 00:21:58.056 A:middle
from the south and measure
whether it's going east or west

00:21:58.056 --> 00:22:01.156 A:middle
and you go north whatever
the angle is, east.

00:22:02.666 --> 00:22:05.196 A:middle
And hopefully that's
enough to get you going

00:22:05.196 --> 00:22:06.696 A:middle
for Homework Number 16.