>> In the last part of the lecture
on descriptive geometry it will cover the topics
in Homework Number 16.
The first part is on angle between two planes.
So you're given two angles in space and you want a view
that will show the angle between the two planes
and such a view would be a view
that shows the two planes simultaneously as edges.
So knowing that in order to find an edge of a plane we need
to find the point view in the plane finding a view
that shows two planes simultaneously
as edge views involves finding a line
that is containing both planes and finding the point view
of that line and of course that line that's common
to both planes is the line of intersection.
So the steps would be first find the line of intersection
and we've talked about this
in previous descriptive geometry lectures.
Once you find the line of intersection you want
to find the point view of that line
but before you find the point view of any line you need
to first find the true length of that line,
so in this case the line of intersection.
How do you do that?
You take a reference plane parallel to the line in any view
and once you have the true length
of the line you take the point view of the line
by taking a references plane perpendicular
to the true length of the line.
The point view of the line of intersection will be shown
and consequently the two planes will be shown
as their edge views.
After you have those planes and the edge views
and you can see the angle between the edge views you need
to show the correct visibility in all the planes
because some part of one plane may be able to cover the other,
the other parts of the plane, the other plane.
The last topic is on skewed lines.
Skewed lines are lines that are not parallel, not intersecting.
So we know that when two lines are parallel those two parallel
lines define a single plane
and when two lines are also intersecting those two lines
also define a single plane.
So any pair of nonparallel, nonintersecting lines are lines
that are not in the same plane
so a skewed length can be defined as lines
that are not co-planar or don't lie
in the same plane, so that's a typo.
The lines do not lie in the same plane.
Topics that will be covered here - finding the shortest distance
between the two skewed lines,
so when two lines are intersecting sometimes we're
interested in the shortest distance
between them also shortest horizontal distance
between the lines and also shortest distance along a
given slope.
So first part, finding the shortest
between skewed lines there's two methods.
The first one is called the line method
and it involves finding the point view of one of the lines.
So the shortest distance between the two skewed lines can be
showing the view wherein one
of the lines is shown as a point view.
So what you need to do is find the point view of one
of the lines and the way to do that is first try
to find the true length of that line
by taking a reference plane that's parallel to the line
as a result of that reference plane parallel
to the line you will get the true length of the line and then
from the view that's showing the tree length
of that line you take the reference plane perpendicular
to that true length and the line becomes point view
and the shortest distance
between the lines is the perpendicular distance
from the point view of the line to the other line.
And then once you find that shortest distance you need
to project those back to all
of the views including the original given views.
That's called the line method
because we're finding the shortest distance
between the skewed lines
by finding the point view of one of the lines.
On the other hand, there's another method
of finding the shortest distance between two lines
by using what's called the plane method.
The idea behind this is that the shortest distance
between two skewed lines can be seen in a view
that shows the two of them being parallel to each other,
so any pair of skewed lines or any pair
of skewed lines it's also possible to find
at least one view where the two lines appear
to be apparently parallel and that's what we're going to do.
This view were in the view parallel
to each other can be constructing a plane containing
one of the lines parallel to the other
and then finding the edge view of the plane.
So the key here is we create a fake plane,
a plane that contains one of the lines and then parallel
to each other and once we have that plane we find the edge view
of that plane and the resulting view will show the two lines
appearing apparently parallel to each other.
So the steps are number one construct a plane containing
let's assume that the two lines given are line A, B and line C,
D. What we do is we construct a plane containing line A,
B and at the same time parallel to line C,
D okay and then once we have that we find a horizontal.
We do that by making sure that the line is parallel to one
of the edges of the plane that we're constructing is parallel
to the line C, D in both of the views that are given.
The length of Ax is arbitrary but x in the top
and front views should align.
Here's an illustration of that.
Okay, so here's the two lines that are given here,
top view and front views.
A, B; C, D in the front; A, B; C, D in the top so according
to this we need to construct a plane containing line A,
B and parallel to C, D and the way we do
that is we construct line Ax that is parallel to line C,
D. So we construct this line Ax parallel to C,
D in the front view and the corresponding Ax parallel
to the same line C, D in the top view so Ax is parallel to C,
D in both views as it does previously.
The length of Ax is arbitrary but the x in the top
and front views should align.
So the length of this Ax that is parallel to C,
D I can end that anywhere I want.
The x could have been somewhere here or here
or I selected it here but once I defined the horizontal position
of x okay, the length of Ax in the front view
that is also parallel to C,D in the front view should correspond
in the same vertical position as horizontal position
as the x in the top here.
So this view here now shows a plane, A, Bx that contains one
of the original lines A, B and parallel to the other line C,
D. We know that the plane A, Bx is parallel to C, D because one
of its edges, namely Ax is parallel to C,
D both in the front and the top views.
It's very important that Ax be parallel to same line Ax,
B parallel to C, D. Once we have that okay,
we focus on the plane here.
We find the edge view of the plane A,
Bx and how do you find the edge view?
We take a horizontal line here let's call this point point y.
Wherever y intersects Bx okay we project that up
and wherever it intersects Bx that will be the y
in the top view and this Ay
in the top view becomes true length.
Once you have the true length
of Ay we take a reference plane here called ARP on this side
and H on this side that is perpendicular
to that true length of Ay
and then we project all the points A, B; C,
D including x. We want x. Okay, but if we project x, A, B and A
and B and we transfer the measurements of height
from the front view so you have the height
of C along the projection of C
that defines the C. Here's D. Measure this height of D
from H will be the same as this height of D from this H
and then do the same thing from A. Along the projection
of A this distance here is the same as this height
of A below the horizontal plane and then B
as well here is the projection, that distance
of B along the projection of B this side of the ARP.
Anyway, that did correctly number 1 plane A,
B and x I didn't plot x here
but x will be somewhere here along this projection, okay.
This distance here.
So x will be here if I did everything correctly A,
B and x will line up.
Now in this view here I'm focusing on A, B and C,
D. Now if I did everything correctly lines A, B and B,
C would be parallel to each other.
Okay, so these two lines now
which are really not truly parallel
in reality we've managed to find a view showing the two
of them parallel to each other because the plane A,
Bx we use to find the edge view of A, Bx is parallel to C,
D therefore the edge view of the plane which contains A,
B as well will be parallel to C,
D. And from here this perpendicular distance
between these two parallel lines will be the shortest distance
but of all of those possible shortest distances between C,
D and A, B, if you want to know exactly where it is it's the one
that was true length in this view and the way we find
which one it is we take an auxiliary reference plane
parallel to the two parallel lines and hence perpendicular
to the true length
of the shortest distance we're looking for.
So we take a reference, it's not shown here.
We take a references plane here, this way parallel to A, B and C,
D and then we project both lines A, B and C, D perpendicular
to that secondary auxiliary reference plane
and we construct A, B and C, D in the second auxiliary view
by transferring this distance away from ARP or ARP 1 to find
where they are in the secondary auxiliary
and then we should see A, B and C,
D in the secondary auxiliary view intersecting each other
and the shortest distance between A, B; C,
D will be the apparent line of intersection
in that auxiliary reference plane, secondary auxiliary view.
I will not show it here because I will
on the next page I will do the problem for Homework Number 16
which is finding the horizontal distance between skewed lines.
So the steps in finding, the first few steps
for finding the shortest horizontal distance
between skewed lines is similar to the first few steps
of finding the distance, shortest distance
between the two lines.
So for the first problem back here is the shortest distance is
along this direction,
the perpendicular distance to the parallel.
On the other hand, if we're looking
for the shortest horizontal distance it has
to be a line that's parallel to the horizontal reference plane.
So the true length
of the shortest horizontal line is going to be going this way
between A, B and C, D parallel to x. We don't know which one
of those infinitely horizontal lines will be the shortest,
so the way we will find it is we are going to take
as our second auxiliary reference plane 1 and ARP 2
that is perpendicular to ARP 1 as opposed
to this problem here finding the shortest distance our ARP 2 is
parallel to A, B okay for the problem
of finding the shortest horizontal distance our ARP 2 is
perpendicular to initial ARP 1.
It can be perpendicular over here or perpendicular over here.
It doesn't matter but we want to find lines that are horizontal
and the horizontal lines will add true lengths
that are parallel to H here.
So the steps are the same for the plane method.
Steps are similar to those
for shortest distance except for step four.
So after step three here okay?
We will see that A, B and C, D parallel to each other
so after step three this will be exactly the same.
But as I said instead of finding the shortest distance,
we're finding the shortest horizontal distance.
So for step four to find a point view
of the shortest horizontal distance the next step is this.
We take an ARP which will be our ARP 2 perpendicular
to the horizontal line which is the H perpendicular
to the previous of the reference plane.
The shortest horizontal distance will appear true length the
resulting you and hence is the apparent intersection
of lines A, B and C, D. What does that mean?
So here's the continuation for that problem.
So Homework Number 6 is this.
We are given 1, 2 and A, B. Number 16 rather.
We're given two skewed lines 1, 2 and A, B as the top view.
1, 2; A, B, there's 1, 2, A, B in the front view
and we're asked to find the shortest horizontal connection
between the two lines.
I think in the problem itself the two lines are center lines
of freeways and we're trying
to find the shortest horizontal on-ramp
that would connect the two freeways.
So just like before the first thing you need
to do is construct a plane containing one
of the lines parallel to the other.
So in this case we're constructing the plane A,
Bx is parallel to 1, 2.
Why? Because we, SO the first step is draw a horizontal,
draw a line parallel to 1, 2 passing through point A
so I draw this line here, Ax parallel to 1, 2.
The length is arbitrary
and in the top view I do the same thing.
I draw one, an Ax parallel to 1, 2, as well.
So the starting point,
the starting lines are draw one Ax parallel to 1, 2 in the front
and Ax parallel to 1, 2 in the top.
As I said earlier the actual position of x doesn't matter
as long as it's along this line but once I've selected x
in the top or the front they should correspond
to each other, okay?
So here's the length that I arbitrarily selected for Ax
so that x must be around here.
Then I will construct my A, Bx, A,
Bx both in the front and the top.
Next thing I do is I will try to find edge view of that A, Bx.
So I'm focusing on A, Bx.
I temporarily ignore 1, 2.
To find the edge view of A, Bx I look for it on the line.
Here's the horizontal line.
Let's call it Ay so point y is on Bx.
I project that to Bx in the top so this point here would be y
so this line A2y will be true length
and what I do next is perpendicular to its true length
and is my first projection plane.
So this projection plane here or reference plane is H
on this side for the top view and ARP 1 on this side.
Then I project all the points including A, B and the height
of A here I borrow from the front view is the same height
here which is distance from H here is the same
as this perpendicular distance of A from H. Do the same thing
for B. This distance here will be the same as this distance
and that will give me A,
B and then I will do the same thing for line 1, 2.
Here are the projections and the actual positions
of 1 will be defined by this height here equal to this height
and then 2, this height here will be equal
to the height here.
Now when I connect 1 and 2 and connect A and B
if I did everything correctly those two lines 1, 2 and A,
B in the primary auxiliary view should be parallel
to each other.
Okay? And as said here in Step 4 the next step is
to take an auxiliary reference plane perpendicular
to a horizontal line.
So take note that this line here is the horizontal reference
plane, H on this side.
So on the first auxiliary view ARP 1 any length parallel
to this will be horizontal,
so any length going this way is horizontal.
So I'm trying to-- there are infinitely many horizontal lines
connecting 1, 2 and A, B. I'm trying to find which of all
of those is the shortest.
Now the shortest of all of those lines is the one
that is already true length here.
I don't know where it is at this point.
O, P is not here initially so what I do is I try to find
that both O by trying-- I know the direction is going this way,
parallel to this horizontal.
I want to make that O, P become a point view and then edge view
by taking a reference plane perpendicular to the horizontal
so this reference plane here ARP 2 is perfectly perpendicular
to ARP 1.
So the next step is take this reference plane ARP 2
perpendicular to ARP 1 then project all the points 1 here,
2 here, A here and B here all the projection lines
or construction lines are perpendicular
to the reference plane.
By the way, there's ARP 1 on this side
and ARP 2 on this side.
This whole view is auxiliary view number 1.
So how do we find say the location
of point 1 away from the ARP 1?
Well we borrow that from this ARP.
We said this ARP 1 corresponds with to ARP 1
so the distance away from ARP 1 were the top view,
this distance here is the distance of point 1
from ARP 1 should be exactly the same as this distance
which is again the distance of 1 from ARP 1.
So I transfer this distance from ARP 1 along the projection
of 1 for that distance.
I do the same thing.
Here is the distance of 2 A from ARP 1.
I transfer that along the projection of 2
so that's this distance.
Same thing, here is the projection of A is the distance
of A from ARP 1, so along the projection of A,
I measure this perpendicular distance are from ARP 1.
That will define A. Same thing here,
this distance here will be the same
as this distance here along the construction line
for B. Then I connect A, B. I also connect 1 to 2.
Now the apparent intersection here
if the shortest horizontal distance are connecting lines
that we're looking for.
So I'm going to call that O, P and this apparent,
this section here I project that here.
That will be the true length of the shortest horizontal distance
from line A, B to line 1, 2 and then if I want to find what
where O, P is in those front view I just project P into 1,
2 and O into A, B and connect them here and find that O,
P in the front view I'd do the same thing.
The O here we'll be projected it to A,
B and the P here will be projected to 1, 2.
Okay, and then the two lengths
as I said will be this while I think the problem also asks
for the compass bearing, so whatever the line is here
in the front view you need
to find its direction the way you find directions of any line,
always from the top view.
All the directions come from the top view so if P or O,
P appears something like this between this measure it's angle
from the south and measure whether it's going east or west
and you go north whatever the angle is, east.
And hopefully that's enough to get you going
for Homework Number 16.