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Cessna Skyhawk C-172, by Textron Aviation
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by
Michael Kennedy
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Updated Mar 31, 2022
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Chapters
00:00
Chapter 1
00:22
Chapter 2
00:55
Chapter 3
01:34
Chapter 4
01:35
Chapter 5
01:43
Chapter 6
01:46
Chapter 7
01:48
Chapter 8
02:05
Chapter 9
02:58
Chapter 10
03:01
Chapter 11
04:11
Chapter 12
04:12
Chapter 13
05:38
Chapter 14
05:40
Chapter 15
05:41
Chapter 16
06:09
Chapter 17
07:12
Chapter 18
07:13
Chapter 19
07:57
Chapter 20
09:49
Chapter 21
09:50
Chapter 22
10:49
Chapter 23
10:51
Chapter 24
12:55
Chapter 25
12:57
Chapter 26
14:13
Chapter 27
14:15
Chapter 28
Audio Transcript
Auto-generated
00:00 - 00:05
Introducing the Cessna Skyhawk C172 by Textron Aviation.
00:06 - 00:12
This is part of a S. C I 309 aerodynamics with Butch Bennett for the March 2022 term.
00:12 - 00:16
This particular module will look at the aircraft selection, airfoils,
00:16 - 00:21
research and the physical laws associated with computing.
00:21 - 00:25
Airfoil performance. Let's introduce the C 1 72 Skyhawk.
00:26 - 00:30
This is a four seat single engine aircraft traditionally used for training here
00:30 - 00:35
in the United States and has been in continuous production since 1948.
00:36 - 00:40
This particular Skyhawk offers a very simplistic fight flight characteristics
00:41 - 00:45
including now updated avionics, including the G 1000 series,
00:46 - 00:47
lower landing speeds,
00:47 - 00:48
lenient stall speeds,
00:48 - 00:52
which will address through some performance calculations later and
00:52 - 00:55
they're the ideal aerodynamics for a training aircraft.
00:55 - 00:55
Today.
00:56 - 01:01
The general aircraft characteristics are very basic. It is a four seat aircraft.
01:01 - 01:02
One crew required.
01:02 - 01:05
The length is relatively small at 27 ft two inches
01:05 - 01:09
with a wingspan of 174 square feet for the wing area
01:10 - 01:15
aspect ratio is 732. Very important to this research is the airfoil type.
01:15 - 01:21
The knack of 24 12 with our empty weight being £1680 although in
01:21 - 01:27
most of our calculations will be using a max gross weight of £2550.
01:27 - 01:29
Making this our Maxwell useful load
01:29 - 01:35
of £878 and a max payload of £870.
01:35 - 01:39
The performance of the aircraft is relatively straightforward maximum cruise.
01:39 - 01:43
About a 124 max cruise speed in a trainer.
01:44 - 01:46
Let's go ahead now and jump into some of
01:46 - 01:50
the resource requirements for airfoil and the behavior.
01:50 - 01:53
The main airfoil data shows this to
01:53 - 01:55
be a relatively straightforward training aircraft.
01:56 - 01:59
The airfoil tools were used to be able to define the cord,
01:59 - 02:04
the camera and the upper and lower surfaces of the 24-12 airfoil service.
02:04 - 02:06
Let's look into the angle of attack.
02:07 - 02:09
Now we selected to generate a data model.
02:09 - 02:12
We used Microsoft sequel server to go ahead and
02:12 - 02:15
load in all the naka data across all aircraft
02:15 - 02:22
types so that we could look at the stall AOA A the CL levels by Naca type.
02:22 - 02:25
When we looked at the max C. L. And the C. L.
02:25 - 02:30
For minimum here to evaluate the AOA and Cl for the Skyhawk.
02:30 - 02:35
We came out with the stall A. O. A at 16.5 degrees.
02:35 - 02:38
This would be the critical angle of attack in which an aircraft
02:38 - 02:43
would stall again and will reference this in our performance calculations later.
02:43 - 02:45
This is a respective of weight
02:45 - 02:46
in speed
02:46 - 02:47
and altitude
02:48 - 02:51
on the critical angle of attack which is at the C. L.
02:51 - 03:00
Max of 1.582 and r zero lift A OA is roughly a negative 2.250 degrees.
03:00 - 03:03
Now let's look at some airfoil comparisons.
03:03 - 03:06
Not part of the assignment but I thought this might be keen
03:06 - 03:09
to look at in terms of assessment products as a whole.
03:09 - 03:10
As we start to define.
03:10 - 03:13
Looking at the airfoils and the AOA
03:13 - 03:16
characteristics around the airfoils for Cessna products.
03:16 - 03:21
But I wanted to bring this one up relatively quick to show the difference between
03:21 - 03:27
transitioning pilots for a jet versus those who come from the piston 1 72 world.
03:27 - 03:31
You can see the overlay of that airfoil with respect to chord line.
03:31 - 03:33
In Canberra is almost identical.
03:33 - 03:35
We went ahead and used the data model
03:35 - 03:39
that we had selected previously which the knack to type here for
03:39 - 03:43
the citation one Cessna product is the 2301 to knock a type
03:43 - 03:46
but you'll see that the stall, the C. L.
03:46 - 03:52
Max and the zero lip Aowei characteristics are almost identical to a Cessna 1
03:52 - 03:58
72 with remarkably small differences noted in red between the stall the C.
03:58 - 04:00
L. Max and the zero lift a away.
04:00 - 04:03
This provides trainers the ability to
04:03 - 04:05
move from training aircraft within the Cessna
04:05 - 04:10
suite of products to aerodynamically similar aircraft
04:10 - 04:12
here represented in their jet sweets.
04:13 - 04:13
Now,
04:13 - 04:16
as the requirements of this assignment requires
04:16 - 04:18
us to look at the symmetrical airfoil,
04:18 - 04:25
I went ahead and loaded a data model again for the 24 12 Naca type on the Cessna 1 72
04:25 - 04:30
and the very first Naca type for a symmetrical airfoil,
04:30 - 04:35
you'll see the airfoil down there below in red which is the 24 12 Cessna 1
04:35 - 04:43
72 and the purple which is the F 1 11 or 006 symmetrical airfoil type.
04:43 - 04:47
You can see in the data model that the stall AOA A. The C. L.
04:47 - 04:50
Max and men are very different based on the type
04:50 - 04:52
of naka selected in the data model.
04:53 - 04:55
There are some rough differences in terms
04:55 - 04:59
of the symmetrical airfoil which distinguishes itself different
04:59 - 05:04
from what we see in the 24-12 with the identical upper and lower services.
05:04 - 05:07
You can notice that here in the lower. Reference in purple
05:07 - 05:10
there is no zero lift a away due to the
05:10 - 05:13
cord line and the mean camera line being identical.
05:14 - 05:17
If you look at the data model to the left and reference the blue,
05:18 - 05:22
excuse me, the purple arrows, you'll see the very bottom one where the zero A.
05:22 - 05:25
Oa indicates zero lift.
05:25 - 05:28
Now the non symmetrical to 4 12 naka provides
05:28 - 05:30
greater lift efficiency at any given a o.
05:30 - 05:31
A.
05:31 - 05:33
With a better lift to drag ratio,
05:33 - 05:37
with improved stall characteristics for a non symmetrical
05:38 - 05:39
airfoil wing.
05:39 - 05:43
Alright, let's continue on and look at some of the calculated behaviors
05:43 - 05:46
that we have for the Cessna 1 72 aircraft.
05:47 - 05:49
Many of the performance and limitations of the
05:49 - 05:52
skyhawk are provided within the flight operations,
05:52 - 05:55
manual of the pilot operating handbook but we're going to confirm
05:55 - 05:57
some of those performance factors by
05:57 - 06:00
looking at specific calculations that address
06:00 - 06:01
speed,
06:01 - 06:06
wait altitude and temperature to arrive at calculated stall
06:06 - 06:10
speeds and lift deficiencies for the Cessna 1 72
06:11 - 06:14
here in the calculated stall speed.
06:14 - 06:21
We're gonna go ahead and address this from a gross weight perspective of £2550.
06:21 - 06:25
The wing area. The Cessna 1 72 is 174 square feet.
06:25 - 06:29
And in this demonstration we're going to look at the calculated stall speed
06:29 - 06:30
At c. l.
06:30 - 06:37
max at 1.58 To this would be the 16.5 critical angle of attack for this aircraft.
06:37 - 06:39
We're going to do this at an altitude of
06:39 - 06:44
5000 MSL which gives us a density ratio of 50000.8617
06:44 - 06:46
And the stadium.
06:46 - 06:48
The standard data points here used at 2992.
06:49 - 06:54
We use the calculated stall stall speed equation here which then gives
06:54 - 06:59
us a stall speed of 56.32 indicated on our air speed indicator.
06:59 - 07:04
Now we're going to reference the calculated stall speed later when we start to look
07:04 - 07:08
at the cl differences between cl cl max and how those might exceed C.
07:08 - 07:09
L. Max
07:09 - 07:12
and what contributions they have to stall speed.
07:12 - 07:13
In a later slide
07:13 - 07:16
Let's go ahead and jump into the lift efficiency.
07:16 - 07:20
We have four conditions in which were required to examine lift deficiency
07:20 - 07:25
for this particular aircraft and demonstrate the capabilities of the Cessna 172.
07:26 - 07:30
Condition one looks at how lift changes with their speed.
07:30 - 07:33
You have a constant AO AO AO and altitude
07:33 - 07:38
are held condition to how lift changes with altitude.
07:38 - 07:40
If the constant AOA and airspeed are held.
07:40 - 07:42
Whereas condition three is,
07:42 - 07:47
how do the required cln ao a for the Skyhawk change with changes in air speed.
07:48 - 07:52
And finally what happens if the required cl is larger than the C. L.
07:52 - 07:55
Max of the Skyhawk airfoil which we referenced just a
07:55 - 07:57
minute ago as we looked at the stall speed.
07:57 - 07:59
Let's look at condition one here.
08:00 - 08:05
Condition one requires us to look at the change with their speed.
08:05 - 08:08
If a constant AOA announced that are held with respect to lift
08:08 - 08:09
here,
08:09 - 08:12
we're going to have the skyhawk traveling initially at 55
08:12 - 08:15
knots and then have its speed up to 110 knots.
08:15 - 08:17
For this particular example,
08:17 - 08:25
I am using max cl cl max at the 1.582 which would be that 16.5 degree angle of attack.
08:25 - 08:27
The critical angle of attack.
08:27 - 08:32
I'm doing this for two purposes to demonstrate here uh that the aircraft is
08:32 - 08:40
pretty much at its full weight value at the 55 knots at £2432.29.
08:40 - 08:47
But when we take that aircraft still at 5000 ft to 110 indicated knots,
08:47 - 08:55
it's 9720 nine's of pressure to try to hold that aircraft at 110 knots
08:55 - 08:59
at 16.5 degrees which this aircraft won't be able to do.
08:59 - 09:03
So this begs the question, can this aircraft handle that?
09:03 - 09:06
Certainly at the 55 knots at 5000 ft,
09:06 - 09:09
the pounds of lift is appropriate for this airfoil.
09:09 - 09:12
However, when you look at this type of category of an aircraft,
09:12 - 09:16
being a normal category rather than a utility
09:16 - 09:18
category of the aircraft,
09:18 - 09:24
we have a part 23 limitation to 3.8 Gs load limit on the airframe,
09:24 - 09:31
which would bring this aircraft's weight load limit at 3.8 to £9690
09:31 - 09:37
in which the 110 knots at 5000 MSL would exceed that. So, with £39 over,
09:38 - 09:39
this shows us that we need to keep
09:39 - 09:44
this aircraft within a limited speed factor to avoid
09:45 - 09:49
structural damage but to stay inside the stall envelope for this given airframe,
09:50 - 09:52
let's look at condition number two,
09:52 - 09:56
condition number two is how does lift change with altitude? If a constant A. O.
09:56 - 09:58
A and airspeeder Hill.
09:58 - 10:00
So similar to condition one will test
10:00 - 10:04
the relationship between lift as altitude increases.
10:04 - 10:07
We should note that there is a direct relationship
10:07 - 10:11
between altitude and density ratio lift will decrease,
10:11 - 10:16
the pounds of lift will decrease as altitude increases.
10:16 - 10:18
So you can see first
10:18 - 10:19
At 100 knots,
10:19 - 10:23
this aircraft will be at 1000 ft MSL and will then
10:24 - 10:31
climb this aircraft to 5000 ft. MSL and you can see that the £2131
10:31 - 10:38
will be at the 1000 MSL rate, but as we increase altitude,
10:38 - 10:43
the lift will decrease and the formula here for that proves as we've
10:43 - 10:48
seen about a £200 decrease between the 404,000 ft that we increased in.
10:48 - 10:49
MSL
10:51 - 10:52
finally, let's look at number three.
10:52 - 10:56
Number three is, how do the required cl in the airway for the sky up,
10:56 - 10:58
Change with changes in air speed.
10:58 - 11:02
And we've done this several ways in which the sky hop will initially travel
11:02 - 11:09
at 100 K. I. S indicated air speed and then reduce the speed to 75 knots
11:09 - 11:12
At the gross weight of 2550 pounds.
11:12 - 11:17
Again using the same given parameters 174 square feet,
11:17 - 11:22
it will then travel at 3000 ft. MSL with the appropriate density ratio.
11:22 - 11:25
So you can see with the formula that we've used here
11:25 - 11:31
that the 30000.472 demonstrated at full weight at 100 knots,
11:31 - 11:36
But in order to achieve the same altitude at a lower air speed.
11:36 - 11:41
In the second formula at 75 knots, it's 0.839,
11:41 - 11:44
we would have to increase the back pressure and increase
11:44 - 11:47
our angle of attack to maintain that same altitude.
11:47 - 11:49
So, at a lower airspeed,
11:49 - 11:52
you're required to increase the AOA to maintain that same altitude.
11:52 - 11:57
The cl must increase at the lower speed to maintain the same altitude as well.
11:57 - 11:59
Some AOA and clb match there.
12:01 - 12:01
Mhm.
12:02 - 12:07
Now the example that we give at the bottom is cautionary as you increase the weight,
12:07 - 12:10
you will stall at a higher indicated airspeed,
12:10 - 12:12
the stall at the speed of which an
12:12 - 12:17
aircraft stalls is not influenced directly by speed,
12:17 - 12:19
weight or temperature.
12:19 - 12:20
In terms of the
12:20 - 12:25
angle in which the aircraft stalls. A critical angle of attack is the only component
12:25 - 12:26
in terms of the
12:27 - 12:30
lift or the airfoil losing all lifting properties.
12:30 - 12:34
But you will arrive at that critical angle of
12:34 - 12:37
attack at different speeds based on your weight.
12:37 - 12:39
And we're looking at the differences here between
12:39 - 12:44
the £1900 aircraft and a 25 £150 aircraft.
12:44 - 12:46
You can see it increases by seven knots.
12:46 - 12:52
You will arrive at the stall speed at the max cl at a higher airspeed
12:52 - 12:57
than you would at a lower airspeed at a lower weight of the aircraft.
12:58 - 13:02
And finally what happens if the required cl is larger than the C. L.
13:02 - 13:06
Max of your of your airfoil and what
13:06 - 13:09
speed regime is usually associated with that condition.
13:09 - 13:12
And again what we're doing here is demonstrating the
13:12 - 13:14
maximum performance of the aircraft in terms of C.
13:14 - 13:17
L. Max and the 16.5
13:17 - 13:20
degree A. O. A. The critical angle of attack
13:20 - 13:26
and what speed you would arrive at that at max gross weight which here is 56.32 knots.
13:26 - 13:32
So in this example if the aircraft CLS assumed it's a 1.61
13:32 - 13:35
that would exceed that which is of 1.58 to the C.
13:35 - 13:39
L. Max of this aircraft here demonstrated at 5000 ft. MSL
13:40 - 13:43
which would mean that the airfoil would be stalled.
13:43 - 13:48
The equation above demonstrate that the skyhawk at gross weight at 5000 MSL
13:48 - 13:53
would reach the critical angle attack at that indicated air speed at 56.32.
13:53 - 13:56
So based on the condition that we used in three
13:56 - 14:01
pilots must fly with a higher AOA at lower speeds to maintain the same altitude
14:01 - 14:06
or to hire a away at lower speeds to avoid a stall but not to the point in
14:06 - 14:09
which they reached the critical angle of attack during
14:09 - 14:12
the increased a a way to maintain altitude.
14:12 - 14:14
I hope you've enjoyed this presentation.
14:15 - 14:19
This has been my approach to aircraft selection and the airfoils,
14:19 - 14:22
research and physical laws associated with calculating
14:22 - 14:24
behaviors of the Cessna 1 72.
14:24 - 14:25
Thank you.
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