# Simple gravity concept question

1. Jul 21, 2007

### Diresu

When an aircraft begins to nose up the effects of gravity begin to slow its velocity.

By dividing 100% by 90 degrees you get the number 1.11. So for each 1 degree incline the force of gravity increases by 1.11%?

So at

1 degree the force of gravity is 1.11%
45 degrees the force of gravity is 50%
90 degress the force of gravity is 100%

-90 degress the force of gravity is -100% meaning it is accelerating the aircraft in the dive.

So at 1 degree the effect would be .0111*Ma?
at 45 degrees it would be .5*Ma?
at 90 degrees it would be 1.0*Ma?

Diving straight down at -90 degrees would be -1.0*Ma?

I just want to make sure I am mathematically and theoretically correct. I don't need to use a sine wave or anything like that? Meaning the Sin of the angle determines the percentage effect.

2. Jul 21, 2007

### xez

Well the sin of the angle of climb does determine the
force component of gravity that is directed
'backward' relative to the nose to tail orientation.

There's no real reason to divide X degrees by X percent
to get an interpolatory linear estimate sort of thing
when you can just say that:

F_rearward = sin(theta)

e.g. in level flight gravity pulls downward SORT-OF
at 90 degrees to the flight direction, and there is no
"tail-ward" gravity component if the head-to-tail line is
taken to be horizontal.

Anyway it's all a little confused and somewhat meaningless
since to achieve level flight you STILL have to have an
force upward counteracting the force of gravity, so you
have an angle of attack and lift effect of the wings to the
airstream that is giving you 9.81*Mass_kg newtons of lift
in level flight to balance the 9.81*Mass_kg newtons of
gravity force.

You have your Lift/Drag ratio that is then the
transformation of the Lift value to a Drag value which
*is* of course a rearward force (Drag) so even in level
flight at constant speed there is so much force
in the FORWARD direction that is needed to overcome
the Drag which, of course is converted to a vertical force
by the wings' lifts/angles of attack to equal the downward
gravity force.

So it seems unnatural and unnecessary to distinguish
the drag, vs. downward vs. forward vs. upward etc.
forces in the way you've mentioned since you can't really
discount the gravitational and drag effects even in level
flight or under any other circumstance...

3. Jul 21, 2007

### xez

i.e. the aircraft is *always* 'nose up' to 'fight' gravity
due to the angle of attack of the wings, even if the 'nose'
may have horizontal orientation while the wings have
a positive angle of attack.

4. Jul 21, 2007

### Diresu

xez, thanks a ton!

So a pure percentage i.e. 45 degress = 50% gravity effect is wrong. Wow, that seems so counterintuitive to me. The Sine of the angle determines the effect of gravity... wow. So according to my calc at 45 degrees the effect of gravity is about 71%. I know you are right but doesn't that sound crazy? I'm glad I asked because I would have never understood that on my own. The only reason I asked about sine was because I saw it in a different equation with gravity. I thought it must have been an error or weird exception.

As for the other forces I know you're right. I'm just talking about a simple, theoretical world with as few variables as possible. Any more difficult and my mind will explode.

5. Jul 21, 2007

### xez

Well it's non-intuitive to think of sin/cos as relating to
forces if you're thinking in terms of sine waves versus
linear slopes.

It will all be much clearer if you look back on the
geometric definition of sin, cos, tan:

Draw a circle of radius 1 circling the origin [x=0,y=0].
Theta is the the angle of any given point on the circle
starting measurement from the positive X axis, i.e.
the circle contains point [x=+1,y=0] and that's the
zero-degree point.

X = COS theta
Y = SIN theta

If you recall that the circle has a RADIUS of 1, that means
there's a length one vector from any point on the
circle to the origin [0,0].

If you recall the pythagorean theorem for the length
of the sides of a right triangle you get:
a^2 + b^2 = c^2 where c is the hypotenuse, and a
and b are the lengths of the other two sides.

So at 45 degrees on the unit-radius circle you have a
line of length 1 to the origin at 45 degrees.
The value of X is the COS, and the value of Y is the SIN.
X = SIN 45 deg = sqrt(2)/2,
Y = COS 45 deg = sqrt(2)/2,
so the length of the triangles sides formed between
the origin, the circle point at 45 degrees, and the
X or Y axis is:
hypotenuse^2 = x^2 + y^2
1^2 = (sqrt(2)/2)^2 + (sqrt(2)/2)^2
1 = 2/4 + 2/4
1 = 1.

So the reason it's 0.707.... is just because that's the
length of the side of the triangle that's needed to reach
a circle of radius 1 at 45 degrees.

If you look at it graphically it'll make sense how to use
SIN and COS in the context of an angle as part of a
circle.

TAN = Sin(theta) / Cos(theta) = Y / X.

So at 45 degrees the X component of the force and the
Y component of the force (SIN and COS) are equal,
but their values aren't 0.5 just because of the radius of
circle used to define SIN, COS, etc.

6. Jul 22, 2007

### Diresu

Thanks xez

I read your post several times and decided to do a test. I used a graphics program and built a 45 degree angle and measured the distances. Holy smokes you are right. I uploaded it here

http://imageigloo.com/images/72145degree.gif

It sounds counterintuitive and yet it is absolutely true.

Thanks a TON!!!! I would have never understood that concept without your explanation.

Last edited by a moderator: Apr 22, 2017
7. Jul 23, 2007

### xez

You're quite welcome; I'm glad that helped you
visualize the geometric aspects of the angles and vector
force components!

I'm sorry that I haven't been able to give a really simple
thrust vs. drag vs. lift vs. final speed vs. time etc.
I kind of see what you're wanting, and it's not that complex
but then again planes aren't rockets. It's easier to look
at rockets since they have no 'lift' and no 'stall' or
variable lift vs. angle vs. speed issues etc. etc.

Take a look at this site; I'm sure they'll go into more
detail than you're interested in in terms of the
physics / mechanics of the flight dynamics, but perhaps
some of the ways of looking at things will resonate with you
and you can find new ways to simplify or think about the
sorts of questions you have vs. the information you have

http://www.av8n.com/how/

8. Jul 24, 2007

### Diresu

Thanks xez,

That site is gold! I'll put it in my favorites.

As for my equation I'm looking for something just that simple, i.e. the rocket equation. I'm trying to create something theoretical with just a few variables. Don't worry though, I'm hot on the trail of that equation. Sadly, I've been sworn off working on it until next week.

Thanks again!!!