G and turn calculations

[Angus]

Hi there all
I have a stupid question to ask.
Imagine an aircraft flying in a circle at 200 mph completing it in a given time with the diameter of 1000 feet. Will 400 mph and  the same 1000 feet precisely double the G load? How is the formula?
Does a constant (say 3)G doing a circle always yield the same time around 360 degrees with the circle growing bigger with more speed?
Im just a bit mixed up here

[Modas]

Reaching WAY back to college...

I think (correct me here if I'm wrong) that G forces are a function of the square of the velocity.  If you double your speed, G's go up by 4.  Course I could be wrong, its been 13 years since I've done this

[eskimo2]

Modas, I think that you may be thinking of energy.  Double the speed 4X the power needed.

Angus, I think that if you double the speed to round the circle, the Gees will also double.

As for the second part, imagine a fly turning a tiny 3" 3G circle.  He's going to get around that 3" circle pretty quick, right?  Probably much less that one second.
Now imagine an F-16 flying at 1000 mph also turning at 3 Gees.  For him to make a 360 at 3 Gees, it's going to take many seconds, if not a minute or so.

eskimo

[Wilbus]

I have a good picture in a book that shows what Eskimo2 says, gotto make it smaller so I can post it.

As he says, the turning circle, although the same G (let's say 3 G) get bigger and bigger the faster you fly.

[bozon]

angus,
G is a measure of acceleration (normalized to units of your own weight).
in a circular motion, the condition the radial acceleration must satisfy:
a=v^2/r

where a - acceleration, v - tangent speed, r - radius.
( ^2 means square)

so if you keep the radius constant and and multiply speed by x, your acceleration will be multiplyed by x^2.

edit:
this assumes you keep a constant bank.

Bozon

[bozon]

this is interesting,

if you add the condition for a horizontal turn only (neither loosing nor gaining alt), then it translates into:
G*cos(b)=1
where b - the bank angle.

using the definition of G: G=lift/mg, and the condition for circular motion:
centripetal force=Gmg*sin(b)=mv^2/r.

from Gcos(b)=1 we get =>   sin(b)=(G^2-1)^0.5/G
(0
combined:
g(G^2-1)^0.5=v^2/r

G=sqrt( v^4/(g*r)^2 + 1 )
sqrt mean the square root.

this give a little unglier expression but for high G, it is still roughly proportional to the square of the speed.

Bozon

[Dawggus]

Well, if the angle of the dangle is perpendicular to the tilt of the ... whoops, I thought you said G-Spot, not G-Force, never mind .

Cya Up!

[ra]

Here's a RC modeling site with a link for a turn radius/G calculator:

http://www.flyingsites.co.uk/downloads/

ra

[senna]

Ok, heres my shot at this one despite the above equations for a some anti gravity stuff.

g=1.226R
    ---------
        Tsquared

where g=gravityx1
R=radious in feet
T=acceleration in feet/per second.

Weight isnt important since the question asks about how many Gs pulled (weight is arbitrary) not how heavy.

Back to watching hotties on MTV

[dtango]

Originally posted by Angus
Does a constant (say 3)G doing a circle always yield the same time around 360 degrees with the circle growing bigger with more speed?

The answer is that not only does the circle grow bigger with more speed, the time to complete the circle increases as well.  Here's why:

Rate of Turn = (360 * Velocity) / (2*Pi*Radius of Turn)

Radius of Turn = Velocity^2 / (gravity*tan(bank angle))

G Load = 1 / cos (bank angle)

[list=1]

• Constant G means that the bank angle is constant.
• If bank angle is constant then the only variable in turn radius is velocity so if velocity increases so does the radius.
• Rate of turn is a function of velocity and turn radius.  Since radius of turn increases with the velocity^2, then the v/radius ratio changes with radius increasing at a rate faster than velocity.  The result is a slower rate of turn.  This becomes obvious when you plug some numbers.

E.g. using the above equations assume a P-51D at 6G's at 260 MPH:
radius=760ft
turn rate=28.6 dps
time around 360 deg=12.5 sec

6G's at 350 MPH would give:
radius=1383ft
turn rate=21.1
time around 360 deg=16.9 sec

(The above example is only hypothetical.  You couldn't sustain a 6G turn for 360 degrees in a P-51D due to E bleed.)

Tango, XO
412th FS Braunco Mustangs

[Nefarious]

Interesting Topic.

Wish I was good in Math.

[Modas]

Originally posted by bozon
angus,
G is a measure of acceleration (normalized to units of your own weight).
in a circular motion, the condition the radial acceleration must satisfy:
a=v^2/r

where a - acceleration, v - tangent speed, r - radius.
( ^2 means square)

so if you keep the radius constant and and multiply speed by x, your acceleration will be multiplyed by x^2.

edit:
this assumes you keep a constant bank.

Bozon

I think this is what I said above.  Guess my college knowledge isn't as rusty as I thought.  Damn, have to go out and drink some more beer .

Thanks for the confirmation guys....  I love beer

[eskimo2]

Originally posted by Modas


I think this is what I said above.  Guess my college knowledge isn't as rusty as I thought.  Damn, have to go out and drink some more beer .

Thanks for the confirmation guys....  I love beer

I think you were right Modas, as Bozon explained, my bad.

Rotation is a form of acceleration!

eskimo