G-Load Calculation Issue

[Zipper730]

As I understand it, the mathematical formula for calculating the radius of a circle based on g-load is A=V^2/R: So, I was curious what would happen if I put 1g in the calculation and, with 32.17405 f/s, I put figures in for an aircraft doing 600 miles an hour

So, I get the following
Step 0: A = V^2/R
Step 1: 32.17405 = (600*1.4666666666666667)^2/R
Step 2: 32.17405 = (880)^2/R
Step 3: 32.17405 = 774400/R
Step 4: 32.17405/1 = 774400/R
Step 5: 32.17405/R = 774400
Step 6: (32.17405/R)32.17405 = 774400/32.17405

R = 24069.08673'
D = 48138.17347'
Circumference = 151230.5321' or 28.64215 statute miles

You'd think that 1g would be straight and level so you'd figure the circle would be the same as the diameter of the earth, but it isn't. Did I make a calculation error?

​

 

[GregP]

For aircraft, the formula for calculating the turn radius is: R = V^2 / (g tan a), where a is the bank angle in radians, g is the g-force, V is velocity in feet per second, and R = radius in feet. Bank angle in a level turn is acos (1/g), where g is the g-force, and acos is the arccosine.

So, if you are doing 300 mph and are pulling 3 g in a level turn, the bank angle is 70.5288°. The radius is 2,127 feet. 300 mph is 440 feet per second. The circumference is 13,367 feet. The distance around a 180° turn is 6,683.5 feet and, at 440 feet per second, it takes 15.19 seconds to complete the 180° turn. I have it all in an Aerodynamics spreadsheet, so calculation is as easy as entering the airspeed and the g force. Turn rate is 12° per second.

PM me if you want some discussion about it. I calculate everything in English units, but I have no trouble converting to or using with metric units. It's just that I'm in the U.S.A. . Since it is aircraft, perhaps we should use knots for true airspeed. The formula is in feet, seconds, and g-force, but changing it for other units is simple. A spreadsheet makes it easy to use feet per second, km/h, m/s, mph, or knots.

About your original post, if you are at 1 g, you are not turning. A level turn will always be at least slightly more than 1 g, by definition.

600 mph at 4 g gives a radius of 6,215 feet, 22.2 seconds to complete a 180° turn, turn rate 8° per second. This is slightly misleading since the formula assumes you are at a constant g-force all the way through the turn but, in real life, you have to roll into the turn, then pull the g-force, and then roll out again when the turn is completed while letting off the back stick. So, the formulas are ideal. Real-world turns will add a slight bit of time due to having to start and stop the turn, particularly if the bank angle is significant. In the case of the 600 mph, 4 g turn, the bank angle is 75.5°, so it can take a bit to roll in, start the turn, then let off and roll out again.

 

[elbmc1969]

And the formula doesn't account for the curvature of the earth. On the "1G" case, you're flying straight and unbanked, maintaining altitude, and only curving due to the Earth being a sphere. Nothing to do with the turn diameter equation.

 

[GregP]

And the formula doesn't account for the curvature of the earth. On the "1G" case, you're flying straight and unbanked, maintaining altitude, and only curving due to the Earth being a sphere. Nothing to do with the turn diameter equation.

The turn diameter equation depends on velocity, g-force, and bank angle. Bank angle and g-force are dependent so, if you have one, you can get the other.

The curvature of the earth is not a factor in any aerodynamic formulas. Gravity is always pointing to the center of mass of the earth. Straight and level flight will always follow the geometric altitude, while constant-altitude flight using the altimeter will always follow isobaric lines. The two altitudes can be significantly different.

These days, if you are using a GPS navigator, I believe you are essentially following geometric altitudes corrected to isobaric lines mathematically by software. Can't have the two being randomly mixed in traffic.

 

[elbmc1969]

The turn diameter equation depends on velocity, g-force, and bank angle. Bank angle and g-force are dependent so, if you have one, you can get the other.

The curvature of the earth is not a factor in any aerodynamic formulas. Gravity is always pointing to the center of mass of the earth. Straight and level flight will always follow the geometric altitude, while constant-altitude flight using the altimeter will always follow isobaric lines. The two altitudes can be significantly different.

These days, if you are using a GPS navigator, I believe you are essentially following geometric altitudes corrected to isobaric lines mathematically by software. Can't have the two being randomly mixed in traffic.

Yeah, but he's trying to the a turning diameter equal to the circumference of the Earth!

 

[pbehn]

As I understand it, the mathematical formula for calculating the radius of a circle based on g-load is A=V^2/R: So, I was curious what would happen if I put 1g in the calculation and, with 32.17405 f/s, I put figures in for an aircraft doing 600 miles an hour

So, I get the following
Step 0: A = V^2/R
Step 1: 32.17405 = (600*1.4666666666666667)^2/R
Step 2: 32.17405 = (880)^2/R
Step 3: 32.17405 = 774400/R
Step 4: 32.17405/1 = 774400/R
Step 5: 32.17405/R = 774400
Step 6: (32.17405/R)32.17405 = 774400/32.17405

R = 24069.08673'
D = 48138.17347'
Circumference = 151230.5321' or 28.64215 statute miles

You'd think that 1g would be straight and level so you'd figure the circle would be the same as the diameter of the earth, but it isn't. Did I make a calculation error?

​

If you fly straight and level that is 1G, eventually you will have flown a circle equal to the earths diameter plus 2 times the altitude.

 

[pbehn]

Further to GregP 's post, pilots have crashed in zero visibility believing they were flying straight and level but in fact were performing a 1 G spiral into the ground

 

[GregP]

Yeah, but he's trying to the a turning diameter equal to the circumference of the Earth!

I must not have read it quite correctly. So, I went back and read it again.

So, once again, the curvature of the earth has nothing to do with turn radius. A "turn," for our purposes means a turn in the yaw axis. That is, the compass moves. If you fly straight and level at exactly 1,000 feet above sea level around the world in a straight line, you will not have completed a turn. The only aerial phenomenon that uses the curvature of the earth is orbit. And it uses it solely because a perfect orbit means equal average geometric altitude above the center of mass of the earth.

Turning means changing horizontal direction, not completing an outside loop with the diameter of the earth plus your altitude.

Cheers!

 

[K5083]

If you see 3g on your g-meter in a level turn, you have a vertical component equal to the 1g weight of the aircraft to stay level and a centripetal component to get you round the turn. Remember the 3g on the meter is directed thorough the vertical related to the aircraft. The 3g will not give you the turn radius, you need to use (via Pythagoras) the square root of eight. Which is 2.8282 if I remember right.


The 1g turn in Zipper's example works to an aircraft banked at 90 degrees pulling 1g but effectively in free fall sideways as far as gravity is concerned. There's no level turn at 1g and the radius of the Earth has nothing to do with it.

 

[GregP]

Hello Zipper 730! This is your thread. What are you trying to ask?

Apparently, we aren't "getting it."

 

[Ivan1GFP]

I believe in this case, Zipper is looking for a centripetal acceleration of 1G which along with 1G of gravity means the pilot / aircraft is actually experiencing about 1.414G (Square Root of 2).

- Ivan.

 

[GregP]

OK, I can buy that.

So, if the centripetal acceleration is 1.000 g and the actual g-load on the aircraft is 1.414 g, AND if the speed is 250 mph, then the turn radius is 4,180 feet. A 180° turn can be completed in 35.8 seconds, and the bank angle is 45° (well, 44.991° anyway). That isn't much of a combat turn. The time to complete a 180° turn gets worse as the speed goes up.

Edit:
Just as an aside, everyone in here who is instrument-rated knows a standard-rate turn is a 30° turn. Supposedly, it takes one minute to complete a 180° turn at standard rate. In my formulas, it doesn't quite work out that way. A 30° banks works out to a g-load of 1.155 g, and a 180° turn takes 61.9 seconds. To get to exactly 60 seconds for a 180° turn, the bank angle needs to be 30.9°. But, on a standard turn and bank indicator, you can't READ down to the tenth of a degree, so the "standard" was made into 30° since you can't tell the difference between the two numbers while in the cockpit. In the airplanes I used to fly, a standard turn and bank indicator had a mark every 10°, and shading to the outside edge of the 30° mark was right about 1 minute for a 180° turn.

 

[GregP]

Sent your PM a reply, Zipper. Cheers.

 

[Zipper730]

If you see 3g on your g-meter in a level turn, you have a vertical component equal to the 1g weight of the aircraft to stay level and a centripetal component to get you round the turn. Remember the 3g on the meter is directed thorough the vertical related to the aircraft.

So there's 1G vertical and 3G horizontal? Or are you talking about 4G for a loop with 1G being just flying level + 3G? Regardless, I'm curious how you got the square root of eight. I figure if you have 4 you'd get the square root of 4...

I believe in this case, Zipper is looking for a centripetal acceleration of 1G which along with 1G of gravity means the pilot / aircraft is actually experiencing about 1.414G (Square Root of 2).

I would have thought it would have been centrifugal acceleration, but you're describing 1G+1G right? I'm curious how the square root gets in there

 

[GregP]

For his example, it would be 1 g vertical and 2.8284 g horizontal. If you square both vertical g and horizontal g, add them up, and take the square root, you get 3 g in a direction straight down through the aircraft's vertical axis, which is inclined at 70.5°to straight and level.

The angle of bank in a 3 g level turn is: angle = arc cos (1/g), or arc cos (1/3) = 70.5°. Actually 70.52878° but you can't read a bank angle anywhere NEAR that accurately with standard instruments.

Nothing K5083 didn't say above. Cheers!

 

[Joe Broady]

So there's 1G vertical and 3G horizontal? Or are you talking about 4G for a loop with 1G being just flying level + 3G? Regardless, I'm curious how you got the square root of eight. I figure if you have 4 you'd get the square root of 4...

The square root of 8 occurs because acceleration is a vector quantity: it has both magnitude and direction. A cockpit accelerometer indicates only magnitude. The direction (in a constant speed level turn) is perpendicular to the cockpit floor. Or, with respect to the horizon, as others have said it's inclined to the vertical by an angle (bank angle) equal to the arc secant of the magnitude. If the meter says 3 g, bank angle is arc sec 3 = arc cos 1/3 = 71°.

As this page explains, a vector can be resolved into two components at right angles to each other. In this case, the 3 g acceleration can be resolved into a 1 g vertical component (which supports the weight of the plane against gravity), and a horizontal component (which forces it into a curved path). The radius of the curve is a function of speed and the horizontal component of the 3 g total. To compute that component, observe that the 3 g total acceleration is equivalent to the hypotenuse of a right triangle. One side (the vertical acceleration) is 1 g and the other side is the unknown horizontal acceleration. According to the Pythagorean formula, the square of 3 equals the square of 1 plus the square of the unknown (call it x). That is, 9 = 1^2 + x^2. Since the square of 1 is 1, the equation simplifies to 8 = x^2, and so the horizontal acceleration is the square root of 8.

 

[Wild_Bill_Kelso]

A two G turn in a (low wing loading) Gloster Gladiator, at 200 miles an hour, banked at 70 degrees (or whatever it is precisely), turns at the same rate as say, a two G turn in a (high wing loading) P-47?

 

[GregP]

Yes, as long as the high wing is going the same speed. The turn radius depends on speed, g-load and bank angle, and the bank angle depends in the g-load, so it all boils down to g-load and speed. Same g-load and same speed is the same bank angle and turn radius.

These are for level turns only. It changes if you are turning while climbing or diving, and the change is not simple to calculate since the g-load and speed change during the turn.

 

[Wild_Bill_Kelso]

Doesn't wing loading and lift have any affect on turn radius? It doesn't make sense to me that a biplane turns at the same radius as a bomber.

How does this compare to say, the Russian 'turn times' testing? Why does an A6M out-turn a Wildcat so easily?

What are the effects of slats and combat flaps? Just reducing the stall speed?

 

[GregP]

Actually no. If you are in a level turn at a certain speed and g-load, then you have enough lift to sustain the turn at that speed. If your wing cannot supply the lift required due to one of the factors you mention above, then you will stall and cannot sustain the turn you are trying to make. The stall speed is the level-flight stall speed multiplied by the square root of the g-load.

So, say you have a level flight stall speed of 90 mph (or 90 knots, etc.) and say you want to turn at 2 g. The square root of 2 is 1.414 and 1.414 time the stall speed is 127.26 mph. So, you cannot make a level turn at ANY speed below 127.26 mph because you will stall first. But, at 130 mpg and 2 g, you CAN turn and the turn radius will be 652 feet using: r = (V^2)/(g tan [bank angle]), and at 2.0 g, the bank angle is 60°. It takes 10.7 seconds to complete a 180° turn and the rate of turn is 17° per second.

You can get these number by knowing the turn radius, calculating the circumference, and then take half of that distance and calculate how long it take to fly that distance at 130 mph. Turn rate is 180° / time to turn, in degrees per second.

Sorry if I went too far into the explanation. Math is fun for me.

Cheers.