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NOTE:. I have resolved my question. I had made a mistake in the algebraic derivation of the Formula I was using to graph this. I have included both the old incorrect formula and the new corrected one, and displayed on the graph the curves resulting from each one.

Without going through the Math, I noticed that my equation was wrong:

WRONG FORMULA: $$R \cong \frac{V^2}{\sqrt{V^2-V_s^2}}$$ CORRECT FORMULA: $$R \cong \frac{V^2 V_s^2}{g (\sqrt{V^4-V_s^4})}$$

where:

1. $R$ .... Turn radius
2. $V$.... Aircraft true air speed
3. $V_S$... Stall speed (TAS)
4. $g$ .... 32.2 $ft/sec^2$

Once I modified the formula properly, the curve indicates exactly what you would expect, so my question is now moot.

Derivation:
Staring with the standard Turn radius formula:

1. $$R \cong \frac{V^2}{G_R}$$

where $G_R$ is the Turning Lift (the horizontal component of the Lift vector), divided by the weight, referred to as Radial G, the G that is actually turning the aircraft and not just holding it up in level flight.

Since the Aircraft Load Factor or total G ($G_T$), radial G ($G_R$) , and the 1 G (god’s G) that is holding the aircraft up in the air form a 90 degree right triangle depicted in the diagram above, they must conform to Pythagoras Theorem that says that in a 90 degree triangle the square of the hypotenuse must be equal to the sum of the squares of the other two sides. so,

$G_R^2 + {1g}^2 = G_T^2$

$G_R = \sqrt{G_T^2 - {1g}^2}$

Back to turn radius formula, substituting expression for $G_R$,

2. $$R \cong \frac{V^2}{\sqrt{G_T^2 - {1g}^2}}$$

Now total G ($G_T$) - assuming we establish the maximum available G at whatever airspeed we are at, is simply the Lift ($L = C_{Lmax} ρ V^2 S$), divided by the aircraft weight

$G_T = g\frac{C_{Lmax} ρ V^2 S}{W} $

Substituting this into our equation gives us

3. $$R \cong \frac{V^2}{\sqrt{{g^2\frac{(C_{Lmax} ρ V^2 S)^2}{W^2}} - {1g}^2}}$$

Now here's the trick, The maximum available lift at stall speed $V_S$ is, by definition, equal to the aircraft weight, so

$W = C_{Lmax} ρ V_S^2 S$

Substituting this, and simplifying,

4. $$R \cong \frac{V^2}{g\sqrt{{\frac{(C_{Lmax} ρ V^2 S)^2}{(C_{Lmax} ρ V_S^2 S)^2}} - 1}}$$

and cancelling,

5. $$R \cong \frac{V^2}{g\sqrt{{(\frac{V^2}{V_S^2})^2} - 1}}$$

Simplifying,

6. $$R \cong \frac{V^2}{g\sqrt{{\frac{V^4}{V_S^4}} - \frac{V_S^4}{V_S^4}}}$$

and finally,

7. $$R \cong \frac{V^2 V_s^2}{g \sqrt{V^4-V_s^4}}$$

NOTE:. I have resolved my question. I had made a mistake in the algebraic derivation of the Formula I was using to graph this. I have included both the old incorrect formula and the new corrected one, and displayed on the graph the curves resulting from each one.

Without going through the Math, I noticed that my equation was wrong:

WRONG FORMULA: $$R \cong \frac{V^2}{\sqrt{V^2-V_s^2}}$$ CORRECT FORMULA: $$R \cong \frac{V^2 V_s^2}{g (\sqrt{V^4-V_s^4})}$$

where:

1. $R$ .... Turn radius
2. $V$.... Aircraft true air speed
3. $V_S$... Stall speed (TAS)
4. $g$ .... 32.2 $ft/sec^2$

Once I modified the formula properly, the curve indicates exactly what you would expect, so my question is now moot.

Derivation:
Staring with the standard Turn radius formula:

1. $$R \cong \frac{V^2}{G_R}$$

where $G_R$ is the Turning Lift (the horizontal component of the Lift vector), divided by the weight, referred to as Radial G, the G that is actually turning the aircraft and not just holding it up in level flight.

Since the Aircraft Load Factor or total G ($G_T$), radial G ($G_R$) , and the 1 G (god’s G) that is holding the aircraft up in the air form a 90 degree right triangle depicted in the diagram above, they must conform to Pythagoras Theorem that says that in a 90 degree triangle the square of the hypotenuse must be equal to the sum of the squares of the other two sides. so,

$G_R^2 + {1g}^2 = G_T^2$

$G_R = \sqrt{G_T^2 - {1g}^2}$

Back to turn radius formula, substituting expression for $G_R$,

2. $$R \cong \frac{V^2}{\sqrt{G_T^2 - {1g}^2}}$$

Now total G ($G_T$) - assuming we establish the maximum available G at whatever airspeed we are at, is simply the maximum available Lift ($L = C_{Lmax} ρ V^2 S$), divided by the aircraft weight

$G_T = g\frac{C_{Lmax} ρ V^2 S}{W} $

Substituting this into our equation gives us

3. $$R \cong \frac{V^2}{\sqrt{{g^2\frac{(C_{Lmax} ρ V^2 S)^2}{W^2}} - {1g}^2}}$$

Now here's the trick, The maximum available lift at stall speed $V_S$ is, by definition, equal to the aircraft weight, so

$W = C_{Lmax} ρ V_S^2 S$

Substituting this, and simplifying,

4. $$R \cong \frac{V^2}{g\sqrt{{\frac{(C_{Lmax} ρ V^2 S)^2}{(C_{Lmax} ρ V_S^2 S)^2}} - 1}}$$

and cancelling,

5. $$R \cong \frac{V^2}{g\sqrt{{(\frac{V^2}{V_S^2})^2} - 1}}$$

Simplifying,

6. $$R \cong \frac{V^2}{g\sqrt{{\frac{V^4}{V_S^4}} - \frac{V_S^4}{V_S^4}}}$$

and finally,

7. $$R \cong \frac{V^2 V_s^2}{g \sqrt{V^4-V_s^4}}$$

NOTE:. I have resolved my question. I had made a mistake in the algebraic derivation of the Formula I was using to graph this. I have included both the old incorrect formula and the new corrected one, and displayed on the graph the curves resulting from each one.

Without going through the Math, I noticed that my equation was wrong:

WRONG FORMULA: $$R \cong \frac{V^2}{\sqrt{V^2-V_s^2}}$$ CORRECT FORMULA: $$R \cong \frac{V^2 V_s^2}{g (\sqrt{V^4-V_s^4})}$$

where:

1. $R$ .... Turn radius
2. $V$.... Aircraft true air speed
3. $V_S$... Stall speed (TAS)
4. $g$ .... 32.2 $ft/sec^2$

Once I modified the formula properly, the curve indicates exactly what you would expect, so my question is now moot.

Derivation:
Staring with the standard Turn radius formula: $$R \cong \frac{V^2}{G_R}$$

where $G_R$ is the Turning Lift (the horizontal component of the Lift vector), divided by the weight, referred to as Radial G, the G that is actually turning the aircraft and not just holding it up in level flight.

Since the Aircraft Load Factor or total G ($G_T$), radial G ($G_R$) , and the 1 G (god’s G) that is holding the aircraft up in the air form a 90 degree right triangle depicted in the diagram above, they must conform to Pythagoras Theorem that says that in a 90 degree triangle the square of the hypotenuse must be equal to the sum of the squares of the other two sides. so,

$G_R^2 + {1g}^2 = G_T^2$

$G_R = \sqrt{G_T^2 - {1g}^2}$

Back to turn radius formula, substituting expression for $G_R$,

$$R \cong \frac{V^2}{\sqrt{G_T^2 - {1g}^2}}$$

Now total G ($G_T$) - assuming we establish the maximum available G at whatever airspeed we are at, is simply the Lift ($L = C_{Lmax} ρ V^2 S$), divided by the aircraft weight

$G_T = g\frac{C_{Lmax} ρ V^2 S}{W} $

Substituting this into our equation gives us

$$R \cong \frac{V^2}{\sqrt{{g^2\frac{(C_{Lmax} ρ V^2 S)^2}{W^2}} - {1g}^2}}$$

Now here's the trick, The maximum available lift at stall speed $V_S$ is, by definition, equal to the aircraft weight, so

$W = C_{Lmax} ρ V_S^2 S$

Substituting this, and simplifying,

$$R \cong \frac{V^2}{g\sqrt{{\frac{(C_{Lmax} ρ V^2 S)^2}{(C_{Lmax} ρ V_S^2 S)^2}} - 1}}$$

and cancelling,

$$R \cong \frac{V^2}{g\sqrt{{(\frac{V^2}{V_S^2})^2} - 1}}$$

Simplifying,

$$R \cong \frac{V^2}{g\sqrt{{\frac{V^4}{V_S^4}} - \frac{V_S^4}{V_S^4}}}$$

and finally,

$$R \cong \frac{V^2 V_s^2}{g \sqrt{V^4-V_s^4}}$$

NOTE:. I have resolved my question. I had made a mistake in the algebraic derivation of the Formula I was using to graph this. I have included both the old incorrect formula and the new corrected one, and displayed on the graph the curves resulting from each one.

Without going through the Math, I noticed that my equation was wrong:

WRONG FORMULA: $$R \cong \frac{V^2}{\sqrt{V^2-V_s^2}}$$ CORRECT FORMULA: $$R \cong \frac{V^2 V_s^2}{g (\sqrt{V^4-V_s^4})}$$

where:

1. $R$ .... Turn radius
2. $V$.... Aircraft true air speed
3. $V_S$... Stall speed (TAS)
4. $g$ .... 32.2 $ft/sec^2$

Once I modified the formula properly, the curve indicates exactly what you would expect, so my question is now moot.

Derivation:
Staring with the standard Turn radius formula:

1. $$R \cong \frac{V^2}{G_R}$$

where $G_R$ is the Turning Lift (the horizontal component of the Lift vector), divided by the weight, referred to as Radial G, the G that is actually turning the aircraft and not just holding it up in level flight.

Since the Aircraft Load Factor or total G ($G_T$), radial G ($G_R$) , and the 1 G (god’s G) that is holding the aircraft up in the air form a 90 degree right triangle depicted in the diagram above, they must conform to Pythagoras Theorem that says that in a 90 degree triangle the square of the hypotenuse must be equal to the sum of the squares of the other two sides. so,

$G_R^2 + {1g}^2 = G_T^2$

$G_R = \sqrt{G_T^2 - {1g}^2}$

Back to turn radius formula, substituting expression for $G_R$,

2. $$R \cong \frac{V^2}{\sqrt{G_T^2 - {1g}^2}}$$

Now total G ($G_T$) - assuming we establish the maximum available G at whatever airspeed we are at, is simply the Lift ($L = C_{Lmax} ρ V^2 S$), divided by the aircraft weight

$G_T = g\frac{C_{Lmax} ρ V^2 S}{W} $

Substituting this into our equation gives us

3. $$R \cong \frac{V^2}{\sqrt{{g^2\frac{(C_{Lmax} ρ V^2 S)^2}{W^2}} - {1g}^2}}$$

Now here's the trick, The maximum available lift at stall speed $V_S$ is, by definition, equal to the aircraft weight, so

$W = C_{Lmax} ρ V_S^2 S$

Substituting this, and simplifying,

4. $$R \cong \frac{V^2}{g\sqrt{{\frac{(C_{Lmax} ρ V^2 S)^2}{(C_{Lmax} ρ V_S^2 S)^2}} - 1}}$$

and cancelling,

5. $$R \cong \frac{V^2}{g\sqrt{{(\frac{V^2}{V_S^2})^2} - 1}}$$

Simplifying,

6. $$R \cong \frac{V^2}{g\sqrt{{\frac{V^4}{V_S^4}} - \frac{V_S^4}{V_S^4}}}$$

and finally,

7. $$R \cong \frac{V^2 V_s^2}{g \sqrt{V^4-V_s^4}}$$

NOTE:. I have resolved my question. I had made a mistake in the algebraic derivation of the Formula I was using to graph this. I have included both the old incorrect formula and the new corrected one, and displayed on the graph the curves resulting from each one.

Without going through the Math, I noticed that my equation was wrong:

WRONG FORMULA: $$R \cong \frac{V^2}{\sqrt{V^2-V_s^2}}$$ CORRECT FORMULA: $$R \cong \frac{V^2 V_s^2}{g (\sqrt{V^4-V_s^4})}$$

where:

1. $R$ .... Turn radius
2. $V$.... Aircraft true air speed
3. $V_S$... Stall speed (TAS)
4. $g$ .... 32.2 $ft/sec^2$

Once I modified the formula properly, the curve indicates exactly what you would expect, so my question is now moot.

NOTE:. I have resolved my question. I had made a mistake in the algebraic derivation of the Formula I was using to graph this. I have included both the old incorrect formula and the new corrected one, and displayed on the graph the curves resulting from each one.

Without going through the Math, I noticed that my equation was wrong:

WRONG FORMULA: $$R \cong \frac{V^2}{\sqrt{V^2-V_s^2}}$$ CORRECT FORMULA: $$R \cong \frac{V^2 V_s^2}{g (\sqrt{V^4-V_s^4})}$$

where:

1. $R$ .... Turn radius
2. $V$.... Aircraft true air speed
3. $V_S$... Stall speed (TAS)
4. $g$ .... 32.2 $ft/sec^2$

Once I modified the formula properly, the curve indicates exactly what you would expect, so my question is now moot.

Derivation:
Staring with the standard Turn radius formula: $$R \cong \frac{V^2}{G_R}$$

where $G_R$ is the Turning Lift (the horizontal component of the Lift vector), divided by the weight, referred to as Radial G, the G that is actually turning the aircraft and not just holding it up in level flight.

Since the Aircraft Load Factor or total G ($G_T$), radial G ($G_R$) , and the 1 G (god’s G) that is holding the aircraft up in the air form a 90 degree right triangle depicted in the diagram above, they must conform to Pythagoras Theorem that says that in a 90 degree triangle the square of the hypotenuse must be equal to the sum of the squares of the other two sides. so,

$G_R^2 + {1g}^2 = G_T^2$

$G_R = \sqrt{G_T^2 - {1g}^2}$

Back to turn radius formula, substituting expression for $G_R$,

$$R \cong \frac{V^2}{\sqrt{G_T^2 - {1g}^2}}$$

Now total G ($G_T$) - assuming we establish the maximum available G at whatever airspeed we are at, is simply the Lift ($L = C_{Lmax} ρ V^2 S$), divided by the aircraft weight

$G_T = g\frac{C_{Lmax} ρ V^2 S}{W} $

Substituting this into our equation gives us

$$R \cong \frac{V^2}{\sqrt{{g^2\frac{(C_{Lmax} ρ V^2 S)^2}{W^2}} - {1g}^2}}$$

Now here's the trick, The maximum available lift at stall speed $V_S$ is, by definition, equal to the aircraft weight, so

$W = C_{Lmax} ρ V_S^2 S$

Substituting this, and simplifying,

$$R \cong \frac{V^2}{g\sqrt{{\frac{(C_{Lmax} ρ V^2 S)^2}{(C_{Lmax} ρ V_S^2 S)^2}} - 1}}$$

and cancelling,

$$R \cong \frac{V^2}{g\sqrt{{(\frac{V^2}{V_S^2})^2} - 1}}$$

Simplifying,

$$R \cong \frac{V^2}{g\sqrt{{\frac{V^4}{V_S^4}} - \frac{V_S^4}{V_S^4}}}$$

and finally,

$$R \cong \frac{V^2 V_s^2}{g \sqrt{V^4-V_s^4}}$$