You have to get into the math a little bit to understand this one, and it has to do with this equation. I put in some dots spacing.

R = V-squared.....
.....g * tan(theta)

where:

R= radius of turn
g= gravity
V= velocity of airplane (speed)
theta= bank angle (zero is no bank)

So, in a level turn weight must equal gravity, and bank angle is constant as well by definition, then what must happen to R with a change in speed (V)?

Here's a practical example for you, as this confused me in ground school, too.

Picture yourself walking down the street. You suddenly see a store you want to go into, and make a sharp turn and go inside.

Now, try doing the same thing, running as fast as you can.

Obviously, the radius is bigger for the runner, and he also takes longer to make the turn. That is radius and rate, respectively.

It confused me in ground school/early flying because I told myself: "But the faster I go, the more 'g' that can be generated!"

True, but it doesn't help. I had this discussion with a student recently, and I proved via the lift formula and turn radius formula that for the same angle of attack, the turn radius is a constant. But that is a different side to this discussion.

I was under the impression that turn performance was based on radial acceleration (vector sum of load factor and gravity).

Ryan:

The first part of your statement is correct: the horizontal component is what counts for turn performance (assuming you are making a horizontal turn, otherwise, a more general and correct statement would be the component in the direction of turn, such as in a loop, immelmann, split-S, pitchback, or sliceback).

But in the horizontal example, the total vector (or 'g'), which is what you feel, is not the horizontal component. Total is vector addition of horizontal and vertical components. So, your equation should say Load Factor = Horizontal + Vertical.

Here's a practical example for you, as this confused me in ground school, too.

Picture yourself walking down the street. You suddenly see a store you want to go into, and make a sharp turn and go inside.

Now, try doing the same thing, running as fast as you can.

Obviously, the radius is bigger for the runner, and he also takes longer to make the turn. That is radius and rate, respectively.

It confused me in ground school/early flying because I told myself: "But the faster I go, the more 'g' that can be generated!"

True, but it doesn't help. I had this discussion with a student recently, and I proved via the lift formula and turn radius formula that for the same angle of attack, the turn radius is a constant. But that is a different side to this discussion.

But in the EM diagram there is that one point where turn rate and rdius is optimized - correct? Or have I been out of the loop too long?

The first part of your statement is correct: the horizontal component is what counts for turn performance (assuming you are making a horizontal turn, otherwise, a more general and correct statement would be the component in the direction of turn, such as in a loop, immelmann, split-S, pitchback, or sliceback).

But in the horizontal example, the total vector (or 'g'), which is what you feel, is not the horizontal component. Total is vector addition of horizontal and vertical components. So, your equation should say Load Factor = Horizontal + Vertical.

Ok... so does that mean that the vertical (over the top 'egg') will have a more positive turn performance while the horizontal will have more negative turn performance?