I am looking for a chart that tells us the lateral distance needed to make a 180 turn. Ie you are on downwind; what distance from the extended centerline must you be, using a constant rate turn from downwind to final, at 80kts, 100kts, 120kts in zero wind conditions? I am thinking 25° bank, but perhaps this is airplane specific and tied to turn rate?

This site shows how to calculate the radius of the turn. Use this and double it for the turn diameter. http://www.flightlearnings.com/2009/08/26/radius-of-turn/ Disclaimer: I just found this from googling

Wayne's page gives you the radius given a bank angle and airspeed. That doesn't tell you what a constant rate turn is at a given airspeed. Code: Radius (NM) = Airspeed Knots) -------------------- 60pi or double it for the diameter of the turn.

Here let me work out the basic equations in the thread. The inputs we need are airspeed, wind (optional, and I will skip it unless requested), and rate of turn (or angle of bank). The reason why the plane turns when it is banked is because a component of the force of lift is horizontal. As a consequence of banking, the vertical component of lift is decreased which means the pilot will need to do some combination of adding power and pulling back on the yoke in order to maintain altitude. I'm going to assume the pilot already has that all figured out and is simply maintaining the desired airspeed while turning and descending as needed. Neglecting vertical motion shouldn't have a significant effect on turn radii. The basic formula for relating forces to circular motion is: Code: Centripetal Force = m v^2 / r Using the definitions of sine and cosine, the horizontal component of lift is L sin (theta). Theta is angle of bank, and the use of sine makes sense since if the bank angle is zero, then sin is zero, and there is no horizontal component to the lift. If theta is 90 degrees, then sin(theta) is 1, all of the lift is horizontal, and none of it is vertical. Since the only horizontal force pulling the plane towards the center of the circle is the horizontal component of the lift, we can combine the two expressions above and get: Code: L sin (theta) = m v^2 / r Unfortunately, we haven't figured out what to do with L yet. Since we've already taken care of the math in the horizontal direction, we can work in the vertical direction next. Although the plane may be descending during the turn, it has a constant vertical speed. Therefore, the upward force (vertical component of lift) has to equal the weight of the plane (m * g). Going back to the definitions of sine and cosine, the vertical component of lift is L cos (theta) and the equation for vertical forces becomes: Code: L cos (theta) = m g This makes sense, because when theta is 0 (no bank), cosine is 1 and L = m g. Using algebra techniques, we can combine the above two questions that are in code blocks. The goals are to make the L's cancel away, m's cancel away, and combine the sine and cosine terms into just one term. We can divide the lower equation into the upper equation. Meaning, create a new equation where the left hand side is L sin (theta) over L cos (theta), and the right and side is m v^2 / r over m * g. Here is the result: Code: sin (theta) / cos (theta) = v^2 / [r g] Using the definition of tangent, the equation can be simplified to: Code: tan(theta) = v^2 / [r g] Sanity checking that equation, for small angle of banks, increasing bank angle increases tan(theta). This means, as you increase bank angle, you need to either increase speed, decrease turn radius, or decrease gravity in order to maintain stable turning flight, which all seems reasonable to me. Rearranging to isolate r: Code: r = v^2 / [g tan(theta)] Essentially, this is the equation on that Flight Learnings page @Somedudeintn linked to before. The last step here is to determine the proper value for g so that you can plug in knots and get back the radius in feet. I will convert the well known 9.8 m/s^2 into nautical miles per hour squared: Code: 9.8 m/s^2 * 0.000539957 nm/m * [3600 s / hr]^2 = 68578.86 nm/hr^2 So if you want the radius of the turn in nautical miles, square the speed in knots, divide by 68578.86, then divide again by the tangent of the bank angle. For example, if I'm turning with a speed of 75 knots and banking at 25 degrees, my radius of turn will be about .175 nm, and the diameter of the turn (how far you should be from the runway centerline to start) will be 0.35 nm. Since there are 6076 feet in a nautical mile we can change that 68578.86 number of a factor of 6076 in order to make the equation spit out feet instead of nautical miles. 68578.86 / 6076 is 11.28, which is almost exactly what @Somedudeintn provided in that link (rounding errors likely caused a very slight discrepancy.) This confirms that the first equation on the linked page is correct. None of this is useful unless you know how to choose a proper bank angle. The last thing I'll do is relate bank angle to rate of turn. If the plane turns 3 degrees in a second (standard rate), it actually swept out 3 degrees of an arc around the center of the circle after one second. The length of that arc is the angle (in RADIANS) times the radius (commonly referred to as S = theta r). We also can take as given that distance equals speed times time. From the first relationship I just discussed: Code: angle = arc-length / r Since the arc-length is a distance, and distance is always speed times time, we can combine those two relationships like this: Code: angle = v * t / r We know rate of turn is usually provided in degrees per second (e.g., 3 degrees per second). We want to isolate a term that is of the form angle per time, as such: Code: angle / t = v / r Technically, that angle is still in radians. What the equation tells us is that the rate of turn, in radians per hour, is the speed in knots divided by the radius in nautical miles. I need to bring back the equation for radius from much earlier in this post. Recall from before: Code: r = v^2 / [g tan(theta)] Combining the above two equations in code blocks: Code: angle / t = [g tan(theta)] / v I will rename "angle / t" ROT, multiply by a factor of 180/pi, and divide by a factor of 3600, to make the units work out: Code: ROT = g tan (theta) * 180 / [pi * 3600 * v] Remember that g is actually 68,578.86. Using the above example, a 75 knot turn at a 25 degree angle of bank results in a ROT of 6.78 degrees per second, well in excess of standard rate. Combining numbers together: Code: ROT = 1,091.467 * tan(theta) / v What angle of bank would be standard rate? Solve for tan(theta): Code: tan(theta) = v * ROT / 1,091.467 Using 3 as ROT, 75 as v, the right hand side is .20614. Taking the arc tangent of that value while in degree mode, the angle of bank is 11.6 degrees. The thumb rule for standard rate turns is to take 10% of the speed (7.5) and add half of that (3.75). The sum is 11.25, which is pretty close to 11.6. Well, in case you were wondering where the equations came from, there it is.

I guess the "too long, didn't read" summary is that I showed where two equations actually come from. I showed that the formula for calculating the radius of a turn, given the speed of the plane and angle of bank, results from a simple centripetal force problem. It isn't a magical equation that can't be validated. It comes from basic physics. I then showed that the formula for relating angle of bank and rate of turn results from basic geometry. I compared it to the popular thumb rule to validate I wasn't way off. I always try to do this when I'm given equations in my pilot training. When I first started cross country flight planning, I learned how to solve the wind triangle by plugging the wind, course, and airplane performance information into my CX-2 flight computer. I then worked the trigonometry out by hand to verify I understood what the calculator was actually doing (basically just solving the law of cosines using numerical methods). My physics background does help me do all this, but I'm not Einstein. Everything I worked out above can be followed if you brush up on relatively basic physics and math, but I don't expect people to unless they are bored.

If y’all had paid attention in high school trig an geometry class, this post would be obvious. I created an Excel spreadsheet years ago to do all this.

The reason a plane turns in, say, a half loop is because it doesn't need a "horizontal component" to turn. It just needs "lift". Just sayin'...

That's a nice diatribe Ringy, but the formula he wants is a lot simpler. All you need is basic geometry, A standard rate 180 turn takes 1 minute. Given the airspeed you know how long of an arc that is. Dividing the half circle length by pi gives you the radius. The rest of the calculation is making sure that you have the units correct (since the speed is NM/hour and we're talking about a minute). You don't need to compute bank angles or forces or anything to make the literal answer to the question "What is the diameter of turn for a constant rate turn at a given speed."

I tend to associate "turning" with horizontal changes in direction. Of course you are right that you can turn vertically or diagonally as well. The requirement is that there be a component of lift (combined with gravity) perpendicular to the direction of motion of the airplane. This of course goes beyond the scope of the original question, but is a good reminder to me that real pilots tend not to think so "horizontally" as I do. As @dmspilot points out, the original post was not specifically focused on standard rate turns. I spent some time on standard rate turns because it was an easy way to validate that my relationship between bank angle and ROT was consistent with an alternate thumb rule, but realistically you aren't actually using standard rate turns in the pattern. "Diatribe" may not mean exactly what you think it does. There was no intended criticism in my post... I was simply showing where the equation that answers the OP's question came from. Then, because it is also potentially useful for using the first equation, I showed how to relate ROT with bank angle and where it came from.

haha just goes to show you, the PhDs among us will fail the simplest exam; ref the first 6 words in the thread. “Why use a large word when a diminutive one will suffice?” lol, just funnin’ ya.

The colored lines are specific to the P-51D but the grey lines including the Xg and Yft (radius) lines are just physics and applicable to all airplanes. For example, a 7g turn at 250 mph will have a radius of ~600 ft and ~35 deg/sec turn rate. That's from the physics @RingLaserGyroSandwich posted. Nauga, from the doghouse

You can download blank ones here: https://evoisard.freeshell.org/doghouse/ Nauga, and Teen Talk Barbie

Cosine, Shmosine. I don't know how to do that stuff. I just eyeball my downwind. I'm at 90 knots abeam at a 1000 feet. I bank 30 degrees, start slowing, kick in in flaps in 10 degree increments and roll out on final at 65 knots, 30 degrees of flaps. Bank stays constant at 30 degrees the entire time. I roll out on centerline and put her on the Numbers. How far out was my downwind? Ain't no wind.

Here is a chart that tells us the lateral distance needed to make a 180 turn. It says what distance from the extended centerline must you be, using a constant rate turn from downwind to final, at 80kts, 100kts, 120kts in zero wind conditions. I used 25° bank. I would have done a little more to make the chart more useful, but I didn't want to provide anything beyond exactly what you asked for. I wouldn't want to fail.

I never took 1 geometry class in HS and never took trig in HS or college... Just eyeball it and it'll be fine

I took that formula and did this to get it on one line and calculate for Diameter. TBA = tangent of bank angle. D = (V^2/(11.26*TBA))2 Now I want to solve for BA, bank angle. I can’t figure it out. I was thinking about Foreflight’s traffic pattern thing in it’s Procedure Advisor. Near as I can tell it puts the downwind 4550 feet from the Runway so this is the Diameter. (yeah, yeah, I know ya don’t have to follow the downwind line on the map.) Often there’s a particular distance you want to be on downwind and speeds you usually want to stay fairly consistent with in the pattern. Bank angle is easily changed. So how do you solve for bank angle when velocity and diameter are the knowns?

Using the general form of your equation (is that last 2 a typo?): D = V^2/(g*tan(bank angle)), where g is the gravitational constant (32.174 ft/s^2, etc...) and V is in like units (e.g. ft/sec) the equation would be: bank angle = arctan(V^2/(g*D)) If speed is in knots and D is in feet then it looks like your version: bank angle = arctan(V^2/11.26*D) Nauga, who just rolls 90 deg and yanks

You are correct about the factor of 11.26... in fact in my first post in this thread I validated the number. Here is a graph which shows what bank angle you will need at various airspeeds assuming you start 4550 feet from the runway centerline:

Thanks. I got no idea where the 11.26 comes from, it’s just the number that was in the formula. Now how about solving for Y, where Y= Yank. Would it be the velocity of the stick during yank or the pressure on the stick during yank?

Radius = Speed / Rate of Turn. Bank angle does not matter. If speed is in knots, then rate of turn must be in radians/hr (which is deg/sec multiplied by 62.8). In other words: Radius (NM) = Speed (knots) / ROT (deg/s) /62.8 At 100 knots, and standard rate (3 deg/sec), you get a radius of 0.53 NM.

Since the OP is not specifically asking about standard rate (just constant rate), bank angle does matter. Your method works if the OP selects a specific rate of turn, perhaps, but the OP would likely need some help selecting a rate of turn. One solution is to use the equation from my first post in the thread which relates rate of turn to bank angle. Another way is to use an equation which accepts bank angle as an input. This has all already been discussed in the thread, though.

Why this question? It's not like you can actually measure the precise offset from the runway when you're in the air or judge the precise point of your turns. So what's the point?

Well, someone needs to say it. You shouldn't be making one continuous turn to final from downwind in the first place. You should make a square pattern, so you can roll wings level on base and look left and right to make sure you aren't the cutter or cuttee out of the pattern.

Can you not, at the midpoint of downwind, glance at the gps distance to the airport and pick off your approximate distance? If not, I can visually tell the difference between 1/2 mi and 3/4 mile so if we knew an approximate answer that would help me get thinking about it. And "what's the OP's point?" Well I am all about practicality so I'm glad you asked. I am planning on Osh. So, I study it. One of the things to look at is crash avoidance. Looking at how others have crashed at Osh. Looky, the FAA has put out a Safety Alert on this: https://ntsb.gov/safety/safety-alerts/Documents/SA_053.pdf (direct PDF download) Aha! It says, "Accidents have occurred when pilots were too slow and stalled, used an excessive bank angle (resulting in an accelerated stall), or overshot the runway (resulting in a cross-control stall) when turning from downwind to base leg or from base leg to final." The airplane I am flying right now <<<avatar is rather intolerant of such maneuvers because of its wing, compared to all other a/c I have flown. So you can see why this interests me. "I know how to fly" "I can handle that" "I do that turn all the time" is baloney to me. I want to know how close I can be to safely make that downwind to base to final turn (assuming no wind and a constant 25° banked turn, as these variables would complicate it too much to use) [ya I could use 30° but I like a buffer]. It's just a baseline that I can modify for later. It's something to help me get thinking about it. It's not something that is taught in transition training. We fly a 'normal' pattern but have never discussed the specific minimum distance. Yes it's variable but we should have somewhere to start. There is a really cool video that talks about the red zone, the area you would enter if you overshot final and needed to crank it over to get back...and how many people have crashed doing that; I will try to find it. That's why! Thanks for askin'. Others may disagree with this type of thought. That's cool. I have my own way of approaching things. Has worked ok for >40 years of flying.

There is considerable disagreement out there about this right now, but we can certainly discuss that in another thread. Bring it up.

Yeah. I was thinking about OSH and other places where they want you ‘right over’ something on downwind. You can get that distance from the runway easily. Now, what bank angle do you need? That’s why my question in post #28.

I like the OP's thought process! That's the thing about OSH. You can't count on a "standard pattern". On 27, for example, you might get sent out over the lake for spacing for slower airplanes ahead on final, or you might get asked to turn inside them (in which instance it won't be a square pattern for any of the faster airplanes). You might have to land on the threshold, or long on another colored dot. And they might change the aiming point when you are on short final. I've dealt with all of these arriving at OSH in the Aztec. When I was there in 2015 someone piloting a piston Malibu overshot the turn to final and tried to save the landing. Stalled it cross-controlled, fortunately low enough to pancake it roughly wings level onto the runway, starting a fire as the engine separated. Bad scene. Just one example.