40g turn?
02-02-2007, 04:41 PM
#1
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40g turn?
Having a intense discussion in "Sport Flying" and wonder if anyone here has a good estimate of the speed and turning radius of a pylon racer that is capable of 170 mph or thereabouts in the straight legs? Another flyer suggests a pylon plane pulls 40g in a 50 ft diameter turn. I say it would break up or skid wide if it attempted such a turn.
What say the racers?
What say the racers?
02-02-2007, 06:13 PM
#3
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RE: 40g turn?
Charlie,
Good math here: http://www.rcuniverse.com/forum/m_52...tm.htm#5304260
Short answer is, at least 30 G for the high-zoot models.
Duane Gall
RCPRO
Good math here: http://www.rcuniverse.com/forum/m_52...tm.htm#5304260
Short answer is, at least 30 G for the high-zoot models.
Duane Gall
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02-02-2007, 06:48 PM
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RE: 40g turn?
Some previous doodles
Well I was doing the math to find out what the shortest way around the course was and how many Gs it took to get there. The plan was to examine 100 ft diameter turns and 50 ft turns. My intuition tells me that a pylon turn is a spiral path turn. But it doesn't tell me whether it's converging or diverging. Neither converging or diverging is, of course, a circle. I chose that. 100 ft turns and on-topping the pylons makes a course 2.71 miles long. Completing this course in 65 seconds requires an average speed of 150 mph. A 100 ft diameter circle at that speed gets you over 30 Gs. I went no further as I was laughing hysterically. 15 Gs of course will produce a 200 ft circle.
Please someone tell me that centripetal acceleration does not equal velocity squared divided by the radius. 150mph is 220fps. Square it then divide it by 50ft then divide it again by 32 ft/secsquared.
30Gs on a 4 lb plane is of course 120lbs. It might seem, upon further math, that a perfect 500 sqin 66012 might produce 240lbs before stall. Given that something well above 120lbs could be a reality and probably is.
Denis
Well I was doing the math to find out what the shortest way around the course was and how many Gs it took to get there. The plan was to examine 100 ft diameter turns and 50 ft turns. My intuition tells me that a pylon turn is a spiral path turn. But it doesn't tell me whether it's converging or diverging. Neither converging or diverging is, of course, a circle. I chose that. 100 ft turns and on-topping the pylons makes a course 2.71 miles long. Completing this course in 65 seconds requires an average speed of 150 mph. A 100 ft diameter circle at that speed gets you over 30 Gs. I went no further as I was laughing hysterically. 15 Gs of course will produce a 200 ft circle.
Please someone tell me that centripetal acceleration does not equal velocity squared divided by the radius. 150mph is 220fps. Square it then divide it by 50ft then divide it again by 32 ft/secsquared.
30Gs on a 4 lb plane is of course 120lbs. It might seem, upon further math, that a perfect 500 sqin 66012 might produce 240lbs before stall. Given that something well above 120lbs could be a reality and probably is.
Denis
02-02-2007, 07:57 PM
#5
Thread Starter
RE: 40g turn?
Thanks guys. I figured 20g was a hairy stress on a model, twice what a human pilot would find possible for even a few seconds. 30g with a model still seems like a lot, but I guess I'll have to condcede could be engineered in a R/C racer (think what a Sidewinder missle must pull).
I'm still wondering what the side slip must be even fully banked and pulling with the elevator. That 200 ft half-circle makes a lot more "intuitive" sense wth its still stout 15 g.
I'm still wondering what the side slip must be even fully banked and pulling with the elevator. That 200 ft half-circle makes a lot more "intuitive" sense wth its still stout 15 g.
02-02-2007, 11:31 PM
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RE: 40g turn?
I find that a very big number to belive but..... On a model there is nothing that weighs anything to get pulled one so it might be possible if this makes any sense to you.
02-03-2007, 07:11 AM
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RE: 40g turn?
I have a calculator at [link=http://www.rcpro.org/rccalc/GForce.aspx]G-Force Calculator[/link].
I got the formulas from an aerodynamicist.
I too found the high numbers a little high to believe. And there are a lot of factors involved in the force and weight distribution. But I was told that if I wanted to be sure that my wings would hold up, I needed to do static "sandbag" testing at the calculated loadings.
I got the formulas from an aerodynamicist.
I too found the high numbers a little high to believe. And there are a lot of factors involved in the force and weight distribution. But I was told that if I wanted to be sure that my wings would hold up, I needed to do static "sandbag" testing at the calculated loadings.
02-03-2007, 08:21 AM
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RE: 40g turn?
If one is realy worried about this then add some weights to the wing tips. In the turn the heavy tips will try and pull the tips outward (downward). This will off set the usual tendency for the wing to fold inwards (Upwards).
Ed S
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02-03-2007, 09:13 AM
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RE: 40g turn?
Don & everyone;
If you were to make constant radius turns at both ends, let's say, around the #2/3 end of the course, entering 10' wide and exiting 10' wide, then the radius would be 60'.
The actual distance around the course in this case is the path around the two 60' radius half-circles at each end plus the two 610' straight-a-way legs which equals:
Distance around the course = (120 x Pi) + (2 x 610) = 1597 ft. and thus for 10 laps distance = 15970 ft. = 3.02 miles (or so)
You can easily estimate the average speed of the plane (within a few percent) by dividing the distance for ten laps (in miles) by the time for ten laps in hours. A 428 plane does the 10 laps in about 65 seconds = 0.01806 hours.
speed = 3.02/0.01806 = 167 mph = 246 ft/sec!
Then using the formula Accelleration = Speed^2/Radius Where Accelleration is in ft/sec^2; speed is in ft/sec; radius is in ft
accelleration = 246^2/60 = 1008.6 ft/sec^2 = 1008.6/32 g's = 31.5 g's. (for a 50' radius turn Accelleration = 37.7 g's)
The number of g's generated has nothing to do with wing area, airfoil type, wing planform, weight, and so forth. It only has to do with the actual curved path the plane takes takes and it's speed through that path.
An interesting concept is that the weight of the wing itself doesn't contribute to any (or at least almost no) wing bending stresses during a turn. This is true because most wings have a uniform weight to area ratio. It is concentrated loads that cause bending stresses in the wing. So, any calculations or tests on the amount of stress the wing can take must look primarily at the fuselage/engine/landing gear/radio/etc loads that are concentrated at the root of the wing. If you had all these things distributed evenly out the wing, then the wing could operate at extremely high g loads.
Example:
Let's say you want to "proof-load" your wing by conducting a sand-bag type test.
Wt of entire plane (428) = 3.75#; wt of wing = 14 oz; therefore weight of concentrated loads at root - 60-14 = 46 oz = 2.88#
Support the wing inverted at the root up off the floor or table with a "rice bag" type cushioned support (to softly spread the load). Put spacers under the wing tips. Then apply weight in 2.88# increments to the bottom of the wing evently distributed pro-rated per area. After placing the load on the wing, remove one of the wing tip spacers to see if the wing holds. If it does, replace the spacer, add more weight and repeat. Therefore, in order for the wing to be able to sustain at least the above calculated 31.5 g's, then you must be able to place a total of 90.7# ( 31.5 x 2.88 NOT 31.5 x 3.75) on the wing in an evenly distributed manner.
These calculations are for a best case secenario. They don't consider turbulence, tail down force, extra hard pulls, etc. These other factors only ADD to the g's and can be minor (in the case of tail down-force) to major (in the case of turbulence and hard pulls).
Another factor in wing strength is how well the wing is secured to the fuselage. The attachment may fail long before the wing itself fails. Actual proof testing should be done with wing attached to the fuselage in the inverted position with the fuselage supporting the wing and its added loads. Also, the attachment method itself may cause what are called "stress risers" (local stress concentrating areas) in the root of the wing which could cause the wing to fail. Adding an extra layer of fiber-glass over the center section will solve this problem. By the way, the front two wing attach screws must withstand three to four times the load than the back two do.
Now when you consider what lift the wing must generate in a turn, you must use the entire weight of the plane multiplied by the g's. For example, the wing of a 3.75# plane flying a 60' radius turn will have to generate at least 3.75 x 31.5 = 118# of lift.
Don, I looked at your G-Force Calculator. It gives the exact same numbers as in my examples above.
Doug Bebensee
If you were to make constant radius turns at both ends, let's say, around the #2/3 end of the course, entering 10' wide and exiting 10' wide, then the radius would be 60'.
The actual distance around the course in this case is the path around the two 60' radius half-circles at each end plus the two 610' straight-a-way legs which equals:
Distance around the course = (120 x Pi) + (2 x 610) = 1597 ft. and thus for 10 laps distance = 15970 ft. = 3.02 miles (or so)
You can easily estimate the average speed of the plane (within a few percent) by dividing the distance for ten laps (in miles) by the time for ten laps in hours. A 428 plane does the 10 laps in about 65 seconds = 0.01806 hours.
speed = 3.02/0.01806 = 167 mph = 246 ft/sec!
Then using the formula Accelleration = Speed^2/Radius Where Accelleration is in ft/sec^2; speed is in ft/sec; radius is in ft
accelleration = 246^2/60 = 1008.6 ft/sec^2 = 1008.6/32 g's = 31.5 g's. (for a 50' radius turn Accelleration = 37.7 g's)
The number of g's generated has nothing to do with wing area, airfoil type, wing planform, weight, and so forth. It only has to do with the actual curved path the plane takes takes and it's speed through that path.
An interesting concept is that the weight of the wing itself doesn't contribute to any (or at least almost no) wing bending stresses during a turn. This is true because most wings have a uniform weight to area ratio. It is concentrated loads that cause bending stresses in the wing. So, any calculations or tests on the amount of stress the wing can take must look primarily at the fuselage/engine/landing gear/radio/etc loads that are concentrated at the root of the wing. If you had all these things distributed evenly out the wing, then the wing could operate at extremely high g loads.
Example:
Let's say you want to "proof-load" your wing by conducting a sand-bag type test.
Wt of entire plane (428) = 3.75#; wt of wing = 14 oz; therefore weight of concentrated loads at root - 60-14 = 46 oz = 2.88#
Support the wing inverted at the root up off the floor or table with a "rice bag" type cushioned support (to softly spread the load). Put spacers under the wing tips. Then apply weight in 2.88# increments to the bottom of the wing evently distributed pro-rated per area. After placing the load on the wing, remove one of the wing tip spacers to see if the wing holds. If it does, replace the spacer, add more weight and repeat. Therefore, in order for the wing to be able to sustain at least the above calculated 31.5 g's, then you must be able to place a total of 90.7# ( 31.5 x 2.88 NOT 31.5 x 3.75) on the wing in an evenly distributed manner.
These calculations are for a best case secenario. They don't consider turbulence, tail down force, extra hard pulls, etc. These other factors only ADD to the g's and can be minor (in the case of tail down-force) to major (in the case of turbulence and hard pulls).
Another factor in wing strength is how well the wing is secured to the fuselage. The attachment may fail long before the wing itself fails. Actual proof testing should be done with wing attached to the fuselage in the inverted position with the fuselage supporting the wing and its added loads. Also, the attachment method itself may cause what are called "stress risers" (local stress concentrating areas) in the root of the wing which could cause the wing to fail. Adding an extra layer of fiber-glass over the center section will solve this problem. By the way, the front two wing attach screws must withstand three to four times the load than the back two do.
Now when you consider what lift the wing must generate in a turn, you must use the entire weight of the plane multiplied by the g's. For example, the wing of a 3.75# plane flying a 60' radius turn will have to generate at least 3.75 x 31.5 = 118# of lift.
Don, I looked at your G-Force Calculator. It gives the exact same numbers as in my examples above.
Doug Bebensee
02-03-2007, 10:50 AM
#10
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RE: 40g turn?
A good analysis of everything Doug, except for a few small points.
You can shorten the course by flying a wider turn at 2-3. Draw out the course, bisect a line between 2-3 and move 50 feet from that point toward pylon 1. You are now standing 71 feet from pylon 2 and 3. That is the radius you want down there. This shortens the perfect course to about 2.69 miles.
Once you have numbers for the loads involved, you can back into the angle of attack for the wing. With that, you can see if that airfoil is still in the drag bucket in the turn.
A perfect course is not 20 turns, it is 39 turns plus what ever it takes to get to the first turn.
You can shorten the course by flying a wider turn at 2-3. Draw out the course, bisect a line between 2-3 and move 50 feet from that point toward pylon 1. You are now standing 71 feet from pylon 2 and 3. That is the radius you want down there. This shortens the perfect course to about 2.69 miles.
Once you have numbers for the loads involved, you can back into the angle of attack for the wing. With that, you can see if that airfoil is still in the drag bucket in the turn.
A perfect course is not 20 turns, it is 39 turns plus what ever it takes to get to the first turn.
02-04-2007, 07:30 AM
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RE: 40g turn?
Highplains;
Of course, you're right, a pilot could fly a shorter course from what I described, especially at #1 pylon. Whether he wants to do a "double-pull" or a "spiral-in" turn at #2/3 is another matter. Also, it appears that at #1, the turn radius is much less than 50' at times. I would suspect that if the average pilot consistently flew the course I described, he would do fairly well.
I like to go through this kind of analysis to "get in the ballpark" with my numbers and to see whether the things that I (or others) are saying are feasible or reasonable (or even possible). Also, I like to share these thought processes and some of the math and theory so that others can plug other numbers into the equations to examine other possibilities. I know I've learned a lot from these forums and I like to occasionally return the favor.
Doug Bebensee
Of course, you're right, a pilot could fly a shorter course from what I described, especially at #1 pylon. Whether he wants to do a "double-pull" or a "spiral-in" turn at #2/3 is another matter. Also, it appears that at #1, the turn radius is much less than 50' at times. I would suspect that if the average pilot consistently flew the course I described, he would do fairly well.
I like to go through this kind of analysis to "get in the ballpark" with my numbers and to see whether the things that I (or others) are saying are feasible or reasonable (or even possible). Also, I like to share these thought processes and some of the math and theory so that others can plug other numbers into the equations to examine other possibilities. I know I've learned a lot from these forums and I like to occasionally return the favor.
Doug Bebensee
