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#1
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Anyone know of a formula for determing the current bank angle, given the
current velocity and rate of change in heading over a particular period of time? I was thinking of adding an artificial horizon to program that I'm working on and figured that by taking the current velocity as reported by the GPS and the change in heading as reported by the last two course-over-ground measurements from the GPS, I should be able to determine a close approximation of the actual bank angle... Of course, this assumes that the current winds are ignored in addition to assuming a coordinated turn... The current system that I am using only gets updates at 1 Hz which might be a bit low, but I'm thinking of getting the Garmin 18 5 Hz unit which should be sufficiently fast in its update rate... |
#2
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![]() "Grumman-581" wrote in message news ![]() Anyone know of a formula for determing the current bank angle, given the current velocity and rate of change in heading over a particular period of time? I was thinking of adding an artificial horizon to program that I'm working on and figured that by taking the current velocity as reported by the GPS and the change in heading as reported by the last two course-over-ground measurements from the GPS, I should be able to determine a close approximation of the actual bank angle... Of course, this assumes that the current winds are ignored in addition to assuming a coordinated turn... The current system that I am using only gets updates at 1 Hz which might be a bit low, but I'm thinking of getting the Garmin 18 5 Hz unit which should be sufficiently fast in its update rate... bank angle = tan ^-1 ( 2*PI*v / g *t ) v= velocity g= gravitational acceleration t = time for 360 degree turn rate PI =3.14159 make sure you use consistant units for v, g and t for SI use v in m/s g = 9.8 m/s^2 t in seconds in imperial g = 32 ft /s^2 ( If I remember correctly) v in ft/sec t in sec Terry PPL downunder |
#3
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Grumman-581 wrote
Anyone know of a formula for determing the current bank angle, given the current velocity and rate of change in heading over a particular period of time? I think that you are going to have to work with TAS instead of GS, but Figure 2.29 on page 179 of Aerodynamics for Naval Aviators has a graph of all the factors involved. Bob Moore |
#4
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On Tue, 09 Jan 2007 13:17:32 +0000, in
28, Bob Moore wrote: I think that you are going to have to work with TAS instead of GS, but Figure 2.29 on page 179 of Aerodynamics for Naval Aviators has a graph of all the factors involved. Yeah, to be perfectly accurate, I would need air speed and not ground speed in addition to wind angle corrections, but I was curious if I could get a good approximation of the bank angle from just what I'm getting from the GPS without having to add a considerably costlier device like an accelerometer / gyro with rs232 outputs... The Garmin 18 5Hz puck type GPS goes for around $150... A bit more than the 1Hz units that are typical for car navigation, but not entirely out of line for this project... The accelerometer / gyro systems that I've seen so far with rs232 outputs go for around $1K-$2K... I understand that there is some new work in accelerometers for PC-based pointing devices that have brought the prices down considerably compared to the previous devices like this: http://www.watson-gyro.com/products/...A_80_spec.html |
#5
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On Wed, 10 Jan 2007 00:03:22 +1100, in ,
d&tm wrote: bank angle = tan ^-1 ( 2*PI*v / g *t ) v= velocity g= gravitational acceleration t = time for 360 degree turn rate PI =3.14159 make sure you use consistant units for v, g and t for SI use v in m/s g = 9.8 m/s^2 t in seconds in imperial g = 32 ft /s^2 ( If I remember correctly) v in ft/sec t in sec Thanks Terry... Couple of questions though... 1. Is the resulting bank angle in radians? 2. Does the formula for the bank angle actually have "2*PI*v / g*t" in it or was it supposed to be "2*PI*v / (g*t)"? |
#6
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Grumman-581 wrote:
The accelerometer / gyro systems that I've seen so far with rs232 outputs go for around $1K-$2K... You can get a 3-axis accelerometer with analog output for around $300. I recently used one like this http://www.xbow.com/Products/product...s.aspx?sid=185 for a project in a ground vehicle. I used a three-axis one that takes power and gives you three analog outputs X, Y, Z. The same company also sells an RS232 A/D board but IMHO it's a little spendy at $300. If USB is OK I know you can get small USB DAQ boxes from several different vendors. I was using an older National Instruments full length ISA card, but then again it was OK for me to have an entire desktop PC case in the back seat of the car. ![]() None of this stuff is official aviation stuff (nice aluminum box, round connectors, etc) but then again it doesn't have an official aviation price tag either. Standard disclaimers apply; I don't get money from any of the companies mentioned. Matt Roberds |
#7
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![]() "Grumman-581" wrote in message news ![]() On Wed, 10 Jan 2007 00:03:22 +1100, in , d&tm wrote: bank angle = tan ^-1 ( 2*PI*v / g *t ) v= velocity g= gravitational acceleration t = time for 360 degree turn rate PI =3.14159 make sure you use consistant units for v, g and t for SI use v in m/s g = 9.8 m/s^2 t in seconds in imperial g = 32 ft /s^2 ( If I remember correctly) v in ft/sec t in sec Thanks Terry... Couple of questions though... 1. Is the resulting bank angle in radians? 2. Does the formula for the bank angle actually have "2*PI*v / g*t" in it or was it supposed to be "2*PI*v / (g*t)"? The bank angle is in what ever units you want. it is the angle expressed as degrees or radians that gives a tangent of 2*PI*v /( g*t) (sorry I should have put in the brackets) check your equation with a standard rate 1 turn. At 100kts a rate 1 turn will have t =120 seconds and give a bank angle of 15.4 degrees. terry |
#8
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On Wed, 10 Jan 2007 18:31:29 +1100, in ,
d&tm wrote: The bank angle is in what ever units you want. it is the angle expressed as degrees or radians that gives a tangent of 2*PI*v /( g*t) (sorry I should have put in the brackets) check your equation with a standard rate 1 turn. At 100kts a rate 1 turn will have t =120 seconds and give a bank angle of 15.4 degrees. I tried it both ways today and pretty much figured that it needed the parentheses around the "g*t"... The atan function was returning things in radians, so that's what gave me the brain fart... I'm getting approximately 15.175 degrees for a bank angle, but that's probably because I'm using a more accurate measurement for pi and g... Not that it is likely to really matter since when I graph it, putting it to the nearest degree will probably be accurate enough... Haven't decided if I want to redraw the gauge each time or have a set of predrawn images that I shift around indexed by the bank angle... Thanks for the help... -- Mike Shelley N581 -- AA5A -- AXH |
#9
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![]() "Grumman-581" wrote in message news ![]() On Wed, 10 Jan 2007 18:31:29 +1100, in , d&tm wrote: The bank angle is in what ever units you want. it is the angle expressed as degrees or radians that gives a tangent of 2*PI*v /( g*t) (sorry I should have put in the brackets) check your equation with a standard rate 1 turn. At 100kts a rate 1 turn will have t =120 seconds and give a bank angle of 15.4 degrees. I tried it both ways today and pretty much figured that it needed the parentheses around the "g*t"... The atan function was returning things in radians, so that's what gave me the brain fart... I'm getting approximately 15.175 degrees for a bank angle, but that's probably because I'm using a more accurate measurement for pi and g... Not that it is likely to really matter since when I graph it, putting it to the nearest degree will probably be accurate enough... Haven't decided if I want to redraw the gauge each time or have a set of predrawn images that I shift around indexed by the bank angle... Its close but is it close enough at greater angles.? what are you using for the conversion of speed.? I used g =9.8 Pi as 3.14159 and v = kts *1852/3600 m/s terry |
#10
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![]() "d&tm" wrote in message ... "Grumman-581" wrote in message news ![]() On Wed, 10 Jan 2007 00:03:22 +1100, in , d&tm wrote: bank angle = tan ^-1 ( 2*PI*v / g *t ) v= velocity g= gravitational acceleration t = time for 360 degree turn rate PI =3.14159 make sure you use consistant units for v, g and t for SI use v in m/s g = 9.8 m/s^2 t in seconds in imperial g = 32 ft /s^2 ( If I remember correctly) v in ft/sec t in sec Thanks Terry... Couple of questions though... 1. Is the resulting bank angle in radians? 2. Does the formula for the bank angle actually have "2*PI*v / g*t" in it or was it supposed to be "2*PI*v / (g*t)"? The bank angle is in what ever units you want. it is the angle expressed as degrees or radians that gives a tangent of 2*PI*v /( g*t) (sorry I should have put in the brackets) check your equation with a standard rate 1 turn. At 100kts a rate 1 turn will have t =120 seconds and give a bank angle of 15.4 degrees. terry As an approximation speed in metres/second is half speed in kts eg 100kts is about 50m/s actually it is 52m/s. |
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