“True airspeed is important because performance charts in your Pilot Operating Handbook (POH) are based on true airspeed.  Remember that indicated airspeed drops off as we climb, due to lower air density at (higher) altitudes not having the same impact on the pitot tube. By using an E6B flight computer, we can arrive at true airspeed. Some airspeed indicators have a true airspeed ring around them that can be set for outside air temperature, pressure altitude, and allow you to read true airspeed off a scale on the ring. Glass cockpit aircraft typically have what is called an air data computer, which processes the above information and displays it for you. Ground speed is used to determine how long to a destination. True airspeed is used with the Pilot Operating Handbook to determine rates of fuel consumption over that period of time.”
Convert between Calibrated Airspeed (CAS), Equivalent Airspeed (EAS), True Airspeed (TAS) and Mach number (M) using the tool below. You need to specify the altitude at which you would like to perform the calculation as well as any one of the four airspeeds.
Use the sliders to select an input speed and whether to apply a temperature deviation offset from the standard atmospheric value. The model is based on the US Standard Atmosphere of 1976. Calculation MethodologyIndicated AirspeedIndicated airspeed is the airspeed reading that the pilots sees on her airspeed indicator (ASI) and is driven by the pitot-static system on board the aircraft. The system uses the difference between the total pressure (measured by the pitot probe) and the static pressure (measured by the static ports) to determine the dynamic pressure which is converted to an airspeed reading. The pitot-static system works on the principle of Bernoulli’s equation which states that an increase in the speed of a fluid must simultaneously result in a drop in the fluid’s pressure, or a decrease in the fluid’s potential energy. The total pressure (also known as the stagnation pressure or pitot pressure) is measured by the pitot probe. The moving air enters the probe and is brought to rest by the geometry of the probe. The measured static pressure is the ambient pressure of the still air which is the barometric pressure of the air at the aircraft’s current altitude. The static pressure is not only used to calculate the airspeed but also the altitude (altimeter) and the vertical speed (VSI) during flight. The static ports are always installed flush which ensures that the port opening is inside the boundary layer where the air is not moving. The airspeed is therefore calculated as follows: The density term in the denominator is not a constant and varies with altitude and temperature. However, the airspeed indicator in the cockpit is always calibrated to sea level density on a standard day. Thus the actual airspeed (true airspeed) will vary considerably from the indicated airspeed as the aircraft flies at higher altitudes and differing temperatures. The big advantage of using IAS in the cockpit is that the aircraft will always stall at the same indicated airspeed (for a given aircraft configuration) regardless of the altitude or ambient temperature. This makes it much easier for a pilot to fly the aircraft as the critical speeds that define the operating envelope remain the same regardless of the ambient conditions. Calibrated AirspeedCalibrated airspeed is the indicated airspeed corrected for instrument and position error. This error is a function of both the quality of the pitot-static system used to calculate the dynamic pressure as well as the location of the probe on the aircraft. Positional errors result from the fact that the local velocity around an aircraft varies as a result of the aircraft's changing geometry. For example, the local velocity over the upper surface of the wing is higher than below the wing in order for lift to be produced. In reality there exists velocity gradients all over the aircraft, especially in regions where there is substantial curvature (forward part of the fuselage, windshield, wing surface). Depending on the location of the pitot-static system, the measured dynamic pressure measured may differ from the actual dynamic pressure due to local induced velocity effects. The offset between indicated and calibrated airspeed is usually published in the aircraft operating manual in the form of a table. Here is an example of the calibration performed for a two-seat light aircraft. Equivalent AirspeedEquivalent airspeed is the calibrated airspeed corrected for compressibility effects. It is also defined as the speed at sea level, under ISA conditions, that would produce the same incompressible dynamic pressure that is produced at the true airspeed for the given aircraft altitude. It is this definition that makes EAS a useful airspeed measurement for aeronautical engineers as it provides a convenient way to calculate loading on the airframe, or handling qualities as the dynamic pressure provided is an equivalent sea level pressure without the need to correct for altitude or temperature. Indicated and Calibrated airspeed is based on the formulation of Bernoulli’s equation, which assumes that the fluid (air in this case) is incompressible. Bernoulli’s experiments were performed in water where this assumption is valid, but compressibility effects in air start to become significant at Mach numbers above 0.3. Divergence between CAS and EAS will be seen at speeds above 200 kts and altitudes above 10 000 ft. CAS must therefore be corrected for compressibility effects to determine EAS as an intermediate step to calculate the True Airspeed (TAS). Compressibility effects can be accounted for through the calculation of the impact pressure, which is a function of the Mach number. True AirspeedThe true airspeed is the speed that the aircraft travels relative to the air mass in which it is flying. The true airspeed is equal to the ground speed in cases where there is no wind, and is used mostly for flight planning and when quoting aircraft performance specifications. True airspeed can be calculated from either the equivalent airspeed, or the Mach number if the outside air temperature (OAT) is known. Mach NumberThe Mach number is the ratio of the True Airspeed to the sonic speed. The speed of sound in undisturbed air is a function only of temperature and not altitude as is often mistakenly assumed. Of course the ambient temperature will decrease as altitude is increased, leading to the reduction in the speed of sound as with increasing altitude.
View Full Version : INSTRUMENT QUESTIONS AVIATROZ 14th Sep 2011, 08:17 Q1. The ISA pressure for 18000feet are: anS. 506hpa plz Someone guide how to cal pressure change with altitude. Q2.An inc. of 0.15Mach result in an inc. of 93kt TAs of an a/c. the local speed of sound is: ANs-620KTs PLZ guide how to solve this problem Q3.An a/c is flying at 400ft from a high temp area to a cold temp area where the temp diff. is 20degree. what will be actual height of a/c: Ans-3680feet {As we are going from low density to high density area........the true height should decrease right!....i get that point but how to calculate barometric error......because QNH is not given:ugh:....only temprature diff of 20degree is given.........plz explain} Q4.An inc. of 0.15Mach number result in an inc. of93kts in TAs.if temp deviation from IsA Is +9DEGREE,the FL is: ANs--FL220{plz explain} Q5.The atmospheric pressure at FL70 in a "standard +10" atmosphere is: ANs---781.85HPa{plz expalin} Q6.Cruising at FL390,M0.84 is found to give a TAs of 499kt.the IsA deviation at this level will be: A. +17---------ans B.+!9 C.-17 {i m getting my ans as +22degree plz help} AVIATROZ 14th Sep 2011, 13:49 Q1. An a/c is flying a rate 1 turn at 480kt TAS. What is the diameter of the turn? Ans-5nm..............{i used formula ROT=Vsquare/g tan{theta}............but did nt able to get the ans................any suggesions} Q2.Assume a perfectly frictionless DI is corrected to give a zero drift on ground at 30N.The DI is set to read 270degree ib ab a/c tracking along 60degree parallel at a grdspeed of 545kts.The Di reading after 80 mins: Ans--283degree{plz explain elitepilot20 14th Sep 2011, 18:47 Q1. An a/c is flying a rate 1 turn at 480kt TAS. What is the diameter of the turn? Ans-5nm..............{i used formula ROT=Vsquare/g tan{theta}............but did nt able to get the ans................any suggesions} you can not calculate radius or diameter using this formula your TAS is 480Kts in rate one turn you need 2 mins to complete your turn therefore you will travel 480/60X2 = 16nm which is also your circumference And circumference = 2 X pie X r therefore 16 = 2 X 22/7 X r r= 2.54nm n diameter = 5nm:8 elitepilot20 14th Sep 2011, 18:59 Q2.Assume a perfectly frictionless DI is corrected to give a zero drift on ground at 30N.The DI is set to read 270degree ib ab a/c tracking along 60degree parallel at a grdspeed of 545kts.The Di reading after 80 mins: Ans--283degree{plz explain is there any option 285/286 because i am getting this but not sure at all how far i am correct on this. n waiting for other explanations as well. AVIATROZ 15th Sep 2011, 07:32 can u plz explain how you got 285as ans magnetic-north 15th Sep 2011, 08:25 (Mach 1) 620 ktas = 100% 6.2 = 1 % + mach 0.15 = 15x 6.2 = 93KTAS keith williams 15th Sep 2011, 10:20 Total wander = Earth Rate + Latitude Nut + Transport Wander Total wander = ER + LN + TW In the Northern hemisphere ER in degrees per hour = -15 degrees x Sin Latitude LN in degrees per hour = + 15 degrees x sin latitude nut setting TW in degrees per hour = (East to West ground speed in knots) x Tan latitude / 60 Latitude nut setting is 30 degrees Latitude is 60 degrees. East – West ground speed = 545 knots Flight time = 80 minutes which is 80/60 hours So ER = (- 15 x sin 60) x 80/60 = - 17.32050808 degrees LN = (+15 x sin 30) x 80/60 = + 10 degrees TW = (545 x tan 60) / 60 x 80/60 = 20.97705978 degrees Adding them all together gives +13.6565517 degrees Adding this to the initial heading of 270 gives 283.6565517 degrees. SuperflyTNT 15th Sep 2011, 11:41 Can anyone please share the formula to calculate RADIUS OF TURN Thank you elitepilot20 15th Sep 2011, 11:44 @superfly for radius use the circle circumference formula i.e. 2.pie.r keith williams 15th Sep 2011, 11:52 Radius Of Turn = TAS squared / g Tan AOB But be careful to ensure that you are consistent with your units. To get the radius in meters you need to use g in m/sec squared and TAS in m/sec. 1 knot = approximatley 0.515 m/s If you want to get the radius in nm by using TAS in knots then you need to convert g = 9.81 m/sec sqaured into nm/hour squared. (Most people choose not to do this) But for the question posted in this thread it is easier to use the circumference method as described by elitepilot. For rate of turn use the following equation. Rate Of Turn in degrees per second = g Tan AOB / TAS AVIATROZ 16th Sep 2011, 09:33 Q.An a/c initiate a rate one turn at a Tas of 270kts.how far will it have travelled after 25 sec. Ans--1.96nm elitepilot20 16th Sep 2011, 18:21 Q.An a/c initiate a rate one turn at a Tas of 270kts.how far will it have travelled after 25 sec. Ans--1.96nm @AVIATROZ you asking or explaining? AVIATROZ 16th Sep 2011, 18:35 @ elite pilot distance travelled by a/c will be TAs* 25sec= 270*(25/3600)=1.87 Which is approx to 1.9nm please advice if u think its not right method thanks elitepilot20 19th Sep 2011, 04:30 it is the right method according to me. hey have you got your roll no. for indigo yet?? AVIATROZ 19th Sep 2011, 04:56 no buddy........hav`nt got the hall ticket yet..........it seem i `ll only be able to give the attemptt in oct{or whenever the next attempt be}..........hoping for the things to turn positive as they are supposed to be...............what about you buddy elitepilot20 19th Sep 2011, 19:34 but you did receive the confirmation from radhika didn't you??? and about me, i didn't even get the confirmation from her. she asked for my details n went dark, haven't heard anything yet. Ques-(from indigo q. bank) echoes are received 30degree to left A/c hdg is 120M, find QTE? no variation given so is this question incomplete or am i not getting the trick? AVIATROZ 19th Sep 2011, 20:22 bro........i m also in same situation bro.........after the detail.........did`nt got any reply from them........anyways.........working hard for the next attempt........ about the question......... as the echoes are being received..........30degree to the left..........so it should 90degree...........now the QTE{true direction from the station} so i think it should be 090 + 180degree=270degree{true........if variation given then apply and get answer} i am not sure if this is right but lets see if someone else give his views on this elitepilot20 20th Sep 2011, 17:49 A (45N 10W) B (48*30*N 15W) SuperflyTNT 22nd Sep 2011, 16:03 Thank you for the explanation and formula. Much appreciated. Shabez 16th Oct 2011, 00:48 Aircraft with pressurized cabin in flight: When switching to the alternate static pressure source, the pointer of the Vertical Speed Indicator: A) indicates a climb, then settles down and reads incorrectly B) indicates a descent, then settles down and reads incorrectly C) indicates a slight continuous descent D) indicates correctly shouldnt this indicate a decent....high pressure inside cockpit so goes to static source indicates decent and then settles downs and indicates incorrectly...but the ans is listed as A Shabez 16th Oct 2011, 11:51 thanks Following 180° stabilized turn with a constant attitude and bank, the artificial horizon indicates: A) too high pitch-up and correct banking. B) too high pitch up and too high banking. C) attitude and banking correct. D) too high pitch-up and too low banking. i think answer ought to be A but it lists it as D i think this is the order in conventional artificial horizon 90 under read 180 correct 270 overread 360 correct pitch always too high except 360 keith williams 16th Oct 2011, 23:10 Your comments are all correct in this one Shabez. Anders S 22nd Feb 2012, 18:10 Hey guys, see if you can solve this one. :ugh: A gyro with no real drift is set to read the correct magnetic heading when the aircraft is heading 090° (T), variation 10°E. After 45 minutes flight eastwards along the parallel of 62N at 545k groundspeed, the gyro reading is: a. 100.5° b. 064.8° c. 075.0° d. 057.3° keith williams 22nd Feb 2012, 22:21 With 10E variation the initial indication when on 090(T) will be 080(M) Earth rate wander in degrees per hour = -15 x Sin Lat 45 minutes = 0.75 hours so total Earth Rate wander over this period is -15 x Lat 62 x 0.75 = -9.933 degrees Transport Wander rate in degrees per hour = E to W groundspeed x Tan latitude /60 When flying W to E the groundspeed is considered negative So over a period of 45 minutes transport wander = -545 x Tan 62 x 0.75 / 60 = -12.812 degrees. Total wander = -9.933 ER + -12.812 TW = -22.745 Final heading indication = initial indication + total wander Final heading indication = 080 - 22.812 = 57.255 degrees The closest option is 57.3 degrees. Anders S 23rd Feb 2012, 08:39 Ah, but of course! It helps if you calculate it the right order. I did 545/60 x tan(62), and obviously got an entirely different answer. Cheers. AVIATROZ 22nd Jul 2012, 13:42 Heelo aviator mates, need some help with couple of questions i am stuck with, Q1.what will be the TAS if cruising altitude is 39000ft,temperature is ISA+5 and CAS200kts: a. 388kts b. 383kts..ans c. 364kts d. 370kts Q2.At FL350 with a JSA deviation of -12,true airspeed when flying at M.78 is: a. 460kts b. 436kts..ans c. 447kts d. 490kts Q3.An a/c is flying AT fl350 with a JSA deviation of +8,The mach no. is.83 and the TAS485KTS.IF A/C descends to FL300 and maintains the same mac no. and TAS,the JSA deviation will be: a. +8 b. -2 c. +2...ans d. -18 {i am getting ans B option,may be i am making some mistake plz correct} Q4.The radius of turn @rate1,and TAS360kts is a. 10nm b. 5nm c. 7.5nm d. 2nm...ans thanks for the guidance. keith williams 22nd Jul 2012, 17:10 Q1. Using the POOLEYS CRP5 39000 ft is above the ISA tropopause so the ISA temperature will be –56.5 degrees Celsius. Adding +5 ISA deviation gives a temperature of –51.5 degrees Celsius. In the airspeed window set -51.5 degrees against 39000 feet. Against 200 knots CAS on the inner scale read off 388 knots TAS on the outer scale. This is greater than 300 knots so we must apply compressibility correction. Compressibility correction = (TAS / 100) – 3 Inserting the 388 knots TAS gives (388/100) – 3 = -0.88. In the compressibility correction window move the pointer 0.88 units to the left. Against 200 knots CAS on the inner scale read off 383 knots TAS on the outer scale. Q2. ISA temperature at 35000 feet = 15 (35 x 1.98) = -54.3 degrees Celsius. Adding –12 temperature deviation gives –66.3 degrees Celsius. Adding 273 to this converts it into 206.7 degrees kelvin LSS = 38.94 x Square root of absolute temperature So LSS = 38.94 x square root of 206.7 = 559.84 knots Mach Number = TAS / LSS so TAS = Mach x LSS So at M0.78 the TAS = 0.78 x 559.84 = 436.68 knots Q3. Mach Number = TAS / LSS so LSS = TAS / Mach So at M0.83 if the TAS is 485 knots we have LSS = 485 / 0.83 = 584.3 knots LSS = 38.94 x Square root of absolute temperature So Absolute temperature = (LSS /38.94) squared Inserting the figures from above gives Absolute temperature = (584.3 / 38.94) squared = 225.2 degrees Kelvin Subtracting 273 converts this into –47.8 degrees Celsius. ISA temperature at 30000 feet = 15 (30 x 1.98) = -44.4 degrees Celsius. Temperature deviation = Actual temperature – ISA temperature Inserting the figures above gives Temperature deviation = ( -47.8 ) – ( -44.4 ) = +3.4 degrees Celsius. The closest option is +2. Q4. A rate 1 turn takes 2 minutes to turn through 360 degrees. Dividing 360 knots by 60 converts into 6 nm per minute. So in 2 minutes the aircraft will fly 12 nm, which is the circumference of the circle. Circumference = 2Pi x Radius So Radius = Circumference / 2Pi Dividing 12 nm by 2Pi gives a radius of 1.9098 nm. The closest option is 2 nm. AVIATROZ 6th Aug 2012, 12:56 Aviator mates need some help with couple of question i am stuck with,your help will be really appreciated. Q1. The time from transmission of interrogation pulse to the receipt of the reply from DME ground is 2000microsecs{ignore delay in dme}The slant range is: a. 330nm b. 185nm c. 165nm...ANS d. 370nm Q2.Flying an ILS approach with 3degree glideslope referenced to 50feet above threshold,an a/c at4.6nm should be at approximate height of: a. 1400ft b. 1380ft c. 1500ft d. 1450ft......ANS {acc. to me the ans should be height=glide path angle*range*100feet=3*4.6*100feet=1380feet i.e. B option but that is not the answer plz explain.} .................................. INSTRUMENT QUESTIONS Q3.An a/c is fitted with 2 altimeters.One is corrected for position error,the other is not corrected for position error. a.Provided that ADC is working normally,there will be no error to either altimeter b.at high speed the non-compensated altimeter will show lower altitude c.at high speed the non-compensated altimeter will show higher altitude (ANS is {c}....but i think ans should be {b}...please explain.) Q4.A low altitude Radio altimeter used in precision approaches,following characteristics: 1. 1540mhz to 1660mhz range 2. pulse transmission 3. frequency modulation 4. height range between 0-5000feet 5.accuracy of +/-2ft between 0-500ft. a. 1,4,5 b 3,4.....ANS... c 3,5 d 2,3,5 {The height range of radio altimeter is from 50-2450 feet}explain,.....acc. to me and should be {c}) thanks Q3. bayblade 6th Aug 2012, 13:10 Q1. The time from transmission of interrogation pulse to the receipt of the reply from DME ground is 2000microsecs{ignore delay in dme}The slant range is: a. 330nm b. 185nm c. 165nm...ANS d. 370nm this is simple formula based. distance= speed * time. = (3*10^8)*(2000*10^-6) =600,000m =(600,000*3.28)/6080 Nm =323.68Nm but note time of transmission and reception is given i.e twice the distance. hence actual distance is 323.68/2 = 161.8Nm. bayblade 6th Aug 2012, 13:18 Q2.Flying an ILS approach with 3degree glideslope referenced to 50feet above threshold,an a/c at4.6nm should be at approximate height of: use the formula: note glideslope is referenced 50 ft above the threshold. but 3 deg takes you down to touchdown. hence you need to add 50 ft to the height obtained. height (in feet) = distance(nm) * glideslope angle *101.3 = 4.6 * 3 * 101.3 =1397.94 ft. hence height of the aircraft is 1397.94 + 50 =1447.9 ft. bayblade 6th Aug 2012, 13:28 An a/c is fitted with 2 altimeters.One is corrected for position error,the other is not corrected for position error. a.Provided that ADC is working normally,there will be no error to either altimeter b.at high speed the non-compensated altimeter will show lower altitude c.at high speed the non-compensated altimeter will show higher altitude position error comes into picture during any of the following events: a) during side slip b) during ground effect c) low speeds d) when aircraft is in a certain attitude. its got nothing to do with high speed. i guess the question means non-compensated for compressibility error, which comes into play during high speeds. the answer should be b according to me. Naveen Kalra 3rd Apr 2013, 18:41 Hi All, need your help for following questions of Radio Aids (GSP), Doppler Radar Chapter, Page no. 167-168. Many thanks in advance. Q3. An aircraft using a single beam Doppler system to measure its groundspeed is travelling at 300 m/s. Depression angle is 60 degrees. Transmission frequency is 10 GHz. What will be Doppler shift? Ans. 10 KHz. Q4. An aircraft uses a single beam Doppler system with depression angle 60 degrees to measure its ground speed. Transmission frequency is 10 GHz. What will the ground speed if the Doppler Shift is 5 KHz. Ans. 300 kt. Q7. An aircraft with a 4-beam Janus array has a depression angle of 60 degrees and an angle of 30 degrees horizontally between the beams and the aircraft longitudinal axis. What will be Doppler shift in the array if the transmission frequency is 11500 MHz and the ground speed is 600 kt.? Ans. 20 KHz. bayblade 3rd Apr 2013, 22:25 doppler shift= 2(relative velocity/wavelength)*Cos(depression angle). here, relative velocity = 300m/s wavelength = (3*10^8)/10G Cos60=0.5 hence doppler shift= 10K Hz the second sum is just the same type, you have to take care to convert the speed in kts to m/s though. for Janus array modify the formula as such: (4*relative velocity*cos dep angle*cos horizontal angle)/wavelength. |
What will be the TAS if cruising altitude is 39 000 ft temperature is ISA +5 and CAS 200 kt?
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