How long would it take to fly 1200 miles in a single engine plane?
Depends on the plane. Single engine planes range in cruise speeds from around 75 knots airspeed (Piper Cub) to 235 knots airspeed (Cessna TTx). So 1200 miles (1931 km or 1042 nautical miles) give us a range of 13 hours 54 minutes to 4 hours 26 minutes.Now you’re well over the maximum range of the Piper Cub, so the pilot’s going to have to stop a number of times for gas, and to get something to eat, stretch their legs, and go to the bathroom. (We have to be practical!) And 13 hours is a LONG time to pilot a plane.The Cessna TTx is right at the limit of it’s range. Although you could technically fly it like that IF you had the right wind, it’s a really bad idea. And if you have a headwind you’ll run out of gas before you get there. So you’ll want to stop for gas, etc. too. But in this plane you only have to stop once.So those numbers above aren’t right. Each stop will lengthen your trip. And that’s without really considering the wind. If you get a tail wind, you’ll go faster relative to the ground, headwind, the opposite. So depends on the wind, which depends on the weather.Airlines get around the variability of weather (most of the time) by knowing the worst case of how long it’ll likely take and scheduling for that. They also have the capability of going faster to make up some time too.
An airplane pilot wishes to fly due west.?
This is a vector problem. You want to fly due west, but you have to fly in another direction to balance out the wind that is pushing you to the south. If you're flying at 305 km/h, and the wind is pushing you 72 km/h, then we have some information for our problem. Picture a right triangle with legs going south and west and the hypotenuse representing the direction you want to go. If you think about it logically, the hypotenuse would be pointing in a northwestern direction - you need to fly north to counter the winds pushing you south. We have the magnitude of the hypotenuse and the "south" leg of the triangle, so we can see that we can use some trig to find the angle north of west and the overall speed. In our triangle, the angle between the hypotenuse and the "south" leg can be found using cos: cos (angle) = adj/hyp cos (angle) = 72/305 angle = 76 degrees But since we're looking for the angle north of west, we need the complement of this angle, or 14 degrees. The ground speed can be found through either trig or the Pythagorean theorem. From looking at the diagram and thinking logically, you would expect the ground speed to be less than the given airspeed, since we're being pushed around by a headwind. I'll solve it using trig: sin (angle) = opp/hyp sin (76) = opp/305 opp = 296 km/h
An airplane pilot wishes to fly due west. A wind of 86.0km/h is blowing toward the south.?
An airplane pilot wishes to fly due west. A wind of 86.0km/h is blowing toward the south. (A)If the airspeed of the plane (its speed in still air) is 360.0km/h , in which direction should the pilot head? (B)What is the speed of the plane over the ground?
If you’re flying in a plane going 500 mph, how long would it take to travel one mile?
7.2 secSpeed of plane is 500 mph which is 500/60 mpmWhich is 500/60x60 mps→ 5/36 mpsNow to find the time in traveling 1 mile distance we use the simple formula d=s x td= 1 miles=5/36 mile per secNow t=d/s=1/(5/36) sec=36/5 sec=7.2 sec
A plane is flying 900 km/h due west and through a hurricane with winds of 300 km/h from the northeast. What is the ground speed of the plane and what is its direction relative to the ground?
Thanks for the question.It is given that the speed of the aircraft is 900 km/h, due West. Since it is asked to find out the ground speed, we will consider the 900 kmph speed as its True Airspeed.Since the aircraft is going towards west its track will be 270 degrees.Now, the wind is blowing from North East to South West at speed of 300 km/h. Therefore, the wind going towards 225 degrees.Ground speed is the sum of the wind speed and the true airspeed of the aircraft.Now when the wind is blowing at an angle to the aircraft, the wind component (headwind and tailwind), can be calculated by using the formula —Tailwind = Windspeed × CosA,where A is the angle between the wind and the aircraft.Wind angle = Direction of aircraft - Direction of wind.In this case, the wind angle is 270-225 = 45 degrees.So, tailwind component is—TW = 300 × Cos (45) => TW = 300 × 0.707Tailwind = 212.1 km/h.Therefore the ground speed of the aircraft will be 900 + 212.1 = 1112.1 km/h.To counteract with blowing wind and maintain the track , the heading of the aircraft has to be changed. So, we have to find out the Wind Correction Angle (WCA).WCA = (Wind speed × Wind angle) ÷ True Airspeed.Therefore, WCA = (300*45)÷900 = 15 degrees.Since the wind is coming ftom the right side, the heading of the aircraft has to be changed by 15 degrees to the right of West.Therefore, new heading will be 270+15 = 285 degrees.Therefore, the aircraft will have —Track = 270 degrees.Heading = 285 degrees.True Airspeed (TAS) = 900 km/h.Ground speed (GS) = 112.1 km/h.Tailwind = 212.1 km/h.Hope you got your answer. Thanks a lot.
An airplane has a ground speed of 350 km/hr in the direction due west. If there is a wind blowing northwest at 40 km/hr, calculate the true air speed and heading of the airplane?
Hello my friend. Thanks for the A2A…Airspeed is the speed of the aircraft relative to the wind blowing around it.Ground speed is the speed of the aircraft relative to the ground.There will be higher ground speed and low true airspeed if there is a tailwind. There will be low ground speed and higher airspeed is a headwind.Now, Ground speed = True Airspeed + Windspeed. So,True Airspeed = Ground speed — Wind speed.Aeronautical Calculations of Headwind and Crosswind.Here it is given that the aircraft is moving towards west. Therefore it is heading 270 degrees at a Groundspeed of 350 kmph.Now, the wind is blowing from Northwest.It means it is blowing from 315 degrees. The wind is blowing from NW to SE.So, the aircraft has both headwind component as well as cross wind components.So, angle between the wind and aircraft is therefore 315 — 270 = 45 degrees.It is to be remembered thatHeadwind = Windspeed * Cos (A).Crosswind = Windspeed * Sin (A).So, here, the headwind will beHW = 40 * Cos 45 = 40 * 0.707 = 28.28 km/h.CW = 40 * Sin 45 = 40* 0.707 =28.28 km/h.True Airspeed (TAS) =350 — ( — 28.28) = 350+28.28 = 378.28 km/h.Windspeed will be negative as it is a headwind. It will be positive as it is a tailwind.Due to the 45 degrees of cross wind, the track of the airplane will change to the below calculated Drift Angle.Drift Angle = (Windspeed * Angle between aircraft & wind) / True Airspeed.Drift angle = (40 *45)/378.28 = 4.75 degrees towards the Wind.So, the drift angle will be also 4.75 degrees.So, the new heading of the aircraft will be 274.75 degrees (West north-West); the true airspeed of the aircraft will be 378.28 km/h.See that the groundspeed is less than the true airspeed. So, it is headwind.Hope you got your answer. Thanks a lot..
Physic Help Please. An airplane is heading due north at 300 km/hr. The wind is blowing northeast at...?
50 sin 45 degrees = 50 * .707106781 = 35.355339 300 + 35.355339 in north direction 35.355339 in east direction find the total magnitude 335.3355339 * 335.3355339 = 112,463.203 35.3355339*35.3355339 = 1250 112,463.203 + 1250 = 113,713.203 SQUARE ROOT 113,713.203 = 337.21 Kilometers/Hour to find direction use arc tangent 35.355339/335.355339 =tan theta theta =6.018 degrees east of north
An airplane flies with a constant speed of 840 km/h. How far can it travel in 3 hours 20 minutes?
840 km/h divided by 60 = 14km/min3 x 60 = 180180 + 20 = 200 minutes total14 x 200 = 2800kmthat is a really awkward way of doing it though. really you can just do:840 x (10/3) = 2800kmenjoy :)
How do you calculate how many miles per hour an airplane travels?
I’ve read the answers posted so far and, without a ton of scrutiny, they are correct. Irrespective of the method you use, to obtain your speed, you have to firmly locate at least two points, called ‘fixes’, to establish the Great Circle distance between them. Once you have the distance, you divide that by elapsed time between the fixes to obtain your ground speed. In the old days, when flying over the ocean where no land or electronic fixes were available, a navigator would have to obtain at least three celestial (star/planet) Lines of Position (LOP’s) to obtain each position. That was a lot of work and wasn’t easy, especially in turbulence or obscured sky. Since one of those fixes could take 20 minutes and the airplane was flying 6 miles per minute, that means you were really figuring out where you were 20 minutes or 120 miles also. When flying from Phuket to Diego Garcia in the Indian Ocean, no fixes were typically available during the day and the navigator had to rely upon a Sun Line LOP to estimate position and resultant speed. This method could be pretty hairy because the next point of land after Diego Garcia was the bottom of the ocean or Antarctica.With that said, if you want to get a little depressed, calculate your speed for a domestic airline trip between two rather close airports such as Portland and Seattle. When you make your calculation, include all the time it takes from the moment you are dropped off at one airport to the moment you leave the other airport. You will find that you spend a significant amount of time combating/tolerating/gnashing your teeth over delays related to infrastructure inefficiencies such as TSA, parking, baggage claim, passenger loading, blah, blah….. For this reason, commercial aviation should probably focus more on infrastructure efficiencies rather than attempt to sequeeze another slight increase in airplane airspeed. If Amtrak ever got their act together with high speed rail, commercial airline short routes would find a real competitor.Note that when you are navigating over water, you typically measure speed in knots (nautical miles per hour) versus (statute) miles per hour. A knot, 1.15 statute miles per hour, is mathematically better suited to Mercator or Lambert Conformal nautical charts.