Quote of the Day
If you are not annoying someone you are not doing anything new.
— Michael Stainer
Introduction
Figure 1: North American X-15 Hypersonic
Rocket-Powered Aircraft. (Source)
I love to read Quora, and I often see interesting factoids there that I inspire me to put pencil to paper and verify them. This week, I read a response to the question "What does the pilot of a supersonic fighter feel when flying at Mach 3 at 40,000 feet?" I found one of the answers particularly interesting because of how the respondent generalized the question to make it more interesting. I love when people take a basic question and turn it into a more interesting question.
The response that intrigued me began by pointing out that an airplane flying at 40,000 feet with a velocity of Mach 3 would probably be melting – the same answer given by other respondents. The author then changed the question by increasing the airplane's speed to Mach 8, its altitude to 70,000 feet, and its location to the equator. I quote part of his response here.
Actually, as the velocity increases, a pilot or passenger would feel slightly less weight, because the Earth's curvature becomes very important. This gets a little unusual because things start to matter that you would not expect are relevant. For example, if you are flying straight east over the equator at Mach 8, you would feel only 87% of your weight, but if you were flying west at the same point, you would feel 93% of your weight. The difference is due to the Earth's rotation.
Since only a few manned research aircraft have even come close to this speed (e.g. Figure 1), I do not expect to be able to perform an actual test. However, a little math will confirm his statement. Let's dig in …
Background
Definitions
- Mach Number (symbol: M)
- A dimensionless quantity representing a fluid's velocity relative to the speed of sound, i.e.
, where vFluid in this case is the air speed of the aircraft, and cSound is the speed of sound at that altitude. (Source)
- Hypersonic
- In aerodynamics, a hypersonic speed is one that is highly supersonic. Since the 1970s, the term has generally been assumed to refer to speeds of Mach 5 and above. (Source)
Approach
Here is my approach to duplicating the respondent's results:
- determine the velocity represented by Mach 8 at 70,000 feet.
Mach number varies with altitude, and we need to determine the velocity in m/s for an airplane at 70,0000 feet with a speed of Mach 8.
- compute the velocity of an object on the Earth's surface at the equator.
Even an object sitting stationary on the equator is actually being carried along by the rotation of the Earth. This movement causes the weight of an object to be measurably less at the equator than at the pole by the factor 288/289, which I will demonstrate in my analysis.
- compute the reduction in apparent gravity on an object due to centrifugal force on an airplane traveling easterly at Mach 8 and 70,000 feet.
When moving easterly, the airplane's velocity adds to the intrinsic velocity of an object being carried along with the Earth's rotation.
- compute the reduction in apparent gravity on an object due to centrifugal force on an airplane traveling westerly at Mach 8 and 70,000 feet.
When moving westerly, the airplane's velocity subtracts from the intrinsic velocity of an object being carried along with the Earth's rotation.
- the weight reduction at the equator while flying at altitude is relative to the person's sea level weight at the pole.
At the poles, a person has no weight reduction due to centrifugal force.
Analysis
Figure 2 shows my mathematical work that duplicates the results cited above (87% and 94%), with the final numbers marked with light-yellow highlights.
Figure 2: My Analysis of the Weight Reduction at the Equator.
Conclusion
I was surprised that high-speed, high-altitude flight could experience enough centrifugal force to have a measurable effect on the weight of a person. At first glance, I found the weight reductions stated to be almost unbelievable – however, a bit of physics shows that the numbers are correct.
Hi, your weight reduction formula for easterly travel included a term of:
(V_Earth)^2/R_Earth. This is the Centrifugal acceleration of a point on the earth's surface which seems to have nothing to do with an aircraft flying at 8 Mach at 70,000 ft. What did I miss?
Thanks in advance
Hi Ronan,
That term should not be in the easterly travel equation. I have made the correction. Thanks for the help in getting things right!
mathscinotes
HI, I sometime feel like that snarky kid that points out where the teacher has not crossed their letter T's or dotted their letter I's. I believe you updated your code but your Figure 2 has not been updated.
I think of you as a great help! The only folks that I struggle with are the flat earth people, who have descended on me in droves.
As far as Figure 2, you were correct. I am having an issue with WordPress. If I replace an image with an image of the same name but later date, WordPress is grabbing the first version. I am now going to give each image a rev number. Thank you for the help. It is greatly appreciated.
mathscinotes
Great to see these calculations. I empathise with your attracting flat earth interest. My whole reason for coming here was the opposite - flying around the equator in a flat earth model would involve similar physics but laterally. I am keen to know how much a person would lean sideways (both East and Westerly) at 500mph (altitude not a factor here). For flat earth adherents to demonstrate using a pendulum and adjustments for track. Of course the resulting 'debate' could become immensely irritating, but for me as a psychologist also rather interesting.
Antigravity is a concept that still people consider a science-fiction utopia. Information on antigravity research has begun to appear in the media lately. In documents preserved in ancient Sanskrit in ancient India over 15,000 years ago, an anti-gravity propulsion method is presented as: a centrifugal force strong enough to counteract all gravitational pull.
In 1992 the scientist Eugen Podkletnov, achivd a very complex device with superconductors (-1970 C), using a high-speed rotation(5000 rot/min) in liquid nitrogen and ultra-powerful electric field (1 M V). Podkletnov claimed a 2% weight reduction. Unfortunately the experiment could never be repeated.
Fig. 1 MAGIC Engine. Scheme of principle
The MAGIC Engine is a device by which rotating, cancels 100% of the weight of the masses. This method was applied when designing, experimenting and making the centrifugal inertial anti-gravitational engine.
Fig. 2 MAGIC Engine. Experimental Model
MAGIC Engine uses a classical electric motor and a drive system to rotate the weights (batteries, fuel, hydraulic oil, etc. ) of the device at a certain angle and a certain speed so that weights should "float". A powerful centrifugal force is generated, which totally cancels the gravitational force of the rotated masses.
Fig. 3 Real MAGIC Engine
By equipping cars, motorcycles, drones, helicopters, planes, missiles, etc., with this MAGIC Engine, a reduction of their weight, a decrease of their fuel consumption, as well as an increase of their payload are achieved. The MAGIC Engine has the following advantages:
- a simple construction and cheap maintenance - high safety in operation
- there is no danger of fire or explosion when operating - low noise in operation (maximum 10 dB)
- it becomes operational in a short time (10 sec)
- it does not produce turbulence or running air
- a technological reduced impact on industry when implemented in production - low sale price.
Some general possible uses of different equiped systems with MAGIC Engine are presented in the following pictures.
Fig.4 Electric car
An electric car weighs 2000 Kg, half of which represents the weight of the batteries. MAGIC Engine may reduce the weight of the car (50%) and double the active driving time.
Fig. 5 Drona
A two people drone weighs is 400 Kg, the weight of the batteries is 200 Kg and the operating time is 30 minutes. MAGIC Engine, will reduce the weight and increase twice the operating time. \
Fig.6 Helicopter
If MAGIC Engine is used, the weight of the helicopter can be reduced up to 30%.
Fig.7 Passenger Airplane
If a passenger airplane weighing 300 tons, uses MAGIC Engine will reduce the weight of the plane and its fuel consumption by about 30%. It will also increase its payload.
Fig. 8 Spaceship
A spacecraft weighs 2000 tons, 80% representing its fuel. If we equip it with MAGIC Engine, the weight will be reduced as well as its fuel consumption, while its payload will be increased.
Dr. Eng. Puiu Caneparu
tel: 0040/722/210443
e-mail: [email protected]
Is an anti-gravitational engine possible?
As fighting gravity seems to have been a futile effort , why not try a more intelligent technique, that of “ bypassing” gravity.
In documents preserved in ancient Sanskrit in ancient India over 15,000 years ago, an anti-gravity propulsion method is presented as: a centrifugal force strong enough to counteract all gravitational pull.
This was the basic idea when starting a series of experiments to “bypass” gravity. After a series of tests, I found out a way to "bypass" gravity, to achieve an anti-gravity effect (Fig. 1), an effect of reduction (even cancellation) of the mass of a body in motion, under certain conditions.Frame 1, bent at an angle ϕ , has the slide 2. By rotating the frame with a certain speed, the slide is risen. When the slide reaches the top of the frame, a sensor (not shown in the figure) controls the reduction of the frame speed, the slide descends slightly, the sensor notices the new position of the slide and commands a higher speed rotation of the frame, resulting in the slide being risen in the side, upper frame, a. s. o. In fact by the controlled movement of the slide (a few centimeters) , a part of the generated centrifugal force Fcf ( a part is normal to the direction of sliding, and a part is parallel to the direction of sliding Fcfx) cancels a part of the weight of the slide G=mg (a part is normal to the sliding direction, and, one part is parallel to the sliding direction Gx).
This way the sliding "floats" without having any weight.
I have created an artificial gravity by achieving an inertial force due to the rotation of the system, which minimizes the effects of a gravitational force.
Threfore the artificial gravity produced (or the gravity of rotation) is the result of a centrifugal force achieved in a rotating reference frame.
Σ Fx = 0; Fcf cos ϕ- mg sin ϕ = 0