ewmon

Wistelee, I hope you realize that (I hope) the good doctor was exercising some jocularity here because 50±50 feet is 50–50 feet to 50+50 feet, better known as 0 to 100 feet. A more realistic range would be roughly ½ to 2 times the craft's length, depending on its application. A 10-foot fuselage length would have a wing span low limit of roughly 10 feet due to a few factors, one being that a wing's lift surfaces would probably not include the fuselage (most wing designs "pierce" the fuselage), which can be assumed to be a minimum of about 2 feet wide. At that extreme, you're flying in something like this: ( )

When you googled "transpulmonary pressure" and read the first few search results, what did you gather from them, and what left you confused?

Merry Christmas! (I thought I'd read a rant here about how "Happy Holidays" is woefully denying Christmas. When I was a kid, we seemed to accept "Seasons Greetings" in stride without any boohooing. I guess it included Hanukkah along with Christmas. Nothing's changed since then.) ... and to all a good night!

Consider the source ... Yahoo's "The Sideshow" blog written by Eric Pfeiffer. This aptly-named tabloid-style blog, which Yahoo choose to categorize as "news", does not focus on science or space aliens, but seems to have the common thread of sensationalism. Furthermore, I find no credentials on Pfeiffer, who unlike Yaho's equally-billed Liz Goodwin (National Affairs Reporter), Dylan Stableford (Senior Media Reporter), Rachel Rosen Hartman (Political Reporter) and others, does not have a Yahoo job title. No mystery there.

If you used a "unit" radius (that is, equaling 1) and scribed a circle at an angle of about 12½° to the drawing surface, the circle's circumference will equal 3, and effectively, "pi" will equal 3/1, ie 3.

I threw the Range (ft) vs Pressure (psi) data into Excel, and assuming a linear relationship, obtained Range = 5.24∙Pressure + 158.7, with an R² of 0.999. Extrapolating, I obtained 683 feet at 100 psi and 761 feet at 115 psi. BTW, I also solved for muzzle velocity from the classic ballistic equation R = V²∙sin2α/g (which required knowing that α was 43°), and even at 29 psi, it's 100 fps (68 mph), so don't stand in front of the muzzle. At 115 psi, it's 157 fps (107 mph), and almost 2½ times the kinetic energy at 29 psi.

Ditto! ~My bigger brothers (when I imitated them). ~A wise friend of mine.

~ Julius Caesar (attributed) Similarly... ~ Field Marshal Herman Göring (Adolph Hitler’s second in command) Yes, it was as though President Dimwit issued an arrest warrant with the name left blank.

~ Benjamin Franklin (1706-1790) ~ Edward Gibbon (1737-1794) ~ Thomas Jefferson (1743-1826) ~ Louis Pasteur (1822-1895)

I always admire and respect an expert at his craft in any field of endeavor.

An excellent question (meaning you may go far in this forum). Let's consider the opposite ... a "kneebow", and that it might point the wrong way. There is a old jackboot style where the forward edge of its top rises above the knee. Such an edge would act like a "kneebow". Consider a animal/person with "kneebows" that is creeping or on "all fours". The "kneebows" might catch on all manner of things (exposed roots, vines, etc) and hinder forward movement, especially escaping. So perhaps the question should be about why are there no "kneebows" (which is the simpler form of the two joints ... and would have evolved first???), instead of why no "elbowcaps".

As a historic note, President Lincoln suspended the right to Habeas Corpus during the Civil War. Can anyone point to any instance in which this suspension was woefully unavailable to apply prior to its inception or to any instance in which it was gratefully applied afterwards? I can't. What makes this power even worse, if I am correct, is that the citizenry be left uninformed of any application of this suspension. So, one day Chris Smith goes missing, and no one — not even the local authorities — knows why.

You (Anilkumar) might want to hear from me, and so, I agree with what has been discussed about these cannonballs/docking scenarios.

An object moving only due to its tangential velocity will travel in a straight line, and thus, cannot orbit the Earth. The Earth's gravity can prevent an object from escaping by keeping it in an orbit or by causing the object to collide with the Earth. The object can move, and it can move in a particular direction. Yes, the object is moving at an uniform acceleration arround the Earth. Well, the satellite is both moving and accelerating. It can't accelerate without moving because δv=∫a (the change in velocity v is the integration of the acceleration a).

Sorry, no. If you do not have a circular orbit, the centripetal force equation is not equal to the gravitational force. If the orbit is elliptical, there is tangential acceleration as well. If the "orbital" motion is linear you can trivially see this is false: at the turning point, the speed is zero, yet there is still a force on the object. I said that the centripetal force is the gravitational force. Okay, let me rephrase it: The gravitational force is centripetal in nature. But you're right, if the orbit is not circular, the centripetal force equation cannot back-calculate to obtain the value of the gravitational force. This is why the centripetal force equation is of such extremely limited use when discussing satellite orbits. Again, the centripetal force equation (that you're using here) is not generally applied to ballistic (aka, "free fall") trajectories that satellites experience. It generally applies to objects restricted to move in a circular path (ie, tethered objects, objects restricted by a circular surface, parts of a spinning wheel, etc). For free-falling satellites in all sorts of simple orbits, the centripetal force is produced by gravity. In the real world, the equation used to compute the force attracting the satellite to the central body is the gravitational force equation. Gravitational force is the only force in the simple orbits (circular, elliptical, linear ,etc) considered in this thread. By definition, and as I said, the gravitational force is the centripetal force (ie, it intersects the central body's center of mass — by definition), independent of whether the object's velocity is normal to it. I'm using the term "centripetal force" generically, and I think you're using it as only the force computed from the centripetal force equation. When the orbit is not circular, then the velocity vector isn't always normal to the gravitational force, and as you say, this gravitational force (which is centripetal in nature) can be broken into a tangential component of force which accelerates and decelerates the object along its line of travel (ie, along its velocity vector) and also a normal component of force (ie, normal to the velocity vector) that does not intersect the center of mass of the central body.

Ok, and in a more general sense, no force needs to exist to maintain any kind of velocity. No, stop talking about a "net force" as if there is more than one force acting on the object. There is only the gravitational force, which produces an incremental velocity (F=ma, a=F/m, and v=∫a=∫F/m) that combines with the existing tangential velocity to produce a new velocity. Velocities are additive. The centripetal force is the net force, and it produces an incremental velocity towards the central body (Earth) that combines with the existing velocity. Velocities are additive. No. No sustaining force is required to maintain velocity. Think of a hockey puck, or an air hockey puck, which maintain their motion without the constant influence of a force. What "net" force as if there's two or more forces combining to produce a "net" force? The term "net force" is misleading. There is only one force: gravitational force. The amount of centripetal force produced by gravity can be back-calculated (F=mv²/r) from the characteristics of the circular orbit(that is, from v) and r) . The velocity does change over time, and there must be an acceleration, and it must conform to that equation. However, this acceleration is a product of gravity, but not from any kind of v²/r effect. The back-calculated value F=mv²/r is not a new type of force. The condition of v²/r exists only because the circumstances (gravity and tangential velocity) produced the circular orbit in the first place. False, the centripetal force will only be the gravitational force, but not "the sum of forces are acting on the body". (Swansont, what "sum of forces"? No other forces exist in the example. Please describe any so-called "other forces".) False. The centripetal force is the gravitational force totally independent from the characteristic of the motion/orbit (circular, elliptical, linear, etc). For example, the Sun's gravitational force alone produces the highly elliptical orbit of Halley's comet. "By definition"? What definition? Certainly not v²/r. The relationship of a=v²/r does not produce the circular orbit. I think I've been clear in this post. The relationship of a=v²/r does not produce the circular orbit, it merely back-calculates the gravitational force that produced the special case of a circular orbit in the first place. And you are correct, a=v²/r does not "apply" (ie, back-calculate) to any other kind of orbit. Yes, Fr = ΣF = Fa + Fb. Actually, the object will accelerate in the direction of the resultant force, so there will be some movement (change in position) and there will be some change in velocity, both in the force's direction. There are no multiple forces to produce a "net" force, there is only the gravitational (centripetal) force (which, okay, is the "net" force) that causes the object to accelerate toward the central body, and yet, the tangential velocity also causes it to move "sideways".

The gravitational force = the centripetal force = the only force = the net force. And this holds true with any orbital motion ... circular, elliptical, linear (ie, straight into the central body), etc. The initial force that provided the initial tangential velocity no longer exists, and if did exist to provide more tangential velocity (such as a rocket engine), it couldn't counterbalance the centripetal force because the two forces are orthogonal. No force is needed to keep the object in motion. Again, the initial force no longer exists. The centripetal force is continuously acting upon the object. Your idea of a force to maintain the tangential velocity is wrong. Again, the initial force no longer exists. It has effectively been converted into the tangential velocity. The initial force no longer exists. Yes, the centripetal force does accelerate the object toward the central body; however, the tangential velocity causes it to also move "to the side", which results in an elliptical or circular orbit. If you would just stop thinking about an initial force still existing, all the physics would become very simple. Additionally, the tangential velocity has absolutely nothing to do with the centripetal force because the centripetal force is the gravitational force, which is not a function of tangential velocity. However, in the extremely unlikely case of a perfectly circular orbit, you can back-calculate the centripetal acceleration using the a=v²/r equation. Please, I'd like to alert everyone here that focusing on the special case of a circular orbit and the a=v²/r relationship that seems to apply (but does not) is causing undue confusion here. The physics involved here is being misrepresented, and I'm sure it's being done sincerely, but it's still being misrepresented.

a = v²/r Another way of looking at it is this equation/concept supposedly computes the a (radial acceleration of a ballistic body) from the v and the r. With satellites, you don't compute the a, because the a derives from the force of gravity. You can't say, for example, that you have a satellite 10,000 km from the center of the earth travelling tangentially at 25 km/sec, therefore the radial acceleration is blah, blah, blah. The radial acceleration is due only to gravity. Therefore, a satellite 10,000 km from the center of the earth has a radial acceleration due to gravity totally independent of its velocity. Instead, the a (radial acceleration) and the v (total velocity) act in concert to produce the satellite's orbital motion. To update the position, you double integrate the acceleration, ∫∫a, and you integrate the velocity, ∫v, and add them to the current position to produce the new position. To update the velocity, you integrate the acceleration, ∫a, and add it to the existing velocity, v. When you do this, you obtain the circular or elliptical orbit.

The equation a = v²/r concerns a tethered object, not a free-falling (ballistic) object. The tangential speed has nothing to do with the centripetal (aka gravitational) force. All orbits are elliptical to some extent (even though a circle is a special case of an ellipse). Perfectly circular orbits only occur when the attracting force varies as the inverse-fifth power of the distance between the objects.

Having been a rocket scientist for 10 years, I choose the force of gravity and tangential velocity answer although the total velocity must be considered. IN terms of a mathematical solution, this "equilibrium" is a matter of the distance from earth converted through its 1/R² relationship to obtain a force resulting in an acceleration (a=F/m) that is double-integrated to produce a change in position combined with integrating the satellite's velocity ("total" not just "tangential") to produce a change in position. If the new position is still outside the earth's surface (and atmosphere), you're still orbiting. The total velocity is then updated with the incremental velocity obtained by integrating the acceleration (v=at). I've done this graphically on paper by hand. I've never heard of a catch me if you can game, and I don't know what it means. (I've even seen the movie and read about Frank Abagnale Jr and his company, including the interview with Australia's Radio National.) The surface of the earth is a spherical ellipsoid, and the escape velocity vector is normal to the surface. It all seems fairly static and non-specific (ie, not relative to a specific situation).

Correct.

Different substances act on different parts of the brain, and in different ways, and cause changes in a person's mood, feelings, perceptions, etc. Most substances either imitate, block or otherwise affect the neurotransmitter chemicals used by the brain's neurons to communicate with each other. A neuron either excites or inhibits other neurons. Excitatory neurons outnumber inhibitory neurons about 10 to 1, but inhibitory signals outweigh excitatory signals about 10 to 1, so a brain without drugs is pretty well balanced. Depressants, such as alcohol and barbiturates, increase the inhibition of brain cells and produce a relaxing effect, but used in large enough amounts, can cause death by suppressing heart beat and/or breathing. Stimulants, such as amphetamines, increase the excitation of brain cells, increase mental function, reduce fatigue, and sharpening one's attention. As far as I know, biochemical addiction to substances occurs because the substance has a short-term effect that triggers a long-term reaction by the brain. When the short-term effect of the substance wears off, the long-term reaction by the brain remains, resulting in feelings of "withdrawal" and the desire/urge/addiction for more drugs to supposedly "cure" those withdrawal symptoms. Nicotine found in tobacco is the most addictive substance. Smoking relaxes a user, but a while later, the internal "need" for another cigarette urges the user to smoke again. The cycle is relatively short and repeats several times a day. Caffeine has a longer cycle. Typically a user has a few cups during the day and feels refreshed and invigorated, but needs caffeine the next morning to "get going again". The same occurs with more powerful and illicit drugs, but the effect can be more pronounced and additive due to the way they affect the brain and the brain's response to them (which is what makes them illegal). Some people are genetically predisposed to addiction to certain substances, but societal, familial, situational, and other influences also exist.

The introverted science/math teacher as exemplified by Ben Stein's Mr Cantwell in the TV sitcom The Wonder Years. Somewhere came the concept that, apart from the other subjects, math and science must be taught in a controlled, methodical, and detached manner. Having tutored students from grade six through graduate level in math and science, I know that there's a time to spell things out plainly or discuss things logically, and there's a time to do something interesting either on paper or in real life. I'm not saying that every math/science teacher must be another Richard Feynman, but to truly register with the students, after theory must come application ... something like MythBusters and definitely NOT like Beakman's World (maybe that's why America has become so scientifically dumbed-down).

Interestingly, in the developing brain, cortical cells sense chemical gradients and crawl relatively great distances to find their proper location.