This web article is reviewing an electric motorbike that can go from 0 to 60 MPH in under one second. In comparing the accelerating of the bike, the article says:
All this gives the driver a G-force three times more than that faced by a skydiver during freefall!
Er.. come again?
G-force, the way it is used for the public, is simply the amount of "reaction" force that we feel. When doing a looping roller coaster, it is the amount of force the seats exert back onto us. That's why at the bottom of a loop, you get G-force greater than "g", the gravitational acceleration at the earth's surface (9.8 m/s^2), while at the top of a bump, you could get "negative G's", meaning it is less than g.
So what's the G-force experienced by a skydiver? If you ignore air resistance, it should be.... er... ZERO! Free falling, by definition, implies that there's no reaction force. That's how the "vomit plane" works when people are training for weightless environment - the plane makes a free fall every few minutes, and every entity in the object will feel no reaction from the floor of the plane since everything is free falling. So equating the bike's acceleration as having three times the G-force experience by a skydiver is rather puzzling.
Now, one could argue that maybe the writer was thinking about the G-force experienced by the skydiver after he/she reaches terminal velocity, which would make the G-force to be equal to g. Assuming that the writer is smart enough to know basic mechanics AND know about terminal velocity, then the comparison is also puzzling. If it is just a simple "g", then why not compare it to the regular person standing still on the ground? After all, it IS the same thing and I'm guessing, A LOT more people are more familiar with standing still on the ground than skydiving.
Nope. I think this is a mistake of comparison.
Zz.
11 comments:
Here is my post on weight and apparent weight. Quite appropriate for this discussion.
http://blog.dotphys.net/2008/09/gravity-weightlessness-and-apparent-weight/
Who but a Physicist ever ignores air resistance? A skydiver who calculated their terminal velocity without air resistance, with or without their parachute open, presumably wouldn't jump. Perhaps a spherical cow would jump. But classic, of course.
Did you even READ the rest of the blog? Look at the scenario I presented in the SECOND case where there is terminal velocity. What do you think is involved THERE? Antigravity?
Oy Freaking Vey!
Zz.
"A skydiver who calculated their terminal velocity without air resistance"
terminal velocity exists because of air resistance. without air there would be no terminal velocity. what a stupid comment. who but a ignoramus would say such a thing.
I think the person meant, how many g's does a person feel when reaching terminal velocity and then PULLING THE CORD. How many G's are felt as the parachute opens up??
But that isn't FREE FALL!
Zz.
It is obvious that no one here has ever been skydiving. I admit that you all seem very intelligent and have great physics vocabularies, but whatever the correct term is I do not know, but when you are falling out of an airplane you feel a great deal of pressure on your body and there must be some way to measure that whether it be in "g" or not
Unfortunately, what you are saying here has very little to do with the blog post. No one is talking about "pressure" or measuring it.
Zz.
Why does everyone get so snippy? A G is just the equivalent of what you feel right now, unaccelerated in Earth's gravity. If you weigh 150 lbs., then 2 G's would make you feel like you weigh 300 lbs. 3 G's + 450 lbs... A skydiver would initially experience 0 G's, rapidly increasing such that at terminal velocity, when fluid resistance is equal to the acceleration of gravity, and his acceleration diminishes to O, he's experience 1 G, the same as a dude sitting at his laptop. If there were no fluid resistance (atmosphere) then he'd accelerate infinitely and maintain 0 G's infinitely. There really is no such thing as "free fall" within an atmosphere. The one falling is impeded and literally lying down on a mattress of compressed air. Until he hits the granite box spring, of course.
Ita actually more to do with your acceleration... When you are standing on the the ground, yes you feel your own weight, but your velocity remains constant. When you jump out of a plane you accelerate towards your terminal velocity, thereby experiencing 'relative' negative g-force. Remember this is just a feeling... The earth's gravity is always acting on us but when we are in the ground we are still we are resisting it and when we fall we are not until we reach terminal velocity. So our relative experience of g-force changes. What the previous comment said about a skydiver experiencing 0 g's is perfectly accurate but I wonder if they've ever jumped of anything and have any understanding of what going from 1 to 0 g's feels like? You feel it. That was the point of the analogy. Apparently it's three times stronger so I would assume you go from 0 (horizontal g's) to 3. Wow. What a bike.
This 'pressure' you experience when jumping out of a plane is actually g force on you changing from 1 (normal everyday g's) to zero. You will experience this while you accelerate up to terminal velocity, at which point you have returned to the normal 1. So what you actually experience, relatively, is negative g-force. (As far as human experience is concerned, and we are very relative beings). Talking about acceleration in terms of g's isn't uncommon and it isn't very difficult to get a gist of what someone means when they say three times more than a skydiver except what they probably really mean is 3g's, as on a horizontal plane we live with 0g's and they are assuming a change of 1g when you jump out of plane, so therefore the change in g force acting on your body is the times more. I'm pretty sure this should be referred to as inertia ? But I'm not a physicist nor have I studied physics.
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