Sunday, January 15, 2023
a machine gun with just a barrel.
In the formula for the g force when throwing a fives ball, it occurred to me that i had cheated a little in suggesting the trajectory was basically on a line to the target. This is not necessarily true. So for instance think of the throwing the hammer competitor. With a very long rope part the forces could be quite gentle, and the distance quite long. Instinctively i think the fives throw is pretty straight, though people using a straight arm will definite have an arc.
This made me think of making big coil of tubing starting at the centre, made into a rigid wheel-like thing. You have it spinning very fast. You place a ball in the tube at the centre and the ball will be thrown out in a direction dependent on the speed of input and rotation of the ball, and the rotation of the wheel. But in principal, this is a machine gun firing out balls as fast as they are fed into the machine.
The balls will emerge rotating very fast as they are forced into close contact with the outer surface of the pipe by the rotation. This could be used, I think to maintain height, by pushing more air down at the front and less up at the rear. The disc would be at right angles to the earth surface. You would have difficulty keeping any oil in this barrel.
There may be a need to have a release mechanism at the exit. You will note that there are two ways of setting this up. Ie you can spin it either way. On exit the speed of the ball relative to the ground will be faster with the opening spinning into the 'wind'.
The thing needs a bit more thought.
The thing needs a bit more thought.
Reminds me slightly of the giroscopic rocket launcher that is being developed.
M
Tuesday, January 10, 2023
acceleration in throwing
People have no idea about acceleration, except perhaps full body acceleration, such as a racing start in a ferrari. But ask them how much acceleration they can achieve throwing a ball, and they have no idea. Even estimating that of a ferrari is almost an impossible talk. Must be much more than throwing a ball, is the normal thought.
They may get the notion that fighter pilots with special clothing might manage to fly in the area of 5 to 10 g.
I have been fascinated by the idea of guessing answers with the help of dimensions such as time, length, mass, dimensionless numbers, etc.
So lets try an example of throwing a fives ball as hard,as you can, and ask the question about the average accelaration in the process.
The key dimensions are
- length of path of the ball before release, taking a straight line between start and finish.
Lets say 6 feet, which might include the thrower stepping forward.
- the speed at time of release. Lets be optimistic and say 90 miles per hour.
Ok, lets get to some easier units. 30 miles an hour is 1 mile in 2 minutes. Or 880 yards in one minute, or 44 yards in 3 seconds, or finally 44 feet in one second. So 90 mph is 132 feet per second.
- 1 g is 32 feet per second per second. that is a length divided by time twice.
Clearly you can do this all in metric if you insist.
That is all we need. The answer we want is a number of g, that is a dimensionless number
Lets build up a formula. The greater the speed the bigger the answer. So lets put the speed in the numerator. Similarly the bigger g is the the smaller the answer will be, so lets have that in the denominator.
Our last figure is just a length, so we need to eliminate time by squaring the speed, which leaves us with a dimension of length.
This we can eliminate by dividing be the distance of the acceleration, namely 6 feet.
So we have arrived at a dimensionless answer of
132 times 132 divided by ( 6 times 32 )
Speed squared / ( length times gravity)
AND
there may be a factor to multiply by. I am going to guess a half, as length times gravity is a force times a distance , that is an energy which is normally half velocity squared, the factor coming from intergration.
So my answer is 132*132/(2*6*32) g
Or 45g
This might seem on the high side, maybe the speed is only 60, giving 20 g.
But the calculation is basically correct. The energy equation is
V**2 / 2 = force * distance
The figures are really amazing, but the effects are only that large at the extremities where the fingers touch the ball, and these must be very tough.
The ferrari is pathetic.
Martin
