Date: Wed, 09 Feb 2005 08:55:40 -0700

Author: Jerry DiMarco

Subject: Re: physics of football

Post:

Thank you, David, they never mentioned this in chalktalk!!!

Jerry


At 10:11 PM 2/8/2005, you wrote:
>Oops! I just noticed that in my last message I inadvertently left out
>a constant factor multiplying x in the logarithm term in the equation
>for the path y(x) in the case where the fluid resistance is linear in
>the velocity. I had written:
>
> >y=h +(tan([phi])+ sec([phi])/e)*x +(v_0^2/g)*ln(1 - e*sec([phi])*x)/e^2
>
>But I ought to have included a factor of (g/v_0^2) multiplying the
>e*sec([phi])*x term inside the ln() function. The equation simplifies
>somewhat if all the distances are written in dimensionless form. If we
>let Y == g*y/v_0^2, and let H == g*h/v_0^2, and let X == g*x/v_0^2, then
>the equation for the path Y = Y(X) boils down to:
>
>Y = H + (tan([phi])+ sec([phi])/e)*X + ln(1 - e*sec([phi])*X)/e^2
>
>where the dimensionless parameter is still given by e == (b*v_0)/(m*g).
>Now if we take the e --> 0 limit we get back to the simple
>(dimensionless) parabola path that occurs when there is no resistance,
>i.e.
>
>Y = H + tan([phi])*X - (1/2)*sec^2([phi])*X^2 .
>
>I think all the rest of the stuff in my previous post is still OK.
>
>David Bowman
>
> >Regarding Jerry's question:
> >
> >> How is the optimum angle affected by the initial velocity?
> >
> >It's complicated if air resistance effects are included. If they are
> >*not* included the optimal angle [phi_o] and maximum horizontal range R
> >at the optimum angle [phi_o] can be exactly solved for, and are given
> >by:
> >
> >[phi_o] = (1/2)*arccos(1/(1 + (v_0)^2/(g*h))) and
> >
> >R = ((v_0)^2/g)*sqrt(1 + 2*g*h/(v_0)^2) .
> >
> >Here v_0 is the initial launch speed, h is the initial height the
> >launch position is above the height of the landing position and g is
> >the planet's surface gravity.
> >
> >>Does it have to do with air resistance?
> >
> >In general, yes. Just *how* has to do with the air resistance depends
> >on the model used for the air resistance. If a linear velocity
> >resistance model (with no lift so the force due to the fluid medium
> >directly opposes the instantaneous velocity of the object through the
> >fluid) then there is a closed form expression for the path y(x) through
> >space. But the equation to be solved for the optimal launch angle
> >giving the maximal range is an implicit one than can be easily solved
> >by any numerical root finding scheme.
> >
> >If x is the horizontal displacement of the object from its launch
> >point, and h is the initial height of the launch point, and y is the
> >vertical position of the object relative to the height of the landing
> >point, and b is the velocity resistance coefficient, (i.e. F_resist =
> > = -b*v), and [phi] is the launch angle, and v_0 is the launch speed,
> >then the equation for the path y(x) is:
> >
> >y=h +(tan([phi])+ sec([phi])/e)*x +(v_0^2/g)*ln(1 - e*sec([phi])*x)/e^2
> >
> >where the dimensionless parameter e == (b*v_0)/(m*g) is the ratio of
> >the initial wind resistance to the weight of the object. Here e
> >measures the relative importance of the resistance for the problem.
> >(If we take the e --> 0 limit we get back to the simple parabola path
> >that occurs when there is no resistance.)
> >
> >If we set h = 0 so the launch height is the same as the landing height
> >the optimal launch angle is given by the solution of:
> >
> >e^2 = (1 + e*csc([phi_o]))*ln(1 + e*csc([phi_o])) - e*csc([phi_o]) .
> >
> >In the case where the wind resistance is small (i.e. the parameter
> >e << 1) we can expand the optimal launch angle and the maximal range R
> >at that launch angle in power series in the small parameter e. The
> >result is:
> >
> >[phi_o] = [pi]/4 - e/sqrt(18) + e^2/9 + order(e^3) and
> >
> >R = ((v_0)^2/g)*(1 - (sqrt(8)/3)*e - (10/9)*e^2 + order(e^3)) .
> >
> >If a more realistic velocity-squared air resistance model is adopted
> >(i.e. F_resist = -b*v^2) the problem becomes *much* harder to solve
> >analytically because then the horizontal motion couples to the vertical
> >motion, and that complicates matters a lot. In this case I don't think
> >that a closed form formula for the path or for the range exists. But
> >it is possible to formally write implicit equations for y and x that
> >each depend on a common parameter (such as the instantaneous local
> >angle or slope of the path) where these implicit functions depend on
> >just integral quadratures that I, unfortunately, can't see any way to
> >perform in closed form. In this case it is still possible to
> >perturbatively expand a series solution in powers of the resistance.
> >But I never got around to actually working it all out.
> >
> >David Bowman
> >
> >> Jerry
> >>At 08:11 PM 2/7/2005, you wrote:
> >>>... For example, for a shotput thrown with an initial
> >>>velocity of 10 m/s from a height of 2 meters, the
> >>>optimum angle is near 40 degrees. 45 degrees will cost
> >>>you 9 cm! :) The higher the initial launch, the
> >>>flatter the optimum trajectory will be.
> >>>
> >>>Marc "Zeke" Kossover
> >>>The Jewish Community High School of the Bay
From MLowry@D115.ORG Wed Feb 9 11:59:40 2005

Back