**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