Doug Goncz
Apr22-04, 02:54 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hey, gang!\n\nCheck this out.\n\nA bicyclist usually maintains constant pressure and cadence on the pedals. So\nthat\'s a constant P.i power input. Through gearing, this is matched to the\nroad. Assuming 100 % transmission efficiency, with F total force on bicycle, m\nmass of bicycle and rider, a acceleration, P.i power input, F.r force of\nrolling resistance, c.d coefficient of drag in units like pounds forced divided\nby mph^2:\n\nF = ma\n\na = F / m\n\na = ( P.i / v + F.r + c.d*v^2 ) /m\n\na = v\'\n\n0 = ( P.i * v^-1 + F.r * v^0 + c.d* v^2 ) / m - v\'\' or\n0 = (P.i * v^-1 + F.r * v^0 + c.d* v^2 - m*v\'\' or\n\nSo the coefficient matrix with origin -1 for this equation is:\n\n\n0 F.r 0 0\nP.i 0 0 c.d\n0 0 m 0\n\nWhere the rows are powers of the differential operator d^n s / dt^n, and the\ncolumns are powers of the differential from -1 to 2.\n\nso far this is just the constant coefficient linear differential operator:\n\nThere are too many zeros in that first row. Yes, the first row is d^0s/dt^0. So\nthis isn\'t quite right.\n\nAnyway, putting the terrain function into this will be easy. It "only" makes it\na second-oreder linear differential operator.\n\nYou see, my text (Finney and Ostberg, p.96) says\n\n"Linear Differential Operators:\n\nThe coefficients in (3-18) need not be constants. Function in C(I) can be used\njust as well, and the expression (3-19) where a sub n, a function of x, is not\nidentically 0 on I, is called an nth-order linear differential operator."\n\nIn my studies I use USGS topographical map data interpolated along survered\nroad segments connecting the vias on the route I pick with the mouse. The\nresult exports as a bitmap route profile and Mathcad reads it in, picking off\nthe profile. I enter the highest charted alititude, the altitude of the lowest\ny axis increment, the terrain distance (not map distance), and bingo, Mathcad\ntells me how much fat I burn going out, then back, or around each time.\n\nSo a sub n of x is not a written out symbolic function. It\'s a lookup table.\nBut I resolve it to a Fourier series by conjugating the one way trip to its\nreversal, taking the Fourier transform, clipping a bit, and inverting the\ntransform, so a sub n of x will always have the form:\n\nMean Altitude + b.1 sin x + c.1 cos x + b.2 sin 2x + c.2 cos 2x ....\n\nand once I know the way to get a solution for distance as a function of time\nwithout terrain, I\'ll have a much faster way to devlop my augmented bicycle\nstudies.\n\nWhat is the technique for getting that negative one out of there?\n\nIt\'s not a proper linear differential operator unless the smallest coefficient\nis zero.\n\n\nYours,\n\nDoug Goncz ( ftp://users.aol.com/DGoncz/ )\n\nMy physics project at NVCC:\nhttp://groups.google.com/groups?q=dgoncz&scoring=d plus\n"bicycle", "fluorescent", "inverter", "flywheel", "ultracapacitor", etc.\nin the search box\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hey, gang!
Check this out.
A bicyclist usually maintains constant pressure and cadence on the pedals. So
that's a constant P.i power input. Through gearing, this is matched to the
road. Assuming 100 % transmission efficiency, with F total force on bicycle, m
mass of bicycle and rider, a acceleration, P.i power input, F.r force of
rolling resistance, c.d coefficient of drag in units like pounds forced divided
by mph^2:F = maa = F / ma = ( P.i / v + F.r + c.d*v^2 ) /ma = v'= ( P.i * v^-1 + F.r * v^0 + c.d* v^2 ) / m - v'' or
= (P.i * v^-1 + F.r * v^0 + c.d* v^2 - m*v'' or
So the coefficient matrix with origin -1 for this equation is:
F.r
P.i c.d
m
Where the rows are powers of the differential operator d^n s / dt^n, and the
columns are powers of the differential from -1 to 2.
so far this is just the constant coefficient linear differential operator:
There are too many zeros in that first row. Yes, the first row is d^{0s}/dt^0. So
this isn't quite right.
Anyway, putting the terrain function into this will be easy. It "only" makes it
a second-oreder linear differential operator.
You see, my text (Finney and Ostberg, p.96) says
"Linear Differential Operators:
The coefficients in (3-18) need not be constants. Function in C(I) can be used
just as well, and the expression (3-19) where a sub n, a function of x, is not
identically on I, is called an nth-order linear differential operator."
In my studies I use USGS topographical map data interpolated along survered
road segments connecting the vias on the route I pick with the mouse. The
result exports as a bitmap route profile and Mathcad reads it in, picking off
the profile. I enter the highest charted alititude, the altitude of the lowest
y axis increment, the terrain distance (not map distance), and bingo, Mathcad
tells me how much fat I burn going out, then back, or around each time.
So a sub n of x is not a written out symbolic function. It's a lookup table.
But I resolve it to a Fourier series by conjugating the one way trip to its
reversal, taking the Fourier transform, clipping a bit, and inverting the
transform, so a sub n of x will always have the form:
Mean Altitude + b.1 sin x + c.1 cos x + b.2 sin 2x + c.2 cos 2x ....
and once I know the way to get a solution for distance as a function of time
without terrain, I'll have a much faster way to devlop my augmented bicycle
studies.
What is the technique for getting that negative one out of there?
It's not a proper linear differential operator unless the smallest coefficient
is zero.
Yours,
Doug Goncz ( ftp://users.aol.com/DGoncz/ )
My physics project at NVCC:
http://groups.google.com/groups?q=dgoncz&scoring=d plus
"bicycle", "fluorescent", "inverter", "flywheel", "ultracapacitor", etc.
in the search box
Check this out.
A bicyclist usually maintains constant pressure and cadence on the pedals. So
that's a constant P.i power input. Through gearing, this is matched to the
road. Assuming 100 % transmission efficiency, with F total force on bicycle, m
mass of bicycle and rider, a acceleration, P.i power input, F.r force of
rolling resistance, c.d coefficient of drag in units like pounds forced divided
by mph^2:F = maa = F / ma = ( P.i / v + F.r + c.d*v^2 ) /ma = v'= ( P.i * v^-1 + F.r * v^0 + c.d* v^2 ) / m - v'' or
= (P.i * v^-1 + F.r * v^0 + c.d* v^2 - m*v'' or
So the coefficient matrix with origin -1 for this equation is:
F.r
P.i c.d
m
Where the rows are powers of the differential operator d^n s / dt^n, and the
columns are powers of the differential from -1 to 2.
so far this is just the constant coefficient linear differential operator:
There are too many zeros in that first row. Yes, the first row is d^{0s}/dt^0. So
this isn't quite right.
Anyway, putting the terrain function into this will be easy. It "only" makes it
a second-oreder linear differential operator.
You see, my text (Finney and Ostberg, p.96) says
"Linear Differential Operators:
The coefficients in (3-18) need not be constants. Function in C(I) can be used
just as well, and the expression (3-19) where a sub n, a function of x, is not
identically on I, is called an nth-order linear differential operator."
In my studies I use USGS topographical map data interpolated along survered
road segments connecting the vias on the route I pick with the mouse. The
result exports as a bitmap route profile and Mathcad reads it in, picking off
the profile. I enter the highest charted alititude, the altitude of the lowest
y axis increment, the terrain distance (not map distance), and bingo, Mathcad
tells me how much fat I burn going out, then back, or around each time.
So a sub n of x is not a written out symbolic function. It's a lookup table.
But I resolve it to a Fourier series by conjugating the one way trip to its
reversal, taking the Fourier transform, clipping a bit, and inverting the
transform, so a sub n of x will always have the form:
Mean Altitude + b.1 sin x + c.1 cos x + b.2 sin 2x + c.2 cos 2x ....
and once I know the way to get a solution for distance as a function of time
without terrain, I'll have a much faster way to devlop my augmented bicycle
studies.
What is the technique for getting that negative one out of there?
It's not a proper linear differential operator unless the smallest coefficient
is zero.
Yours,
Doug Goncz ( ftp://users.aol.com/DGoncz/ )
My physics project at NVCC:
http://groups.google.com/groups?q=dgoncz&scoring=d plus
"bicycle", "fluorescent", "inverter", "flywheel", "ultracapacitor", etc.
in the search box