Davis equation
Amenadiel
According to the book Railroad Engineering (Hay), the internal resistance of
a train can be calculated by this
formula:
Ru= 1.3+29/w +b*V + CAV^2/wn
where Ru is unit resistance in pounds per ton
w =weight per axle
b= experimental coefficient which express flange friction, shock, sway and
concussion
A=cross sectional area of the car or locomotive
C= drag coefficient based in the shape of the front end of the car or
locomotive and the overall configuration.
n = number of axles
Well, my doubt is:
I can easily figure out what the resistance for a single car is, but I don't
quite understand how to modify the equation when there is a train set which
consist on a locomotive, 10 passenger cars and 5 freight cars.
the resistance of all the set is:
- the sum of the individual resistance of each component?
- the average of the individual resistance of each component?
- the weighted sum of the individual resistance of each component according
to the weight?
or:
must I add or get the average of every "b" of each component and every "C"
of each component and use that as arguments for the equation? In that case,
the values of w and n must be of the total of the train?
Thanks in advance
--
______________________________________________
Felipe Figueroa Fagandini ffig...@libra.cl
Ingeniero Civil de Industrias PUC ffig...@puc.cl
Ingeniero de Proyectos Libra Ingenieros Consultores Ltda.
James Robinson
>
> I can easily figure out what the resistance for a single car is, but I don't
> quite understand how to modify the equation when there is a train set which
> consist on a locomotive, 10 passenger cars and 5 freight cars.
You add up the resistances for each of the individual vehicles in the
train.
Recognize that since the basic equation is of the form A + BV + CV^2,
you can simply add up the coefficients for all of the vehicles in the
train to arrive at a single equation for the calculation. That is, if
one car is A + BV + CV^2, and the next is D + EV + FV^2, then the
combination can be expressed as (A+D) + (B+E)V + (C+F)V^2
For the example you give above, you would calculate the coefficients for
the locomotive, then the coefficients for an individual passenger car,
and multiply them by 10, then the coefficients for an individual freight
car, and multiply them by 5. Finally, add up all the "A" coefficients,
"B" coefficients, and "C" coefficients to end up with a total equation
for the train.
> the resistance of all the set is:
> - the sum of the individual resistance of each component?
Yes. You need to add the resistances of each individual vehicle in the
train, as calculated using the equation.
Amenadiel
"James Robinson" <was...@212.com> escribió en el mensaje
news:3F732742...@212.com...
R. J. Sutherland
>Amenadiel wrote:
>>
>> I can easily figure out what the resistance for a single car is, but I don't
>> quite understand how to modify the equation when there is a train set which
>> consist on a locomotive, 10 passenger cars and 5 freight cars.
>
>You add up the resistances for each of the individual vehicles in the
>train.
>
>[...] the basic equation is of the form A + BV + CV^2,
While mathematically-speaking that's quite correct, physically-speaking it's not
quite so simple, at least where the V^2 term is concerned. This represents
aerodynamic resistance, and the coefficient C of a single unit is determined by
(some relevant physical constants and) its effective cross-sectional area
(frontal and side). Linking units together provides a degree of slipstreaming
(in the same way that eg. a racing cyclist can save energy when riding right
behind a line of team-mates), so these coeffficients don't simply add. A train
consisting of a group of box cars followed by a group of flat cars has less wind
resistance, for example, than a train of the same composition with alternating
box cars and flat cars, and both will have less resistance than the simple sum
of the individual resistances of the individual units. Unfortunately, there's no
simple formula for combining aerodynamic resistances (it's still a very
poorly-understood subject), although some approximation to reality might be
achieved by discounting the coefficients of units according to their positions
in the train (least discount at the front, slightly more at the rear, most in
the middle), and possibly even according to the detailed arrangement of the
various units.
Since frontal aerodynamic resistance isn't affected in any way by gross weight,
unlike frictional resistance, I also have serious fundamental reservations about
any train resistance formula that expresses aerodynamic resistive force relative
to gross weight or mass, even though it seems to be an all-too-common practice
in the railway world. (A suitable choice of coefficients can make the equation
correct for a given train of a given weight, of course, but the practice still
implies a non-existent change in aerodynamic resistance due to a change in unit
loads...)
James Robinson
>
> James Robinson wrote:
>
> >You add up the resistances for each of the individual vehicles in the
> >train.
> >
> >[...] the basic equation is of the form A + BV + CV^2,
>
> While mathematically-speaking that's quite correct, physically-speaking it's not
> quite so simple, at least where the V^2 term is concerned. This represents
> aerodynamic resistance, and the coefficient C of a single unit is determined by
> (some relevant physical constants and) its effective cross-sectional area
> (frontal and side).
Since most of the resistance equations are typically curve fits of
empirical data, there is no guarantee that the V squared term is
restricted to air resistance. While most descriptions suggest that, the
term could include the effect of such things as truck instability, which
might cause a resistance proportional to the square of speed.
> Linking units together provides a degree of slipstreaming
> (in the same way that eg. a racing cyclist can save energy when riding right
> behind a line of team-mates), so these coeffficients don't simply add.
The effect of slipstreaming is taken into account by using a lower "C"
coefficient, which is the streamlining coefficient for the effect of air
resistance. In most cases, to simplify calculation, groups of cars will
be considered together. It is also possible to calculate resistances
for individual cars, but that is usually mathematical overkill. Once
you have the resistance, there is no reason why the coefficients can't
be added together. Mathematically, that is exactly what you would do
for a simulation program.
> Unfortunately, there's no simple formula for combining aerodynamic
> resistances (it's still a very poorly-understood subject), although
> some approximation to reality might be achieved by discounting the
> coefficients of units according to their positions in the train ...
Why not? Just make an assumption of the average coefficient for a
train, and make a simple calculation. The results are typically
adequate for most simulation purposes. There is no reason to
overcomplicate things.
> Since frontal aerodynamic resistance isn't affected in any way by gross weight,
> unlike frictional resistance, I also have serious fundamental reservations about
> any train resistance formula that expresses aerodynamic resistive force relative
> to gross weight or mass, even though it seems to be an all-too-common practice
> in the railway world.
I can't think of a single resistance formula that varies the V squared
term by weight. The original Davis equations of the 1920s, as an
example, assume the C coefficient is independent of weight. The other
equations I can think of all assume weight only affects the constant
term (A), and the coefficient for the term that varies proportional to
V, (B). Some recent testing suggests that the B coefficient in fact
approaches zero, so only the constant term (A) is affected by weight.
Amenadiel
> "R. J. Sutherland" wrote:
> >
> > James Robinson wrote:
> >
> > >You add up the resistances for each of the individual vehicles in the
> > >train.
> > >
> > >[...] the basic equation is of the form A + BV + CV^2,
> >
>
> > Linking units together provides a degree of slipstreaming
> > (in the same way that eg. a racing cyclist can save energy when riding
right
> > behind a line of team-mates), so these coeffficients don't simply add.
>
That is correct, but the weight of the V^2 component of the equation is
kinda small in comparison to A+BV.
I was concerned about... let's say a train made only of identical
locomotives. The total internal resistance, in pounds, should grow in
proportion to the number of locomotives. In that case, the unitary
resistance, in pounds per ton, should be nearly the same for one locomotive
as for ten locos, because the total resistance, in the second case, would be
ten times larger when multiplied ten times the weight of a single loco.
So it seems right to me to simply add the total resistance of every single
component, and then, to get the unitary resistance (lb/ton) of all the train
set, divide by its total weight... which is a weighted average of the
individual resistance of every component.
Of couse this ignores the little error generated by the term CV^2, but
indeed this gives us a more pesimistic result, so we keep on the safe side.
Excuse the poor english, by the way. :(
R. J. Sutherland
>[...] the weight of the V^2 component of the equation is
>kinda small in comparison to A+BV.
>
That all depends on the train's speed (and shape)! It will indeed be relatively
unimportant for (say) a heavy, slow-moving freight train, but it won't be
trivial at all for (say) a modern high-speed train like the French TGV. The
force is proportional to v^2, so the motive power - which is the main
determinant of locomotive capability - is proportional to v^3. That power
requirement escalates very quickly with increase of speed - a change from 20 kph
to 200 kph increases the required motive power 1000 times, so a train that only
needs 2kW to overcome aerodynamic resistance @ 20 kph will need 250kW to do the
same @ 100 kph and a whopping 2MW to do the same at 200 kph! These values are
not actual values for any particular train, but are fairly representative of
reality, I believe. Maybe someone with more knowledge can quote actual
frictional & aerodynamic power values at different speeds for a high-speed train
like the TGV....
>I was concerned about... let's say a train made only of identical
>locomotives. The total internal resistance, in pounds, should grow in
>proportion to the number of locomotives. In that case, the unitary
>resistance, in pounds per ton, should be nearly the same for one locomotive
>as for ten locos, because the total resistance, in the second case, would be
>ten times larger when multiplied ten times the weight of a single loco.
>
>So it seems right to me to simply add the total resistance of every single
>component, and then, to get the unitary resistance (lb/ton) of all the train
>set, divide by its total weight... which is a weighted average of the
>individual resistance of every component.
Yes, if you discount the aerodynamic force, that's quite right. You can even
include the aerodynamic force provided you don't change the gross train weight
and try to use the existing "unitary" resistance coefficients to calculate a new
absolute total resistance, since aerodynamic resistance doesn't scale directly
by weight, even in the case where the weight is due (as in your instance) to a
number of identical units such as locos or wagons.
I believe that the train resistance equation should actually be an absolute one
rather than a specific ("unitary") one, of the form
F= W.(a + b.v) + A.c.v^2
where F is the total motive force (eg. in N), W is the gross train weight (eg.
also in N), A is an area factor (eg in m^2) related in some way to the physical
dimensions of the train, and the coefficients a, b, c are dependent only on
physical constants and not on any physical property (eg. mass, weight) of the
train.
I would welcome anyone quoting any well-respected train resistance equation,
whether it supports my above contention or not. (I've seen some "wierd and
wonderful" ones myself! I suspect most were developed empirically by railway
engineers before we had reached even our present imperfect understanding of
aerodynamic resistance.)
It could be, of course, that in the "real world" the above equation might be
pedantically correct but swamped by a multiplicity of incidental practical
factors such as the state of maintenance of the axleboxes and even the state of
cleanliness of the rolling stock(!), so that a simple specific force equation
like the Davis is good enough for everyday use....
>Of course this ignores the little error generated by the term CV^2, but
>indeed this gives us a more pessimistic result, so we keep on the safe side.
It depends on what your criteria are! If the criterion is "will the available
tractive power cause the train to exceed a particular speed limit?", discounting
the aerodynamic resistance will indeed keep us on the safe side. On the other
hand, if the criterion is "what is the least motive power needed to ensure that
the train can operate at a certain (fast) minimum speed?", we are definitely not
keeping on the "safe" side (in engineering terms at least)!
>Excuse the poor english, by the way. :(
It's not that bad really. Infinitely better than my Spanish...!
James Robinson
>
> That is correct, but the weight of the V^2 component of the equation is
> kinda small in comparison to A+BV.
As another poster has pointed out, the V^2 component can be significant,
particularly at higher speeds. In some equations, like that recommended
by the AAR, the BV component is practically zero, so the CV^2 component
is the largest part of the total resistance pretty well throughout the
speed range.
> ...the unitary resistance, in pounds per ton, should be nearly the
> same for one locomotive as for ten locos, because the total
> resistance, in the second case, would be ten times larger when
> multiplied ten times the weight of a single loco.
Correct. The lead locomotive will have a higher resistance, but as
locomotives are added, the unitary resistance will drop asymptotically
toward the average of the trailing locomotives.
> So it seems right to me to simply add the total resistance of every single
> component, and then, to get the unitary resistance (lb/ton) of all the train
> set, divide by its total weight... which is a weighted average of the
> individual resistance of every component.
You could do that, but I don't see why you want to. Since you have
calculated the total resistance, isn't that what you are trying to
find? Dividing by weight after that doesn't add any additional
information.
> Of couse this ignores the little error generated by the term CV^2, but
> indeed this gives us a more pesimistic result, so we keep on the safe side.
>
> Excuse the poor english, by the way. :(
I don't see anything wrong with it. If you hadn't mentioned it, I
wouldn't have paid any attention to the way you wrote your message,
since no mistakes jump out when I read it.
James Robinson
>
> Amenadiel wrote:
>
> I believe that the train resistance equation should actually be an absolute one
> rather than a specific ("unitary") one, of the form
>
> F= W.(a + b.v) + A.c.v^2
It doesn't matter, since you have to multiply by the total weight at
each speed to arrive at the total resistance. Any application will
recast the equation to achieve the desired end result.
> I would welcome anyone quoting any well-respected train resistance equation,
> whether it supports my above contention or not.
Any of the equations I've seen will work. The only reason they were
written as lb./ton was to demonstrate the effect of increasing axle
loads on the resistance of a train. You can recast the equations to
provide any desired result. As an example, when you apply the equation
basic unitary equations, you multiply the appropriate components by the
total weight of the vehicle, and the weights will cancel out in the V^2
term, so you will end up with a contribution to the total resistance
that is independent of weight.
> I suspect most were developed empirically by railway engineers
> before we had reached even our present imperfect understanding of
> aerodynamic resistance.)
They were. There are some that have melded empirical tests with
theoretical aerodynamics to allow users to create their own resistances
for different styles of vehicle, but they still typically end up as
quadratic equations.
> It could be, of course, that in the "real world" the above equation might be
> pedantically correct but swamped by a multiplicity of incidental practical
> factors such as the state of maintenance of the axleboxes and even the state of
> cleanliness of the rolling stock(!), so that a simple specific force equation
> like the Davis is good enough for everyday use....
A railroad is not a particularly exact operation. Trying to refine a
resistance equation to many significant figures is mostly an exercise in
futility. There can be some benefit in defining components to rank
appropriate research effort, where several options are being considered,
but for typical operational analysis, such accuracy is unnecessary.
