The seeder problem
A class exercise suggested by one of the students.
(This is a very rough summary of the calculations,
not intended as a formal presentation)
David C, 2014
Worst case scenario:
A seeder blade cutting through hard-packed, wet soil with lots of weed roots
interlocking to form a net that must be cut through. This is how it looks in the
Therefore the hardest force to overcome by dragging a blade through the dirt
(it seems to me) is to cut through the weed roots. Pressure due to contact
with the soil itself is secondary.
Therefore again, the model of dense fluid flow around a cylindrical object is
not an accurate model. What we need to know from the beginning is how
much pressure it will take to cut through a network of weed roots in the soil.
This can be estimated by driving a spade head vertically through grass. My
own experience of this is that my body weight alone on top of a spadehead is
not sufficient to penetrate grass, but I can drive the spade into the grass if I
apply a bit of force. This I normally achieve by hopping onto the spade, by as
little as 5cm. This has the effect of amplifying my natural body weight.
Okay, so a little physics:
My body mass is about 80kg. Dropped from a height of 5cm, my body will
acquire a small amount of kinetic energy which then is consumed as work
done driving the spadehead into the ground.
mgh ½ mv2 force x distance.
The distance penetrated into the ground could be as little as 3cm.
Therefore (80)(10)(0.05) .. force x 0.03 = (m’)(10)(0.03)
A lot of cancellations allow me to see that the equivalent mass (m’) = 5/3 m
I can use this equivalent mass to calculate the pressure on the tip of the
Assume the dimensions of the spade are: 20cm long x 3mm thick.
That gives a cross-sectional area of 0.2 x (3 x 10E-3) m2 = 0.6 x 10E-3 m2.
The pressure on the spade head required to penetrate the weed roots is...
P = m’ x g / area = 5/3(80)(10) / (0.6x10E-3) Pa
P = (4000/3) (10E+3) / (0.6) Pa = (4/1.8) Mpa = 2.22 Mpa.
This, if you like is the yield stress of the network of weed-roots in the soil.
The nineteen blades will have to match this same yield stress with a much
greater combined cross sectional area.
So say the frontal area of one of these blades as it moves horizontally through
the soil is
10cm deep x 1cm wide = 0.1 x 0.01 = 10E-3 m2.
Now there has to be some effect due to the shape of the blade. Carey thinks
the blade is shaped like a tear-drop but is not sure. A shape like this would
slice the roots more easily than a blunt front surface, and so the required force
would be significantly lower than the one I’m about to calculate. But we can
use this as a guide.
Max stress = 2.22 Mpa = applied force / combined frontal area.
Applied force (N) = (2.22x10E+6)(Pa) x 19 x (10E-3) (m2)
= 2.22 x 19 x 10E(+6-3) (N) = 42 kN
So in order to pull nineteen very blunt blades through the dirt, the tractor will
need to apply about 42 kN of force.
What is this in horse-power? Well, to do that we have to equate horse-power
with the nearest metric unit which is Watts of power.
The ‘power’ is a combination of the force applied and the speed at which the
tractor is moving.
Assume the tractor’s speed is 10 kph, which converts to about 3 m/s.
Required power = 42 x 3 kW = 126 kW.
That seems very high. The conversion to horsepower (from the internet)
appears to be simply a matter of dividing the metric answer by 1000.
That means the tractor needs about 120-130 horsepower to tow the seeder at
Remember that this is a worst case scenario in several respects:
1: Tightly packed, weed-packed soil
2: Blunt profile presented to the direction of motion. Any form of sharpening
to the shape will reduce the required force.
Speaking purely intuitively now, without reference to practical data, I would
expect that a sharp-edged blade would reduce the required force by half, and
that loose, weed-free soil would reduce that force again by half.
Therefore in dense, weed-packed soil, with a teardrop-shaped blade, the
required horsepower might be only 60hp, and in loose soil the required
horsepower might only be 30hp.