What’s the g-force in a rally car?

Photograph by RX-Guru at German Wikipedia - Transferred from de.wikipedia to Commons., CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=2080303

Photograph by RX-Guru at German Wikipedia – Transferred from de.wikipedia to Commons., CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=2080303

An FIA rally-cross car can accelerate from 0-60 mph in 2 seconds.

Lets check equations of motion, first some definitions.
u – initial velocity
v – final velocity
a – acceleration
t – time
s – displacement

v = u + at
s = ut + ½at2
v2= u2+ 2as

BBC Bitesize resource on equations of motion.
—-

v is 60 mph is 26.8224 m/s
and average acceleration = 26.8224 / 2 = 13.4112 m/s^2
= 13.4112 / 9.80665 g
= 1.368 g
~ 1.4 g

Back calculating….

v = u + at
v = 0 + 13.4112 * 2

OK, in agreement with what we expected.

Distance travelled
s = ut + 1/2 * a t2
= 0 + 0.5 * 13.4112 * 4
= 2 * 13.4112
= 26.8224 m
~ 26.8 m

Rally car

By Foto: Stefan Brending, Lizenz: Creative Commons by-sa-3.0 de, CC BY-SA 3.0 de, https://commons.wikimedia.org/w/index.php?curid=35347288

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Octave/ Matlab is fun – Enter the Matrix

f = rot90((diag((ones(5,1)))) + hankel(zeros(5,1),2*(ones(5,1))),1)
warning: hankel: column wins anti-diagonal conflict
f =

0 2 2 2 3
0 0 2 3 2
0 0 1 2 2
0 1 0 0 2
1 0 0 0 0

hess(diag((ones(4,1))))
ans =

1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1

eig(diag((ones(4,1))))
ans =

1
1
1
1

diag([1,2,3],4)
ans =

0 0 0 0 1 0 0
0 0 0 0 0 2 0
0 0 0 0 0 0 3
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0

rot90(shift(rot90(rand(4),1),1),3)
ans =

0.141505 0.100645 0.200515 0.552900
0.163904 0.081520 0.505508 0.356016
0.458187 0.275491 0.472148 0.434868
0.988882 0.267269 0.680475 0.424024

Emergency Number

My department has two emergency numbers, one for office hours (8.45-5.15) and one for the rest of the time. The numbers are 67760 for office hours and 31818 for the rest of the time. In lunch time I’m not sure what we should do.

If we add the number we get 67760 + 31818 = 99578

which is 421 short of 99999. 99999 and 421 are easy to remember. So if you can remember this helpful other numbers and one of the emergency numbers you should easily be able to work out the other one. So now you only have to remember either 31818 or 67760.

Puzzling Socks

Maths

Here is a puzzle about socks;

You have a drawer in your bedroom that contains an assortment of black, blue and brown socks. You start randomly taking individual socks out of the drawer, but it’s too dark to see what colour socks you have taken. How many socks do you need to take out in order to guarantee that you have got a matching pair?

The answer is Four socks. Taking out one more than the number of colours of sock in the drawer guarantees at least one pair.

Reality

Unfortunately this doesn’t help me. Each of my socks have one matching sock, so to find a pair I have to remove almost all of them.

Mtdata Script example 1

Here is an example macro for use with thermodynamics software mtdata. This script sets up a stepped calculation of equilibrium for a steel composition, it then output a text file of the mass fraction of each phase. Reading the script is made difficult by the use of contracted version of the commands. The first command selects the multiphase modules could have been written ‘mu’ rather than ‘multiphase’. The next command starts ‘def sys…’ is the same as ‘define system’. Note most commands end with an exclamation mark, this is the way to tell mtdata you really mean what you typed.


:Macro file to calculate equilibrium phase fractions
:Mathew Peet and Hala Salman Hasan 20th June 2007
multiphase
def sys "Fe,C,Si,Mn,Cr,Mo,Co,Al" source plus sub_sgte !
:this is a comment
:uncomment following line to use tcfe database instead
:def sys "Fe,C,Si,Mn,Cr,Mo,Co,Al" source tcfe !
classify absent phase(*) !
classify normal phase(FCC_A1,BCC_A2,M23C6,M6C,cementite,m7c3) !
classify misc(FCC_A1) 1 !
set w 100 !
set w(2) 0.78 w(3) 1.6 w(4) 2.02 w(5) 1.01 w(6) 0.24 w(7) 3.87 w(8) 1.37 !
step temp 773 1373 5 !
:this line is a comment
:following 3 lines output a spreadsheet of phase fractions with temperature
com pri gra !
ord mass phase !
plot tabu spread !
:uncomment line/ type following to see compositions of phases
:compute print brief print mol !

Example output of the above calculation was archived.

Simple Math: Quadratic Equation

quadratic equations

Gnuplot script

How do trees raise water?

Capillarity can raise liquid against gravity, I was interested to know how much this can account for the movement of water in trees, is capillarity alone enough to lift water up a tree trunk?

For water in a tube there is an equation like this;

h = 2T cos (theta) / pgr
where; T = 0.0728 J/m^2 at 20 degrees C, the contact angle theta = 20 degrees, density of water, p = 1000 kg/m^3, and the local acceleration due to gravity can usually be taken to be 9.8 m/s^2.

So the height is given by;

h = 1.4x10^{-5) / r

So tubes of the following diameters can lift water to the following heights.

Diameter (mm) Height (m)
1 0.014
0.1 0.14
0.01 1.4
0.001 14
0.0001 140

If we assume perfect contact these heights become only slightly larger. (cos(theta)=1)
1 micrometre diameter tube can lift water 14.85 m, it’s necessary to have a tube 0.1 micrometres or 100 nanometers to lift water a height of 148 meters by capillary action alone, if the above equations are valid. Looking around on the internet gave a size of xylem in a tree to be around 20-30 micrometres (0.02 – 0.03 mm) in size, which if true enough to lift water around half a meter. It would be interesting to see if this value is correct.

If you are interested to find more there is a 2004 nature article on the limits of tree height here.