Arduino Break Beam Gravity

GOAL: Measure acceleration due to Earth's gravity using Arduino microsecond timer circuit

Drop a ball at L = 10, 20, 30, 15, 25 inches 5x each.

 O -> ball at L height
 |
 |
 (L-4) inches
 |
 |
 + -> sensor 1
 | -> 4 in
 + -> sensor 2

Measure time in microseconds to fall between sensor 1 and sensor 2 when it breaks the light beam.
RESULT: my estimate is $g = 9.5 m / s^{2}$ . This is within 3% of true value. Air resistance probably affected my estimate.
OVERALL RESULT: you can use an Arduino Uno micro-second timer circuit to accurately estimate acceleration due to gravity.

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Data

The data right of the serial port:
--- Miniterm on /dev/ttyS4  9600,8,N,1 ---
--- Quit: Ctrl+] | Menu: Ctrl+T | Help: Ctrl+T followed by Ctrl+H ---
setup
elapsed_time 54112
elapsed_time 52152
elapsed_time 51488
elapsed_time 51888
elapsed_time 50536
elapsed_time 36364
elapsed_time 35464
elapsed_time 35548
elapsed_time 33984
elapsed_time 36768
elapsed_time 27836
elapsed_time 28036
elapsed_time 27056
elapsed_time 28044
elapsed_time 28972
elapsed_time 39532
elapsed_time 40480
elapsed_time 38668
elapsed_time 41200
elapsed_time 44512
elapsed_time 29268
elapsed_time 30200
elapsed_time 28488
elapsed_time 30252
elapsed_time 29868

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What Do the Laws of Physics Say Should Happen?

$F = m a$ # newton's law
$a = F / m$
Force of earth ( $E$ ) gravity at surface on object with mass M. The acceleration is $F / m = g$ for any mass!
- $F = G m_{E} m / r_{E}^{2} = g m$
Knowing the acceleration g we can get the velocity and position by integrating twice.
- $v = g t + v_{0}$
- $x = 1 / 2 g t^{2} + v_{0} t + x_{0}$
Starting at height L, 0 velocity and triger sensors S apart:
- fall L-S to sensor1 in t1 to v1.
- fall S to sensor2 with initial velocity v1 in time t2
$(L - S) = 0.5 g t_{1}^{2}$ # from x equation
$t_{1} = s q r t (2 (L - S) / g)$ # rearrange: time to get to sensor 1
$v 1 = g t_{1}$ # velocity at sensor 1 from v equation
$v 1 = g s q r t (2 (L - S) / g)$ # substitute t_1 from a above
$v 1 = s q r t (2 g (L - S))$ # simplify

$S = 0.5 g t_{2}^{2} + v_{1} t_{2}$ # sensor fall S with initial v1 using x equation
$0.5 g t_{2}^{2} + v_{1} t_{2} - S = 0$ # rearrange
quadratic formula $t = [- b + / - s q r t (b^{2} - 4 a c)] / 2 a$ . $a = 0.5 g$ . $b = v 1$ . $c = - S$
$t_{2} = - v_{1} + / - s q r t (v_{1}^{2} + 4 \cdot 0.5 g S) / (2 \cdot 0.5 g)$
$t_{2} = - v_{1} + s q r t (v_{1}^{2} + 2 g S) / g$ # positive root, time has to be positive
Substitute v1:
$t_{2} = [- s q r t (2 g (L - S)) + s q r t ((2 g (L - S)) + 2 g S)] / g$
$t_{2} = [- s q r t (2 g (L - S)) + s q r t (2 g L)] / g$
$t_{2} = [s q r t (2 g L) - s q r t (2 g (L - S))] / g$
$t_{2} = s q r t (2) [s q r t (L) - s q r t (L - S)] / s q r t (g)$
$t_{2} = s q r t (2 / g) [s q r t (L) - s q r t (L - S)]$

RESULT: t_2 is the calculated time as measured by sensors S apart falling from a height of L

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Statistical Data Analysis

Pull the data into R

# observed sensor times in microseconds
t = c(54112, 52152, 51488, 51888, 50536, 36364, 35464, 35548, 33984, 36768, 27836, 28036, 27056, 28044, 28972, 39532, 40480, 38668, 41200, 44512, 29268, 30200, 28488, 30252, 29868)/1.0E+06

# the heights the experiment was performed at
l = c(rep(10,5),rep(20,5),rep(30,5),rep(15,5),rep(25,5))

S=4

myx = (sqrt(l)-sqrt(l-S)) # what the linear model inputs. t2 equation

# make a linear estimate of the relationship (least squares: t = A*myx+B)
mylm = lm(t ~ myx)
summary(mylm)
Residuals:
       Min         1Q     Median         3Q        Max 
-0.0021731 -0.0007420 -0.0003242  0.0008086  0.0037078 
Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) 0.0001866  0.0012439    0.15    0.882    
myx         0.0730061  0.0023870   30.58   <2e-16 ***


# make a linear estimate of the relationship forcing 0 intercept
mylm = lm(t ~ myx+0)
summary(mylm)
Residuals:
       Min         1Q     Median         3Q        Max 
-0.0021436 -0.0007521 -0.0003316  0.0008306  0.0037004 
Coefficients:
     Estimate Std. Error t value Pr(>|t|)    
myx 0.0733548  0.0005332   137.6   <2e-16 ***

RESULT: the empirical constant estimate is between 0.0730 and 0.0733: $t_{2} = 0.073 * [s q r t (l) - s q r t (l - S)]$
From model:
$t_{2} = s q r t (2 / g) [s q r t (l) - s q r t (l - S)]$
$s q r t (2 / g) = 0.073$
$2 / g = (0.073 * 0.073)$
$g = 2 / (0.073 * 0.073)$
$g = 375.30493526$ inches/s/s
$g = 9.5 m / s^{2}$ # 1 m = 39.37 inches
True: $g = 9.8 m / s^{2}$

RESULT: My estimate of the acceleration due to gravity 9.5 m/s^2 is within 3% of the true value 9.8 m/s^2. Air resistance?

Plot data

# what the model predicts given correct constants
S=4
modelTgivenL = function(L){
gtrue = 9.8*39.37 # inches/s^2
return(sqrt(2/gtrue)*(sqrt(L) - sqrt(L-S) ))
}
modelEstTgivenL = function(L){
constestimate = 0.0733 # from linear fit
return(constestimate*(sqrt(L) - sqrt(L-S) ))
}

png("drop.png",width=1024,height=1024)
plot(t ~ l,pch="x",xlab="L height",ylab="sensor time",cex=2)
points(seq(10,30), modelTgivenL(seq(10,30)),type="b", col="red",cex=2)
points(seq(10,30), modelEstTgivenL(seq(10,30)),type="b", col="green",cex=2)
title("drop times. black=measured red=model green=estimated model")
dev.off()

RESULT: modeling and empirical data line up pretty well. Times are a bit high for 20 inches (did I consistently hold it too low?)
TODO: I plugged in the measured distance between sensors = 4 inches. Could I have estimated this also to get two estimates, g and S, from the data? g can compensate S so you can't detangle.

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Timer Construction

For timer construction see:

Arduino Break Beam Gravity

Data

What Do the Laws of Physics Say Should Happen?

Statistical Data Analysis

Plot data

Timer Construction

README_RESULT_arduino-break-beam.html