I tried using nls() in R to fit the following expression to a set of data:

enter image description here

where g=9.8, alpha & B_0 are unknown, a = 0.01, z_0 = 0.3 such that:

theory <- as.formula(V~-(9.8)ab*(pi)(0.02^2)(T)(0.3-(0.5)(9.8)(T^2))exp(-a(0.3-(0.5)(9.8)*(T^2))^2))

enter image description here where T = t and V = E(t) from the theory model

I was expecting a cubic-like curve but it resulted in a relatively straight line. Is there another way to try and fit the curve, such that the result is closer to: enter image description here

EDIT: here are the results I got when using nls:

theory <- as.formula(V~-(9.8)ab*(pi)(0.02^2)(T)(0.3-(0.5)(9.8)(T^2))exp(-a(0.3-(0.5)(9.8)*(T^2))^2))

mydata <- read.csv("voltage1.csv") nls.1 <- nls(theory,start=list(a=.8,b=.01),data=mydata) overview(nls.1) plotfit(nls.1,smooth=TRUE)

enter image description here

  • 2
    $\begingroup$ It would be much better if you typed out your data and formulas, otherwise we have to do it (which decreases the chances of you getting an answer). You say that you have fitted a line successfully however it is not to your liking. 1. Can you show us a plot of your data? 2. How much do you trust your given constants? $\endgroup$ Mar 8, 2022 at 8:42

1 Answer 1


The two peaks arise from this term $(z-0.5gt^2)$ and its square $(z-0.5gt^2)^2$.

Moving away from the point where $(z-0.5gt^2) = 0$, the peaks increase from the increase in the magnitude of $(z-0.5gt^2)$ (which can be negative/positive) and the peaks decrease when the terms $e^{-(z-0.5gt^2)^2}$ start to become dominant.

In the plot below you see that this term $(z-0.5gt^2)^2$ is zero around the point $t = \sqrt{2z/g} \approx 0.247$. The negative peak will be on the left of it and the positive peak will be on the right of it.

Your data has the both peaks on the left of $t = 0.25$, and that is why there is no good fit. Probably your value of $z$ is wrong or the time $t$ needs to be shifted.

plot of exponential term

If you allow flexibility for either $z$ or if you let the parameter $t$ be flexible with a shift, then you are able to get a fit that is not a straight line because now there is are two peaks possible. However, compared to the data the peaks are more broad (the red curve in the plot). It can be fixed by using a slightly different formula (the green curve in the plot)

$$v = -g \alpha b_0 \pi a^2 (t) (z-0.5 g (t)^2)^3 e^{-\alpha(z-0.5g(t)^2)^2}$$

The strange thing about your formula without the power $(z-0.5 g (t)^2)^3$ is that the derivative $dv/dt$ is not zero when this term $(z-0.5 g (t)^2)$ is zero (the point in between the two peaks. But in the plots we see that in between the peaks the slope is zero, or at least there is a strong inflection point.

plot of fits

t = c(seq(0,0.15,0.01), 0.155, seq(0.16,0.25,0.01))
v = c(0,-0.1,-0.2,-0.3,-0.4,-0.8,-1.3,-2.6,-2.4,-1.6,-0.5,-0.2,0,0.4,1,3.6,4.6,3.6,1,0.2,rep(0,7))

g = 9.8
a = 0.01
z = 0.3
mod = nls(v ~ -g*alpha*b*pi*a^2*(t+t0)*(z-0.5*g*(t+t0)^2)*exp(-alpha*(z-0.5*g*(t+t0)^2)^2) , start = c(alpha = 200, b = 500, t0 = 0.15))
mod3 = nls(v ~ -g*alpha*b*pi*a^2*(t+t0)*(z-0.5*g*(t+t0)^2)^3*exp(-alpha*(z-0.5*g*(t+t0)^2)^2) , start = c(alpha = 200, b = 50000, t0 = 0.15))

ts = seq(0,0.25,0.0001)
lines(ts, predict(mod, newdata = list(t=ts)), col = 2)
lines(ts, predict(mod3, newdata = list(t=ts)), col = 3)
  • $\begingroup$ Whoa this is insanely helpful, thank you so much for the work and thought put into your response! And yes, I believe you are right about t needing to be shifted, to be honest I am new to Physics and using oscilloscopes, so I simply shifted the horizontal axis (time) when I got a frame of the data I wanted to conveniently always start at t=0 when the curve starts, granted I believe t is supposed to represent when the magnet is dropped initially from a height of z_0 which I had as 30 cm, or 0.3 m. Either way, accounting for a shift in t seems like the best approach. Thank you! $\endgroup$ Mar 8, 2022 at 18:44
  • $\begingroup$ @Will I get now that this equation relates to the voltage generated by a magnate falling through a solenoid. The difference in the shape is due to the length of the magnet and the length of the solenoid. $\endgroup$ Mar 9, 2022 at 7:52

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