On July 19th, 2008, my friend Dr. Matthias Koch brought a small mathematical problem to my birthday party: Suppose you are driving a car with an initial speed of 100 km/h. You are continuously loosing speed while driving  to be exact, you loose 1 km/h for every kilometer you drove. The question now is, how long does it take to drive 50 km with this style of driving? The first observation is that dv / ds = 1 since the graph of your speed plotted against the distance is linear, beginning with v = 100 km/h at s = 0 km and ending with v = 0 km/h at s = 100 km. Of course, you will never reach the point s = 100 km, but this is not the problem here since we are interested in the time it takes to reach a distance of 50 km. Thinking about the graph resulting from plotting the speed versus time it is obvious that v(t) must be of the form exp(t) since the speed at any distance s is exactly 100  s. Since the distance traveled at any time t is just the time integral over v(t) which is of the form 100 * exp(x), one could write As nice as an this analytical solution is, it would (of course) be nicer to have a model of the problem on an analog computer to confirm the result obtained above. All we need is a computer setup for this problem. Assuming a scaled model where 100 km are represented by a machine value of 1, we can write v(t) = 1  s(t), where s(t) is the time integral over v(t). The resulting computer setup is extremely simple: The lower integrator just yields the time integral over the constant 1. Its output directly corresponds to the time spent traveling. (By the way, the analog computer used is a Telefunken RA 742.)
The time required to reach 50 km is then computed as 

20JUl2008, 21DEC2008, 19FEB2016, ulmann@analogmuseum.org 