Let me start with the additional problems that arise:
* The trajectory of the suns is not stable. If they are slightly off-center, gravity will pull them towards the side of the cylinder that they're closer to, analogous to the ringworld stability issue. You could work around this by using stellar engines of some sort to keep the stars centered, or making the cylinder a tiny bit flexible and using motors to change its shape dynamically....
* There is nowhere for the heat that is generated by fusion in the cores of the suns to escape, apart from conduction through the crust towards outside space. You could work around this by putting large holes in your cylinder through which outside space is visible, by making your crust very thin (on the order of meters), or by making it very conductive (by adding an active cooling system that pumps heat outside). This oversimplified illustration shows the relevant mechanisms that keep earth's surface at its equilibrium temperature, and how the inside of your cylinder would heat to over 2 million Kelvins without any countermeasures:
Now, to your actual question.
The only relevant parameter is the distance between the suns, in AU. 24
The speed at which they move follows automatically from your requirement that one sun should pass every 24 hours. It will be rather high, though :)
You will, of course, always see an infinite number of suns, but most of them will be very dim and very close to the horizon. Here's what the sky will look like, with the apparent brightness of the suns (= the area they occupy in the sky) written next to the dots.
suns spaced at 1AU: suns spaced at 20AU:
To calculate the total illumination, some math is required. You need to calculate the infinite sum of the contributions of each sun. In this formula, d is the distance between the suns in AU, and o is the offset from mid-day, where o=0 means mid-day, and o=1 means mid-day tomorrow.
This gives the following equation for the momentary strength of illumination (assuming that the power output of one sun at 1AU distance is 1):
-(π sinh((2 π)/d))/(d (cos(2 o π) - cosh((2 π)/d)))
To find your preferred value of d, just plot this formula for various values.
Here's a quick python snippet that does exactly that, since I couldn't get nice plots out of Wolfram Alpha:
```python
!/usr/bin/env python3
from argparse import ArgumentParser from math import sqrt, sinh, cos, cosh, pi import numpy from matplotlib import pyplot as plt
cli = ArgumentParser() cli.add_argument('--distance', type=float, default=1) cli.add_argument('--average-illumination', type=float, default=0.25) args = cli.parse_args() power = 0.31831 * args.average_illumination * args.distance
hours = numpy.arange(0, 24, 1/60) illuminations = [] for hour in hours: offset = hour / 24 - 0.5 illuminations.append( -power * pi * sinh((2 * pi)/args.distance) / (args.distance * (cos(2 * offset * pi) - cosh((2 * pi)/args.distance))) )
fig = plt.figure() ax = fig.add_subplot(1, 1, 1) ax.set_xticks(range(25)) ax.set_xlim(0, 24) ax.set_yscale('log') ax.grid() ax.plot(hours, illuminations) ax.set_title(f'spacing: {args.distance} AU, ' f'luminosity: {power} L0, ' f'min: {min(illuminations):.5g}, ' f'max: {max(illuminations):.5g}')
from https://en.wikipedia.org/wiki/Lux#Illuminance
ax.annotate("moonless clear sky with airglow", (0.5, 0.002/100e3)) ax.annotate("full moonlight", (0.5, 0.3/100e3)) ax.annotate("dark limit of civil twilight", (0.5, 3.4/100e3)) ax.annotate("family living room lighting", (0.5, 50/100e3)) ax.annotate("very dark overcast day", (0.5, 100/100e3)) ax.annotate("sunrise or sunset on clear day", (0.5, 500/100e3)) ax. 371