To illustrate the difficulty of building a beam-target fusion system, we will consider one promising fusion fuel, the proton-boron cycle, or p-B11.
Boron can be formed into highly purified solid blocks, and protons easily produced by ionizing hydrogen gas. The protons can be accelerated and fired into the boron block, and the reactions will cause several alpha particles to be released. These can be collected in an electrostatic system to directly produce electricity without having to use a Rankine cycle or a similar heat-driven system. As the reactions create no neutrons directly, they have many practical advantages for safety also.
The chance of a collision is maximized when the protons have an energy of about 675 keV. When they fuse, the alphas carry away a total of 8.7 MeV. Some of that energy, 0.675 MeV, must be recycled into the accelerator to produce new protons to continue the process, and the generation and acceleration process is unlikely to be much more than 50% efficient. This still leaves ample net energy to close the cycle. However, this assumes every proton causes a fusion event, which does not occur. Considering the probability of a reaction, the resultant cycle is:
where  and  are the probabilities that any given proton or boron will undergo a reaction. Re-arranging, we can show that:
That means that to break even, the system needs at least  of the particles to undergo fusion. To ensure that a proton has a chance to collide with a boron, it must travel past many boron atoms. The collision rate is:
where  is the nuclear cross section between a proton and boron,  is the density of boron, and  is the average distance the proton travels through the boron before undergoing a fusion reaction. For p-B11,  is 0.9 x 10 cm,  is 2.535 g/cm, and thus  ~ 8 cm. However, travelling through the block causes the proton to ionize the boron atoms it passes, which slows the proton. At 0.675 MeV, this process slows the proton to sub-keV energies in about 10 cm, many orders of magnitude less than what is required.