We break the barrier of $3/2$ for the problem of online load balancing with known makespan, also known as bin stretching. In this problem, $m$ identical machines and the optimal makespan are given. The load of a machine is the total size of all the jobs assigned to it and the makespan is the maximum load of all the machines. Jobs arrive online and the goal is to assign each job to a machine while staying within a small factor (the competitive ratio) of the optimal makespan. We present an algorithm that maintains a competitive ratio of $139/93<1.495$ for sufficiently large values of $m$, improving the previous bound of $3/2$. The value 3/2 represents a natural bound for this problem: as long as the online bins are of size at least $3/2$ of the offline bin, all items that fit at least two times in an offline bin have two nice properties. They fit three times in an online bin and a single such item can be packed together with an item of any size in an online bin. These properties are now both lost, which means that putting even one job on a wrong machine can leave some job unassigned at the end. It also makes it harder to determine good thresholds for the item types. This was one of the main technical issues in getting below $3/2$. The analysis consists of an intricate mixture of size and weight arguments.