Comment by snickerbockers 6 years ago The article doesn't mention this, but 51 days is approximately 2^32 milliseconds... 3 comments snickerbockers Reply wolfgke 6 years ago 2^32 ms is about 49.71 days ( (2^32)/(1000 * 3600 * 24) ), so less than the reboot cycle of 51 days. SAI_Peregrinus 6 years ago I mentioned this in another thread, but 2^32 * 1024us is 50.9 days. So it's probably a systick at 1.024ms overflowing a uint32_t. If you've got a 1us timer it's a lot cleaner for the CPU to make the tick happen at 1024us than at 1000. rightbyte 6 years ago On the 51th day you need to power cycle, counting from day one.
wolfgke 6 years ago 2^32 ms is about 49.71 days ( (2^32)/(1000 * 3600 * 24) ), so less than the reboot cycle of 51 days. SAI_Peregrinus 6 years ago I mentioned this in another thread, but 2^32 * 1024us is 50.9 days. So it's probably a systick at 1.024ms overflowing a uint32_t. If you've got a 1us timer it's a lot cleaner for the CPU to make the tick happen at 1024us than at 1000. rightbyte 6 years ago On the 51th day you need to power cycle, counting from day one.
SAI_Peregrinus 6 years ago I mentioned this in another thread, but 2^32 * 1024us is 50.9 days. So it's probably a systick at 1.024ms overflowing a uint32_t. If you've got a 1us timer it's a lot cleaner for the CPU to make the tick happen at 1024us than at 1000.
2^32 ms is about 49.71 days ( (2^32)/(1000 * 3600 * 24) ), so less than the reboot cycle of 51 days.
I mentioned this in another thread, but 2^32 * 1024us is 50.9 days. So it's probably a systick at 1.024ms overflowing a uint32_t. If you've got a 1us timer it's a lot cleaner for the CPU to make the tick happen at 1024us than at 1000.
On the 51th day you need to power cycle, counting from day one.