Why can't the Atmel ATtiny family delay over a minute?
This is absolutely nothing to do with "why can't the atmel attiny family delay over a minute?". It is all to do with the way that C++ works.

See my page about Integer arithmetic and overflow.

To excerpt from that:



On the Arduino (Uno) platform, what do you think will be printed here?

void setup ()
  {
  Serial.begin (115200);
  Serial.println ();

  Serial.println (30000 + 30000); // twice 30000
  Serial.println (60 * 60 * 24);  // seconds in a day
  Serial.println (50 / 100 * 1000); // half of 1000
  }  // end of setup

void loop () { }


Did you guess:

60000
86400
500


Nope!

It prints:

-5536
20864
0


This is because of integer arithmetic. If the compiler can, it treats an numeric literal (like 60) as an int type, which means it has the range -32768 to +32767.

And, arithmetic is done using the type of the largest argument, which means the arithmetic in each case is done as 16-bit arithmetic, and thus it overflows once it reaches 32767.

For example, 30000 + 30000 = 60000 which is 0xEA60 in hex. Unfortunately, 0xEA60 is exactly how -5536 is stored in an int type, which is why it prints -5536.

Meanwhile, 60 * 60 * 24 = 86400 which is 0x15180 in hex. As that doesn't fit in 16 bits, it is truncated to 0x5180 which is 20864 in decimal (as printed).

Finally, in integer arithmetic 50/100 is zero, multiply zero by 1000 and you still get zero, which is why the final result is zero.



Solution

First, you can add a suffix to numeric literals (eg. L for long, or UL for unsigned long), and add a decimal place to floats, like this:

void setup ()
  {
  Serial.begin (115200);
  Serial.println ();

  Serial.println (30000L + 30000); // twice 30000
  Serial.println (60L * 60 * 24);  // seconds in a day
  Serial.println (50.0 / 100 * 1000); // half of 1000
  }  // end of setup

void loop () { }


Now we get:

60000
86400
500.00


You only need to help out with the first literal, once the compiler knows we are using longs (or floats) it sticks with them for the expression.*



* It's actually somewhat more complex than that as this link explains: Understand integer conversion rules.

The compiler "promotes" a value in an expression to match another "higher-ranked" type, under certain circumstances. For example, adding an int and a long will result in the int being promoted to a long (regardless of whether it appears first in the expression or not). However if an int is being added to another int, it will not promote them to a long, even though the result may not fit into an int.