Question: I have a plug in leaf blower/vacuum and would like to convert it to battery power. here is a link to the equipment in question. [http://www.amazon.com/Factory-Reconditioned-Ryobi-ZRRY42110-Electric-Mulcher/dp/B004NC014Q/ref=cm\_cr\_pr\_product\_top?ie=UTF8](http://rads.stackoverflow.com/amzn/click/B004NC014Q). 12amp is its maximum output and i don't necessarily need it running that high but would like to know if it is possible to simply wire it to a 24v battery and still use it for at least an hour?

Answer: 
Providing the circuit with a larger capacitors is a good start. To determine the required capacity you need the energy used during the spike. To do this one can measure voltage and current as a function of time with an oscilloscope. 
The energy the capacitor has to provide during the spike is roughly \$W\_S \approx \int\_{t\_{s0}}^{t\_{s1}}u(t)(i(t)-I\_{USB,max}) dt\$

where \$I\_{USB,max}\$ is the current the USB Port can provide. The energy in a capacitor is

\$W= \frac{1}{2}C\cdot V\_{USB}^2 \$ and the capacitor needs to keep at least the voltage \$V\_\mu\$ required to supply the microcontroller. Thus the minimal energy in the capacitor is \$W\_\mu=\frac{1}{2}\cdot V\_{\mu}^2\cdot C\$. Consequently we have \$ W = W\_s + W\_\mu \$ which solves to a minimal capacity of \$C=2\frac{\int\_{t\_{s0}}^{t\_{s1}}u(t)(i(t)-I\_{USB,max}) dt}{V\_{USB}^2-V\_{\mu}^2}\$
Alternatively set a very large capacitor in parallel to the supply. Measure the current over time during the spike. The minimal capacity would then be \$C= \frac{(\overline{I}-I\_{USB,max})\cdot T}{V\_{USB}}\$ where \$\overline{I}\$ is the arithmetic average of the current flown during the spike.

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If those current spikes result from switching the LEDs and the switching timing is not critical (e.g. no optical data transmission) you can add an inductance in series to the LEDs. Sized large enough compared to the needed energy and duration of the spikes this will even the spike over a larger time interval reducing the required capacitor size by using the current supplied via USB. This is not recommended for the time critical wireless communication.

Edit: After a spike there also has to be enough time to the next spike in which the capacitor can be recharged, which depends on \$I\_{USB,max}\$.
There are 3 solutions to your problem as far as I can see:

1. Turn off the low voltage drop detection function in your MCU.
2. Limit the current draw, i.e. remove some of the LEDs, or add higher value resistors.
3. Use super-capacitor such as 1F rated at 5V.