I’ve been cycling a lot this season, but off-leash dogs are the rule here rather than the exception. They like to chase me on the bike, and even the ones that don’t seem likely to directly harm me are liable to cause an accident.
To combat this, I use pepper spray. To keep the pepper spray handy, I use this simple device I found on Amazon.com. I was skeptical when I first saw it that it would be stable enough, but actually it has worked quite well. I use Fox OC spray with it, and it has never failed to stop an aggressive dog.
The spray is a bit pricey, but if you ride the same route or routes regularly the dogs eventually start to learn and you don’t need it that much.
I have a dog and I don’t like pepper spraying dogs, but irresponsible owners have created dangerous conditions for me and other cyclists and pedestrians, so it has come to this.
I am studying for the General license exam and came across question G8B07 (as of April 2015, it will change at some point), asking the following:
What is the frequency deviation for a 12.21-MHz reactance-modulated oscillator in a 5-kHz deviation, 146.52-MHz FM-phone transmitter?
The answer is given as 416.7 Hz, but I looked and couldn’t find an explanation. Finally I found one here, sort of. After reading about how the FM phone modulation is done and puzzling a bit at the numbers it became clear.
To understand the question, you must understand that the reactance-modulated oscillator produces a carrier frequency of 12.21 MHz (as specified). With no input, this is multiplied by some circuits aptly called multipliers that result in the output frequency, given here as 146.52 MHz. Simply dividing 146.52/12.21 shows you that a multiplier of 12x is in effect in this question.
Now, what the question is asking (and this took me a while to figure out) is what change in modulation of the reactance-modulated oscillator will result in a 5 kHz deviation of the output? Since we know that whatever change happens in the oscillator is going to be multiplied 12x, we can divide the output deviation (5 kHz) by the multiplier (12) to get 416.7 Hz. In other words, a 416.7 Hz deviation in the reactance-modulated oscillator will be multiplied 12x to 5 kHz by the multipliers before being output.
My wireless ISP (WISP) uses the Mikrotik hotspot feature with RADIUS on the back end to authenticate our users. This implements a captive portal that redirects all DNS requests so that the user is taken to a login page if they’re not logged in. Once they log in once, the system associates their radio with their account, and they don’t have to log in anymore under normal circumstances.
However, once logged in, users still have all their DNS requests proxied through the routers. A lot of users want to use their own DNS (like OpenDNS or Google Public DNS), and that’s fine with me, but a user ran the
namebench utility and found that their DNS was being forcibly proxied.
It took some hunting, but I finally found this post on the Mikrotik forums which details how to get around this. Basically:
- The hotspot adds dynamic DNS redirect rules. If you go to
/ip firewall nat and just
print, these rules don’t show up. If you do
print dynmic they do. The relevant lines are:
2 D chain=hotspot action=redirect to-ports=64872 protocol=udp dst-port=53 log=no log-prefix=""
3 D chain=hotspot action=redirect to-ports=64872 protocol=tcp dst-port=53 log=no log-prefix=""
- We still want non-logged-in-users to have their DNS redirected, so we need to add something here that will enable authenticated hotspot users through. The magic incantation here (because it’s entries 2 and 3) is
set 2,3 hotspot=!auth, which results in the following:
2 D chain=hotspot action=redirect to-ports=64872 protocol=udp hotspot=!auth dst-port=53 log=no log-prefix=""
3 D chain=hotspot action=redirect to-ports=64872 protocol=tcp hotspot=!auth dst-port=53 log=no log-prefix=""
And now namebench works as expected.
And it’s looking good today. 🙂
I am not an expert on this, I just wanted to document a problem I had and a solution I found today, in a concise way. Comments correcting me or suggesting better ways are very welcome.
I have a network running OSPF internally, and advertising routes to the upstream ISP over BGP at two separate edge routers (multi-homed, single ISP). We discovered last night that internally bringing down any of the subnets we advertise results in the dropping of those routes from the tables of the edge routers (as expected). This drops the advertisements. What we did NOT expect was that flap damping from upstream of us then null-routes that subnet for up to a few hours.
So, how do we retain our adaptive internal routing (OSPF) while avoiding route flap? I was a bit stumped about this, but I found a more complex article that describes a multi-homed BGP setup. A key part of that setup was a little trick to avoid this problem. Nameley, set up a static, black hole route for the subnet on the edge router, with maximum distance. This way, even if the OSPF route disappears, the router still “knows” a route to the subnet and won’t drop the advertisement.
For example, if you want to advertise the subnet 18.104.22.168/24, you should add a static route like
/ip route add dst-address=22.214.171.124/24 type=blackhole distance=254 comment="prevent flapping of the route over BGP"
I’ve tested it and it seems to work as expected. The route is not active as long as the OSPF route is in the routing table. If it disappears, the black hole route becomes active.