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Ask HN: Can phone apps have monthly fees?

3 points by dfragnito 15y ago

I have been unable to find a phone app with a monthly fee. Is it possible, and if so what would be the major hurdles? I see this 30% in app purchase nonsense, I assume that would apply?

CptCodeMonkey 15y ago
I use the jailbreak only Groove Shark app plus the Pandora music app. Both have annual/monthly fee's attached which is fine because I am paying for the service they provide.
- dfragnito 15y ago
  Thanks for the reply. I am interested in the issues involved in developing a phone app with monthly fees (I should of made that more clear). I will take a closer look at Pandora and see how they implemented their subscription model in their phone app.
aespinoza 15y ago
I think it makes more sense to charge for a service, more than an an application. I think applications are seen as 'Products you own'. It might not be true. Evernote for exmaple charges for the cloud storage and services provided by their web app and cloud implementation, the iphone app is really just another channel.
revorad 15y ago
One way to bypass the app store restricitons is to charge separately on the web, like Instapaper does.
AznHisoka 15y ago
Yep apps can have monthly fees. Think about diet programs like Weight Watchers where you the tools come with the monthly program. No major hurdles except coming up with that product in the first place.
gte910h 15y ago
Apple doesn't allow autorenewing fees for "just apps". You have to be basically a magazine that makes episodic content the end user ends up owning. You can use their non-renewing subscription, 30% still applies, and renewing is a bit of a hassle.
dfragnito 15y ago
I understand what's involved now. This separation of an "app" from it's service component is a step backwards. It's my belief that "software" is a service. But we will play this game for now until eventually everyone comes to their sense and all apps become web apps. I am with Mozilla on this one. In the end I just want to serve markup to a browser and then all will be good.

3 points by dfragnito 15y ago

I have been unable to find a phone app with a monthly fee. Is it possible, and if so what would be the major hurdles? I see this 30% in app purchase nonsense, I assume that would apply?

CptCodeMonkey 15y ago

I use the jailbreak only Groove Shark app plus the Pandora music app. Both have annual/monthly fee's attached which is fine because I am paying for the service they provide.

dfragnito 15y ago
Thanks for the reply. I am interested in the issues involved in developing a phone app with monthly fees (I should of made that more clear). I will take a closer look at Pandora and see how they implemented their subscription model in their phone app.

aespinoza 15y ago

I think it makes more sense to charge for a service, more than an an application. I think applications are seen as 'Products you own'. It might not be true. Evernote for exmaple charges for the cloud storage and services provided by their web app and cloud implementation, the iphone app is really just another channel.

revorad 15y ago

One way to bypass the app store restricitons is to charge separately on the web, like Instapaper does.

AznHisoka 15y ago

Yep apps can have monthly fees. Think about diet programs like Weight Watchers where you the tools come with the monthly program. No major hurdles except coming up with that product in the first place.

gte910h 15y ago

Apple doesn't allow autorenewing fees for "just apps". You have to be basically a magazine that makes episodic content the end user ends up owning. You can use their non-renewing subscription, 30% still applies, and renewing is a bit of a hassle.

dfragnito 15y ago

I understand what's involved now. This separation of an "app" from it's service component is a step backwards. It's my belief that "software" is a service. But we will play this game for now until eventually everyone comes to their sense and all apps become web apps. I am with Mozilla on this one. In the end I just want to serve markup to a browser and then all will be good.