If you'll follow the second link you'll see that the FSF agrees with my position. If the source wasn't included with the distribution, anyone who has the written offer can request it and it must be supplied.
Who has the written offer? People who have received the binaries from you, either directly or indirectly. If you commercially distribute binaries not accompanied with source code, the GPL says you must provide a written offer to distribute the source code later. When users non-commercially redistribute the binaries they received from you, they must pass along a copy of this written offer. This means that people who did not get the binaries directly from you can still receive copies of the source code, along with the written offer.
So unless you've been given the binary, they don't have to give you the code.
Did anyone who purchased one of these phones ask for the source? They don't have to put the source out there for the general public.
Actually, according to the GPL, if they don't provide the source with every phone then they DO have to give it to anyone in the general public upon request - until they've sold or otherwise "distributed" the last one and for a period of time thereafter.
Actually, according the the GPL, they don't. They just have to give it to anyone who uses the binary. However, most of the time anyone in the general public could be a user, but that's not assured.
The numbers do add up. Here's the math: Note - I am not writing out all the digits, but I kept them when doing the math to avoid rounding error.
Given: x = x0 + v0*t + 1/2*a*t^2 v = v0 + a*t Started from rest 1st segment: 1.4 miles in 4.65 seconds 2nd segment: 1.8 miles in 1.30 seconds
Assumption: Constant acceleration during each segment, although different acceleration for the segements.
Solution: First segment acceleration: x = x0 + v0*t + 1/2*a*t^2 x = 1.4 miles; x0 = 0 (starting from here); v0 = 0 (starting from rest); t = 4.65 sec 1.4 = 1/2*a*4.65^2 solving: a = 1.4/10.81125 ~= 0.129 miles/sec^2
Final speed at end of first segment: v = v0 + a*t v0 = 0 (starting from rest); t = 4.65 sec; a ~= 0.129 miles/sec^2 v ~= 0.129*4.65 solving: v ~= 0.602 miles/sec or 2167 mph
Second segment acceleration: x = x0 + v0*t + 1/2*a*t^2 x = 1.8 miles; x0 = 0 (starting from here, or we could add 1.4 to this and x); v0 = 0.602 miles/sec; t = 1.3 sec 1.8 = 0.602*t + 1/2*a*1.3^2 solving: a = 1.0172/0.845 ~= 1.204 miles/sec^2
Final speed at end of second segment: v = v0 + a*t v0 = 0.602 miles/sec; t = 1.3 sec a ~= 1.204 miles/sec^2 solving: v ~= 2.17 miles/sec, or 7800 mph
Thus assuming constant acceleration, we actually achieve a velocity greater than 6400 mph. With decreasing acceleration (a real-world condition), 6400 mph is a believable result.
Slashdot Mirror
If you'll follow the second link you'll see that the FSF agrees with my position. If the source wasn't included with the distribution, anyone who has the written offer can request it and it must be supplied.
Who has the written offer? People who have received the binaries from you, either directly or indirectly.
If you commercially distribute binaries not accompanied with source code, the GPL says you must provide a written offer to distribute the source code later. When users non-commercially redistribute the binaries they received from you, they must pass along a copy of this written offer. This means that people who did not get the binaries directly from you can still receive copies of the source code, along with the written offer.
So unless you've been given the binary, they don't have to give you the code.
Did anyone who purchased one of these phones ask for the source? They don't have to put the source out there for the general public.
l #GPLRequireSourcePostedPublic>l #WhatDoesWrittenOfferValid>
Actually, according to the GPL, if they don't provide the source with every phone then they DO have to give it to anyone in the general public upon request - until they've sold or otherwise "distributed" the last one and for a period of time thereafter.
Actually, according the the GPL, they don't. They just have to give it to anyone who uses the binary. However, most of the time anyone in the general public could be a user, but that's not assured.
http://www.fsf.org/licensing/licenses/gpl-faq.htm
and
http://www.fsf.org/licensing/licenses/gpl-faq.htm
The numbers do add up. Here's the math:
Note - I am not writing out all the digits, but I kept them when doing the math to avoid rounding error.
Given:
x = x0 + v0*t + 1/2*a*t^2
v = v0 + a*t
Started from rest
1st segment: 1.4 miles in 4.65 seconds
2nd segment: 1.8 miles in 1.30 seconds
Assumption:
Constant acceleration during each segment, although different acceleration for the segements.
Solution:
First segment acceleration:
x = x0 + v0*t + 1/2*a*t^2
x = 1.4 miles; x0 = 0 (starting from here); v0 = 0 (starting from rest); t = 4.65 sec
1.4 = 1/2*a*4.65^2
solving: a = 1.4/10.81125 ~= 0.129 miles/sec^2
Final speed at end of first segment:
v = v0 + a*t
v0 = 0 (starting from rest); t = 4.65 sec; a ~= 0.129 miles/sec^2
v ~= 0.129*4.65
solving: v ~= 0.602 miles/sec or 2167 mph
Second segment acceleration:
x = x0 + v0*t + 1/2*a*t^2
x = 1.8 miles; x0 = 0 (starting from here, or we could add 1.4 to this and x); v0 = 0.602 miles/sec; t = 1.3 sec
1.8 = 0.602*t + 1/2*a*1.3^2
solving: a = 1.0172/0.845 ~= 1.204 miles/sec^2
Final speed at end of second segment:
v = v0 + a*t
v0 = 0.602 miles/sec; t = 1.3 sec
a ~= 1.204 miles/sec^2
solving: v ~= 2.17 miles/sec, or 7800 mph
Thus assuming constant acceleration, we actually achieve a velocity greater than 6400 mph. With decreasing acceleration (a real-world condition), 6400 mph is a believable result.
IEEE Spectrum also has an article dealing with the future flash technologies in the current issue