Thermal Time
29
The formal transforms which convert one time scale to the other, for
an organism whose development rate depends only on temperature, is
where t is the thermal time and R is the rate of development at temperature
T (which, in turn, depends on time). The function g is the inverse of R
and allows, in principle, the conversion of thermal time back to clock
time.
In practice, the integral in Eq. (2.6) is always approximated as a sum
because temperature generally is not a predictable function of time. For
the usual calculation of thermal time we assume a straight line relationship between development rate and temperature, such as that shown in
Fig. 2.7. We also assume that temperatures are always within the range
Tb . . . T, , where Tb is the base temperature (low temperature at which
development stops) and T, is the temperature at which the development
rate is maximum. Thermal time, and therefore organism development, is
then directly proportional to the sum of products of (I;: - Tb) and the
length of the time increment, where I;: is the temperature at a particular
time, with the condition that I;: - Tb > 0. Given these assumptions, the
equation for thermal time increments is
Ati = (I;: - Tb)At when I;: > Tb; otherwise Ati = 0. (2.7)
The time step, At, is chosen so that temperature is fairly constant during
one time increment. The units of A t are day-degrees, or hour-degrees,
depending on the units of At. No thermal time is accumulated when I;:
is at or below the base temperature. Thermal time is computed as:
From Fig. 2.7, it can be seen that the rate is 1.35 d-' when the temperature
is T, = 33" C. The base temperature is Tb = 10" C. At 33" C, the time
for completion is 111.35 = 0.74 days. The thermal time for completion
at this constant temperature is 0.74 days, or (33" C - 10" C)/1.35 = 17.0
day-degrees. When the temperature of the melon fly eggs varies during
germination, we can use the varying temperature, with Eq. (2.8), to find
z , since the start of the stage. Once z , reaches 17.0 day-degrees, the
stage will be complete.
The inverse operation indicated by the second of Eqs. (2.6) is used
to find the calendar or clock time required to complete a developmental
stage. An analytical form of the inverse is not possible except in the trivial
case where temperature is constant. To find the calendar time required
for completion of the egg stage in the example just presented, we would
construct a table of t, and the corresponding tn . We would then enter the
29
The formal transforms which convert one time scale to the other, for
an organism whose development rate depends only on temperature, is
where t is the thermal time and R is the rate of development at temperature
T (which, in turn, depends on time). The function g is the inverse of R
and allows, in principle, the conversion of thermal time back to clock
time.
In practice, the integral in Eq. (2.6) is always approximated as a sum
because temperature generally is not a predictable function of time. For
the usual calculation of thermal time we assume a straight line relationship between development rate and temperature, such as that shown in
Fig. 2.7. We also assume that temperatures are always within the range
Tb . . . T, , where Tb is the base temperature (low temperature at which
development stops) and T, is the temperature at which the development
rate is maximum. Thermal time, and therefore organism development, is
then directly proportional to the sum of products of (I;: - Tb) and the
length of the time increment, where I;: is the temperature at a particular
time, with the condition that I;: - Tb > 0. Given these assumptions, the
equation for thermal time increments is
Ati = (I;: - Tb)At when I;: > Tb; otherwise Ati = 0. (2.7)
The time step, At, is chosen so that temperature is fairly constant during
one time increment. The units of A t are day-degrees, or hour-degrees,
depending on the units of At. No thermal time is accumulated when I;:
is at or below the base temperature. Thermal time is computed as:
From Fig. 2.7, it can be seen that the rate is 1.35 d-' when the temperature
is T, = 33" C. The base temperature is Tb = 10" C. At 33" C, the time
for completion is 111.35 = 0.74 days. The thermal time for completion
at this constant temperature is 0.74 days, or (33" C - 10" C)/1.35 = 17.0
day-degrees. When the temperature of the melon fly eggs varies during
germination, we can use the varying temperature, with Eq. (2.8), to find
z , since the start of the stage. Once z , reaches 17.0 day-degrees, the
stage will be complete.
The inverse operation indicated by the second of Eqs. (2.6) is used
to find the calendar or clock time required to complete a developmental
stage. An analytical form of the inverse is not possible except in the trivial
case where temperature is constant. To find the calendar time required
for completion of the egg stage in the example just presented, we would
construct a table of t, and the corresponding tn . We would then enter the